Markov Musings 3: Spectral Theory

July 13, 2023 by Ethan N. Epperly 1 Comment

This post is part of a series, Markov Musings, about the mathematical analysis of Markov chains. See here for the first post in the series.

In the previous two posts, we proved the fundamental theorem of Markov chains using couplings and discussed the use of couplings to bound the mixing time, the time required for the chain to mix to near-stationarity.

In this post, we will continue this discussion by discussing spectral methods for understanding the convergence of Markov chains.

Mathematical Setting

In this post, we will study a reversible Markov chain on $\{1,\ldots,m\}$ with transition probability matrix $P \in \real^{m\times m}$ and stationary distribution $\pi$ . Our goal will be to use the eigenvalues and eigenvectors of the matrix $P$ to understand the properties of the Markov chain.

Throughout our discussion, it will be helpful to treat vectors $f\in\real^m$ and functions $f:\{1,\ldots,m\}\to \real$ as being one and the same, and we will use both $f_i$ and $f(i)$ to denote the $i$ th entry of the vector $f$ (aka the evaluation of the function $f$ at $i$ ). By adopting this perspective, a vector is not merely a list of $m$ numbers, but instead labels each state $i=1,2,\ldots,m$ with a numeric value.

Given a function/vector $f$ , we let $\expect[f]$ denote the expected value of $f(X)$ where $X$ is drawn from $\pi$ :

$\expect[f] \coloneqq \expect_{X \sim \pi} [f(X)] = \sum_{i=1}^m \pi_i f(i).$

The variance is defined similarly:

$\Var(f) \coloneqq \expect [(f - \expect[f])^2] = \sum_{i=1}^m (f(i) - \expect[f])^2 \pi_i.$

As a final piece of notation, we let $\delta_i$ denote a probability distribution which assigns 100% probability to outcome $i$ .

Spectral Theory

The eigenvalues and eigenvectors of general square matrices are ill-behaved creatures. Indeed, a general $m\times m$ matrix with real entries can have complex-valued eigenvalues and fail to possess a full suite of $m$ linearly independent eigenvectors. The situation is dramatically better for symmetric matrices which obey the spectral theorem:

Theorem (spectral theorem for real symmetric matrices). An $m\times m$ real symmetric matrix $A$ (i.e., one satisfying $A^\top=A$ ) has $m$ real eigenvalues $\lambda_1,\ldots,\lambda_m$ and $m$ orthonormal eigenvectors $u_1,\ldots,u_m$ .

Unfortunately for us, the transition matrix $P$ for a reversible Markov chain is not always symmetric. But despite this, there’s a surprise: $P$ always has real eigenvalues. This leads us to ask:

Why does are the eigenvalues of the transition matrix $P$ real?

To answer this question, we will need to develop and a more general version of the spectral theorem and use our standing assumption that the Markov chain is reversible.

The Transpose

In our quest to develop a general version of the spectral theorem, we look more deeply into the hypothesis of the theorem, namely that $A$ is equal to its transpose $A^\top$ . Let’s first ask: What does the transpose $A^\top$ mean?

Recall that $\real^m$ is equipped with the standard inner product, sometimes called the Euclidean or dot product. We denote this product by $(\cdot,\cdot)$ and it is defined as

$(f,g) \coloneqq f^\top g = \sum_{i=1}^m f(i) g(i).$

We can think of the standard inner product as defining the amount of the vector $f$ that points in the direction of $g$ .

The transpose of a matrix $A$ is closely related to the standard inner product. Specifically, $A^\top$ is the (unique) matrix satisfying the adjoint property:

$(f,Ag) = (A^\top f, g)\quad \text{for every }f,g\in\real^m.$

So the amount of $f$ in the direction of $Ag$ is the same as the amount of $A^\top f$ in the direction of $g$ .

The Adjoint and the General Spectral Theorem

Since the transpose is intimately related to the standard inner product on $\real^m$ , it is natural to wonder if non-standard inner products on $\real^m$ give rise to non-standard versions of the transpose. This idea proves to be true.

Definition (adjoint). Let $[\cdot,\cdot]$ be any inner product on $\real^m$ and let $A$ be an $m\times m$ matrix. The adjoint of $A$ with respect to the inner product $[\cdot,\cdot]$ is the (unique) matrix $A^*$ such that
$[f,Ag]=[A^*f,g]\quad\text{for every }f,g\in\real^m.$
A matrix $A$ is self-adjoint if it equals its own adjoint, $A=A^*$ .

The spectral theorem naturally extends to the more abstract setting of adjoints with respect to non-standard inner products:

Theorem (general spectral theorem). Let $A$ be an $m\times m$ matrix which is set-adjoint with respect to an inner product $[\cdot,\cdot]$ . Then $A$ has $m$ real eigenvalues $\lambda_1,\ldots,\lambda_m$ and $m$ eigenvectors $u_1,\ldots,u_m$ which are $[\cdot,\cdot]$ –orthonormal:
$[u_i,u_j]=\begin{cases}1, & i=j, \\0,& i\ne j.\end{cases}$

Reversibility and Self-Adjointness

The general spectral theorem offers us a path to understand our observation from above that the eigenvalues of $P$ are always real. Namely, we ask:

Is there an inner product under which $P$ is self-adjoint?

Fortunately, the answer is yes. Define the $\pi$ -inner product

$\langle f, g \rangle \coloneqq \expect[f\cdot g] = \sum_{i=1}^m f(i)g(i) \pi_i.$

This inner product is very natural. If $f$ and $g$ are mean-zero ( $\expect[f] = \expect[g] = 0$ ), it reports the covariance of $f(x)$ and $g(x)$ where $x \sim \pi$ is drawn from $\pi$ .

Let us compute the adjoint of the transition matrix $P$ in the $\pi$ -inner product $\langle \cdot,\cdot \rangle$ :

$\begin{align*}\langle f, Pg \rangle&= \sum_{i=1}^m f(i) \left(\sum_{j=1}^m P_{ij} g(i) \right)\pi_i= \sum_{i,j=1}^m f(i) g(j) \pi_iP_{ij} .\end{align*}$

Recall that we have assumed our Markov chain is reversible. Thus, the detailed balance condition

$\pi_i P_{ij} = \pi_j P_{ji} \quad \text{for } i,j=1,2,\ldots,m$

implies that

$\langle f, Pg \rangle= \sum_{i,j=1}^m f(i) g(j) \pi_jP_{ji} = \sum_{j=1}^m \left( \sum_{i=1}^m f(i) P_{ij} \right) g(j) \pi_j = \langle Pf, g\rangle.$

Thus, we conclude that $P$ is self-adjoint.

We cannot emphasize enough that the reversibility assumption is crucial to ensure that $P$ is self-adjoint. In fact,

Theorem (reversibility and self-adjointness). The transition matrix of a general Markov chain with stationary distribution $\pi$ is self-adjoint in the $\pi$ -inner product if and only if the chain is reversible.

Spectral Theorem for Markov Chains

Since $P$ is self-adjoint in the $\pi$ -inner product, the general spectral theorem implies the following

Corollary (spectral theorem for reversible Markov chain). The reversible Markov transition matrix $P$ has $m$ real eigenvalues $\lambda_1 \ge \cdots \ge \lambda_m$ associated with eigenvectors $\varphi_1,\ldots,\varphi_m$ which are orthonormal in the $\pi$ -inner product:
$\langle\varphi_i,\varphi_j\rangle=\begin{cases}1, &i=j\\0,& i\ne j.\end{cases}$

Since $P$ is a reversible Markov transition matrix—not just any self-adjoint matrix—the eigenvalues of $P$ satisfy additional properties:

Bounded. All of the eigenvalues of $P$ lie between $-1$ and $1$ .¹
Eigenvalue 1. $\lambda_1 = 1$ is an eigenvalue of $P$ with eigenvector $\varphi_1 = \mathbf{1}$ , where $\mathbf{1}$ is a vector of all one’s.

For a primitive chain, we can also have the property:

Contractive. For a primitive chain, all eigenvalues other than $\lambda_1$ have magnitude $< 1$ .²

Distance Between Probability Distributions Redux

In the previous post, we used the total variation distance to compare probability distributions:

Definition (total variation distance). The total variation distance between probability distributions $\sigma$ and $\pi$ is
$\norm{\sigma - \pi}_{\rm TV} = \max_{A \subseteq \{1,\ldots,m\}} |\sigma(A) - \pi(A)| = \frac{1}{2} \sum_{i=1}^m |\sigma_i - \pi_i|.$

The total variation distance is an “ $\ell_1$ “ way of comparing two probability distributions since it can be computed by adding the absolute difference between the probabilities $\sigma_i$ and $\pi_i$ of each outcome. Spectral theory plays more nicely with an “ $\ell_2$ ” way of comparing probability distributions, which we develop now.

Densities

Given two probability densities $\sigma$ and $\pi$ , the density of $\sigma$ with respect to $\pi$ is the function³ $h = d\sigma/d\pi$ given by

$h(i) = \frac{d\sigma}{d\pi}(i) = \frac{\sigma_i}{\pi_i} \quad \text{for } i=1,2,\ldots,m.$

The density $h = d\sigma/d\pi$ satisfies the property that

(1) $\expect_{X \sim \sigma} [g(X)] = \expect_{Y \sim \pi}[g(Y)h(Y)] = \sum_{i=1}^m g(i)h(i) \pi_i.$

To see why we call $h = d\sigma/d\pi$ a density, it may be helpful to appeal to continuous probability for a moment. If $0\le X \le 1$ is a random variable with distribution $\sigma$ , the probability density function of $\sigma$ (with respect to the uniform distribution $\mathrm{Unif}[0,1]$ ) is a function $h$ such that

$\expect_{X \sim \sigma} [g(X)] = \expect_{Y\sim \mathrm{Unif}[0,1]} [g(Y)h(Y)] = \int_0^1 g(x) h(x) \, dx.$

The formula for the continuous case is exactly the same as the finite case (1) with sums $\sum_{i=1}^m (\cdots) \pi_i$ replaced with integrals $\int_0^1 (\cdots) \, dx$ .

The Chi-Squared Divergence

We are now ready to introduce our “ $\ell_2$ ” way of comparing probability distributions.

Definition ( $\chi^2$ -divergence). The $\chi^2$ -divergence of $\sigma$ with respect to $\pi$ is the variance of the density function:
$\chi^2(\sigma \mid\mid \pi) \coloneqq \Var \left(\frac{d\sigma}{d\pi} \right) = \expect \left[\left( \frac{d\sigma}{d\pi} - 1 \right)^2\right] = \expect \left[\left(\frac{d\sigma}{d\pi}\right)^2\right] - 1.$

To see the last equality is a valid formula for $\chi^2(\sigma\mid\mid\pi) \coloneqq \Var(d\sigma/d\pi)$ , note that

(2) $\expect\left[\frac{d\sigma}{d\pi}\right] = \sum_{i=1}^m \frac{d\sigma}{d\pi}(i) \pi_i = \sum_{i=1}^m \frac{\sigma_i}{\pi_i} \pi_i = \sum_{i=1}^m \sigma_i = 1.$

Relations Between Distance Measures

The $\chi^2$ divergence always gives an upper bound on the total variation distance. Indeed, first note that we can express the total variation distance as

$\norm{\sigma - \pi}_{\rm TV} = \frac{1}{2} \expect \left[ \left| \frac{d\sigma}{d\pi} - 1 \right| \right].$

Thus, using Lyapunov’s inequality, we conclude

(3) $\norm{\sigma - \pi}_{\rm TV} = \frac{1}{2} \expect \left[ \left| \frac{d\sigma}{d\pi} - 1 \right| \right] \le \frac{1}{2} \left(\expect \left[ \left( \frac{d\sigma}{d\pi} - 1 \right)^2 \right]\right)^{1/2} = \frac{1}{2} \sqrt{\chi^2(\sigma \mid\mid \pi)}.$

Markov Chain Convergence by Eigenvalues

Now, we prove a quantitative version of the fundamental theorem of Markov chains (for reversible processes) using spectral theory and eigenvalues:⁴

Theorem (Markov chain convergence by eigenvalues). Let $\rho^{(0)},\rho^{(1)},\rho^{(2)},\ldots$ denote the distributions of the Markov chain at times $0,1,2,\ldots$ . Then
$\chi^2\left(\rho^{(n)} \, \middle|\middle| \, \pi\right) \le \left( \max \{ \lambda_2, -\lambda_n \} \right)^{2n} \chi^2\left(\rho^{(0)} \, \middle|\middle| \, \pi\right).$

This raises a natural question: How large can the initial $\chi^2$ divergence between $\rho^{(0)}$ and $\pi$ be? This is answered by the next result.

Proposition (Initial distance to stationarity). For any $i \in \{1,\ldots,m\}$ ,
(4) $\chi^2(\delta_i \mid\mid \pi) \le \frac{1}{\pi_i}.$
For any initial distribution $\rho^{(0)}$ , we have
(5) $\chi^2\left( \rho^{(0)} \, \middle|\middle| \, \pi \right) \le \frac{1}{\min_{1\le i\le m} \pi_i}$

The condition (4) is a direct computation

$\chi^2(\delta_i \mid\mid \pi) = \Var(d\delta_i/d\pi) \le \expect[(d\delta_i/d\pi)^2] = 1/\pi_i.$

For (5), observe that $\chi^2(\rho^{(0)} \mid\mid \pi)$ is a convex function of the initial distribution $\rho^{(0)}$ . Therefore, its maximal value over the convex set of all probability distribution is maximized at an extreme point $\rho^{(0)} = \delta_i$ for some $i$ . Consequently, (5) follows directly from (4).

Using the previous two results and equation (3), we immediately obtain the following:

Corollary (Mixing time). When initialized from a distribution $\rho^{(0)}$ , the chain mixes to $\varepsilon$ -total variation distance to stationarity
$\norm{\rho^{(n)} - \pi}_{\rm TV} \le \varepsilon$
after
(6) $n = \left\lceil \frac{1}{\log(\max\{\lambda_2,-\lambda_n\})}\log \left( \frac{1}{2\varepsilon \min_{1\le i \le m}\sqrt{\pi_i}} \right) \right\rceil \text{ steps}.$

Indeed,

$\begin{align*}\norm{\rho^{(n)} - \pi}_{\rm TV} &\le \frac{1}{2} \sqrt{\chi^2(\rho^{(n)} \mid\mid \pi)} \\&\le \frac{1}{2} (\max\{\lambda_2,-\lambda_n\})^n \sqrt{\chi^2(\rho^{(0)} \mid\mid \pi)} \\&\le \frac{(\max\{\lambda_2,-\lambda_n\})^n}{2\min_{1\le i \le m} \sqrt{\pi_i}}.\end{align*}$

Equating the left-hand side with $\varepsilon$ and rearranging gives (6).

Proof of Markov Chain Convergence Theorem

We conclude with a proof of the Markov chain convergence result.

Recurrence for Densities

Our first task is to derive a recurrence for the densities $d\rho^{(n)}/d\pi$ . To do this, we use the recurrence for the probability distribution:

$\rho^{(n+1)}_j = \sum_{i=1}^m \rho_i^{(n)} P_{ij}.$

Thus,

$\left( \frac{d\rho^{(n+1)}}{d\pi}\right)_j = \frac{\rho_j^{(n+1)}}{\pi_j} = \frac{\sum_{i=1}^m \rho^{(n)}_i P_{ij}}{\pi_j}= \sum_{i=1}^m \rho^{(n)}_i\frac{P_{ij}}{\pi_j}.$

We now invoke the detailed balance $P_{ij}/\pi_j = P_{ji} / \pi_i$ , which implies

$\left( \frac{d\rho^{(n+1)}}{d\pi}\right)_j = \sum_{i=1}^m\rho^{(n)}_i \frac{P_{ji}}{\pi_i} = \sum_{i=1}^m P_{ji} \frac{\rho^{(n)}_i}{\pi_i} = \left( P \frac{d \rho^{(n)}}{d\pi} \right)_j.$

The product $P(d\rho^{(n)}/d\pi)$ is an ordinary matrix–vector product. Thus, we have shown

(7) $\frac{d\rho^{(n+1)}}{d\pi} = P \frac{d\rho^{(n)}}{d\pi} \quad \text{for } n =0,1,\ldots.$

Spectral Decomposition

Now that we have a recurrence for the densities $d\rho^{(n)}/d\pi$ , we can understand how the densities change in time. Expand the initial density in eigenvectors of $P$ :

$\frac{d\rho^{(0)}}{d\pi} = c_1 \mathbf{1} + c_2 \varphi_2 + \cdots + c_m \varphi_m.$

As we verified above in (1), we have $\expect[d\rho^{(0)}/d\pi] = 1$ . Thus, we conclude $c_1 = 1$ . Since the $\varphi_j$ are eigenvectors of $P$ with eigenvalues $\lambda_j$ , the recurrence (7) implies

$\frac{d\rho^{(n)}}{d\pi} = \mathbf{1} + c_2 \lambda_2^n \varphi_2 + \cdots + c_m \lambda_m^n \varphi_m.$

Conclusion

Finally, we compute

$\chi^2(\rho^{(n)} \mid\mid \pi) = \sum_{i=2}^m c_i^2 \lambda_i^{2n} \le (\max \{ \lambda_2, -\lambda_m \})^{2n} \sum_{i=2}^m c_i^2 = (\max \{ \lambda_2, -\lambda_m \})^{2n} \chi^2(\rho^{(0)} \mid\mid \pi).$

The theorem is proven.

Markov Musings 2: Couplings

July 5, 2023 by Ethan N. Epperly Leave a comment

This post is part of a series, Markov Musings, about the mathematical analysis of Markov chains. See here for the first post in the series.

In the last post, we saw a proof of the fundamental theorem of Markov chains using the method of couplings. In this post, we will use couplings to provide quantitative bounds on the rate of Markov chain convergence.

Mixing Time

Let us begin where the last post ended by recalling some facts about the mixing time of a Markov chain. Consider a Markov chain $x_0,x_1,x_2,\ldots$ with stationary distribution $\pi$ . Recall that the mixing time is

$\tau_{\rm mix} \coloneqq \min \left\{ n \ge 1 : \max_{\rho^{(0)}} \norm{\rho^{(n)} - \pi}_{\rm TV} \le \frac{1}{2e} \right\}.$

Here, $\rho^{(n)}$ denotes the distribution of the state $x_n$ of the chain at time $n$ and $\norm{\cdot}_{\rm TV}$ is the total variation distance. For more on the total variation distance, see the previous post.

At the end of last post, we saw the following theorem, showing that the mixing time controls the rate of Markov chain convergence.

Theorem (mixing time as convergence rate). For any starting distribution $\rho^{(0)}$ ,
$\norm{\rho^{(n)} - \pi}_{\rm TV} \le e^{-\lfloor n / \tau_{\rm mix}\rfloor}.$
Consequently, if $n\ge \tau_{\rm mix}\cdot \lceil \log(1/\epsilon)\rceil$ , then $\norm{\rho^{(n)} - \pi}_{\rm TV} \le \varepsilon$ .

This result motivates us to develop techniques to bound the mixing time $\tau_{\rm mix}$ .

Couplings and Convergence

In the previous post, we made essential use of couplings of probability distributions. In this post, we will extend this notion by using couplings of entire Markov chains.

Definition (coupling of Markov chains). A coupling of Markov chains with transition matrix $P$ are two processes $x_0,x_1,x_2,\ldots$ and $y_0,y_1,y_2,\ldots$ such that (A) each process is individually a Markov chain with transition matrix $P$ : In particular,
$\prob\{x_{n+1} = j \mid x_n = i \} = \prob\{y_{n+1} = j \mid y_n = i \} = P_{ij}$
and (B) if $x_n = y_n$ , then $x_{n+1} = y_{n+1}$ .

A coupling of Markov chains thus consists of a pair of Markov chains which become glued together should they ever touch.

There are several ways we can use couplings to bound the mixing time of Markov chains. Here is one way. Let $x_0,x_1,x_2,\ldots$ and $y_0,y_1,y_2,\ldots$ be a coupling of Markov chains for which $x_0$ is initialized in some initial distribution $x_0 \sim \rho^{(0)}$ and $y_0$ is initialized in the stationary distribution $y_0 \sim \pi$ . Let $\tau_{\rho^{(0)}}$ be the time at which $x$ and $y$ meet:

$\tau_{\rho^{(0)}} \coloneqq \min\{ n \ge 0 : x_n = y_n\}.$

The following theorem shows that the tails of the random variable $\tau$ control the convergence of the Markov chain $x_0,x_1,x_2,\ldots$ to stationarity.

Theorem (convergence from couplings). Then $\norm{\rho^{(n)} - \pi}_{\rm TV} \le \prob\{\tau_{\rho^{(0)}} > n\}$ .

This result is an immediate application of the coupling lemma from last post. Using this bound, we can bound the mixing time

Corollary (mixing time from couplings). $\tau_{\rm mix} \le 2e \max_{\rho^{(0)}} \expect[\tau_{\rho^{(0)}}]$ .

The maximum is taken over all initial distributions $\rho^{(0)}$ .

Indeed, by the above theorem and Markov’s inequality,¹

$\norm{\rho^{(n)} - \pi}_{\rm TV} \le \prob\{\tau_{\rho^{(0)}} > n\}\le \frac{\expect[\tau_{\rho^{(0)}}]}{n}.$

Take a maximum over initial distribution $\rho^{(0)}$ and set $n \coloneqq \tau_{\rm mix}$ . Then the left-hand side is at most $1/2e$ , so

$\frac{1}{2e} \le \max_{\rho^{(0)}}\norm{\rho^{(\tau_{\rm mix})} - \pi}_{\rm TV} = \prob\{\tau_{\rho^{(0)}} > n\} \le \frac{\max_{\rho^{(0)}}\expect[\tau_{\rho^{(0)}}]}{\tau_{\rm mix}}.$

Rearranging gives the corollary.

Example: Bit Strings

There are a number of examples² to prove bounds on mixing times using the method of couplings, but we will focus on a simple but illustrative example: a lazy random walk on the set of bit strings.

A bit string is a sequence of 0’s and 1’s. Bit strings are used to store and represent information on digital computers. For instance, the character “a” is stored as “1100001” using the ASCII encoding.

The Chain

Our example is a Markov chain whose states consist of all bit strings of length $N$ . For each step of the chain,

With 50% probability, we do nothing and leave the state unchanged.
With 50% probability, we choose a (uniform) random position from $1$ to $N$ and flip the bit in that position, changing 0 to 1 or 1 to 0.

One can think of this Markov chain as modeling a random process in which a bit string is corrupted by errors which flip a single bit. The stationary distribution for this chain is the uniform distribution on all bit strings (i.e., each bit string of length $n$ has the same probability $2^{-n}$ ). Therefore, one can also think of this Markov chain as a very inefficient way of generating a string of random bits.³

Designing the Coupling

Let us use the method of couplings to determine how fast this chain mixes. First, observe that there’s an alternative way of stating the transition rule for this Markov chain:

Pick a (uniform) random position from $1$ to $N$ and set the bit in that position to $0$ with 50% probability and $1$ with 50% probability.

Indeed, under this rule, half of the time we leave the state unchanged because we set the bit in the selected position to the value it already had. For the other half of the time, we flip the selected bit.

Now, we construct a coupling. Initialize $x_0$ in an arbitrary distribution $\rho^{(0)}$ and $y_0$ in the stationary distribution (uniform over all bit strings). The key to construct a coupling will be to use the same randomness to update the state of both the “ $x$ ” chain and the “ $y$ ” chain. For this chain, there’s a natural way to do this.

At time $n$ , pick a (uniform) random position $1\le p_n \le N$ and a (uniform) random bit $b_n \in \{0,1\}$ . Set $x_{n+1}$ by changing the $p_n$ th bit of $x_n$ to $b_n$ , and set $y_{n+1}$ by changing the $p_n$ th bit of $y_n$ to $b_n$ .

We couple the two chains by setting the same position $p_n$ to the same bit $b_n$ .

Bounding the Mixing Time

To bound the mixing time, we need to understand when these two chains meet. Observe that if we pick position $p$ to set at some point, the two Markov chains have the same bit in position $p$ at every time later in the chain. Consequently,

If all of the positions $1,\ldots,N$ have been chosen by time $n$ , then $x_n = y_n$ .

The two chains might meet before all of the positions have been chosen, but they are guaranteed to meet after.

Set $\tau_{\rm pick}$ to be the time at which all positions $1,\ldots,N$ have been selected at least once. Then our observation is equivalent to saying that the meeting time $\tau_{\rho^{(0)}}$ is smaller than $\tau_{\rm pick}$ ,

$\tau_{\rho^{(0)}} \le \tau_{\rm pick}.$

Thus, to bound the mixing time, it is sufficient to compute the expected value of $\tau_{\rm pick}$ .

Collecting Coupons

Computing the expected value of $\tau_{\rm pick}$ is actually an example of a very famous problem in probability theory known as the coupon collector problem. The classic version goes like this:

A cereal company is running a promotion where each box of cereal contains a coupon, which is equally likely to be any of one of $N$ different colors. A coupon of each color can be traded in for a toy. What is the expected number of boxes we need to open in order to qualify for the toy?

The coupon collector is exactly the same as our problem, with the $N$ different colors of coupons playing the same role as the $N$ different positions in our bit strings. Going forward, let’s use the language of coupons and boxes to describe this problem, as its more colorful.

When tackling a hard question like computing $\expect[\tau_{\rm pick}]$ , it can be helpful to break into easier pieces. Let’s start with the simplest possible question: How many picks do we need to collect the first coupon? Just one. We always get a coupon in our first box.

With that question answered, let’s ask the next-simplest question: How many picks do we need to collect the second coupon, differently colored than the first? There are $N$ colors and $N-1$ of them are different from the first. So the probability of picking a new coupon in the second box is $(N-1)/N$ . In fact,

Until we succeed at picking the second unique coupon, every box has an independent $(N-1)/N$ chance of containing such a coupon.

The number of boxes needed is therefore a geometric random variable with success probability $p = (N-1)/N$ , and its mean is $1/p = N/(N-1)$ . On average, we need $N/(N-1)$ picks to collect the second coupon.

The reasoning is the same for the third coupon. There are $N-2$ coupons we haven’t collected and $N$ total, so the probability of picking the third coupon in each box is $(N-2)/N$ . Thus, the number of picks is a geometric random variable with success probability $(N-2)/N$ with mean $N/(N-2)$ .

In general, we need an expected $N / (N-k)$ picks to pick the $k$ th coupon. Adding up the expected number of picks for each $k = 1,2,\ldots,N$ , we obtain that the total number of picks required is to collect all the coupons is

$\expect[\tau_{\rm pick}] = 1 + \frac{N}{N-1} + \frac{N}{N-2} + \cdots + \frac{N}{N-(N-1)} = N \left( \frac{1}{N} + \frac{1}{N-1} + \cdots + \frac{1}{2} + 1 \right).$

More concisely, if we let $H_N$ denote the $N$ th harmonic number

$H_N = 1 + \frac{1}{2} + \cdots + \frac{1}{N}$

then $\expect[\tau_{\rm pick}] = NH_N$ . Since the $N$ th harmonic number is at most $H_N \le \log(N) + 1$ , we have

$\expect[\tau_{\rm pick}] \le N \log (N) + N.$

Conclusion

Putting all of these ingredients together, we obtain

$\tau_{\rm mix} \le 2e \, \expect[\tau_{\rho^{(0)}}] \le 2e \, \expect[\tau_{\rm pick}] \le 2e \, N \log(N) + 2e \, N.$

The mixing time is (at most) proportional to $N\log N$ .

More on Bit Strings and Collecting Coupons

Using more sophisticated techniques, it can be shown that the random variable $\tau_{\rm pick}$ satisfies

$\prob\{\tau_{\rm pick} > N \log N + cN\} \le e^{-c}$

for every $N\ge 1$ and $c\ge 0$ , from which it follows that

$\norm{\rho^{(n)} - \pi}_{\rm TV} \le \varepsilon \quad \text{provided that} \quad n \ge N \log N + N \log(1/\varepsilon).$

Using a more sophisticated analysis—also based on couplings—we can improve this result to

$\norm{\rho^{(n)} - \pi}_{\rm TV} \le \varepsilon \quad \text{provided that} \quad n \ge \frac{1}{2} N \log N + c \,N \log(1/\varepsilon)$

for some constant $c > 0$ . The constant $1/2$ in this inequality cannot be improved. See section 5.3.1 of Markov Chains and Mixing Times.

Markov Musings 1: The Fundamental Theorem

June 29, 2023 by Ethan N. Epperly 3 Comments

For this summer, I’ve decided to open up another little mini-series on this blog called Markov Musings about the mathematical analysis of Markov chains, jumping off from my previous post on the subject. My main goal in writing this is to learn the material for myself, and I hope that what I produce is useful to others. My main resources are:

The book Markov Chains and Mixing Times by Levin, Peres, and Wilmer;
Lecture notes and videos by theoretical computer scientists Sinclair, Oveis Gharan, O’Donnell, and Schulman; and
These notes by Rob Webber, for a complementary perspective from a scientific computing point of view.

Be warned, these posts will be more mathematical in nature than most of the material on my blog.

In my previous post on Markov chains, we discussed the fundamental theorem of Markov chains. Here is a slightly stronger version:

Theorem (fundamental Theorem of Markov chains). A primitive Markov chain on a finite state space has a stationary distribution $\pi > 0$ . When initialized from any starting distribution $\rho^{(0)}$ , the distributions $\rho^{(0)},\rho^{(1)},\rho^{(2)},\ldots$ of the chain at times $0,1,2,\ldots$ converge at an exponential rate to $\pi$ .

My goal in this post will be to provide a proof of this fact using the method of couplings, adapted from the notes of Sinclair and Oveis Gharan. I like this proof because it feels very probabilistic (as opposed to more linear algebraic proofs of the fundamental theorem).

Here, and throughout, we say a matrix or vector is $> 0$ if all of its entries are strictly positive. Recall that a Markov chain with transition matrix $P$ is primitive if there exists $n$ for which $P^n > 0$ . For this post, all Markov chains will have state space $\{1,\ldots,m\}$ .

Total Variation Distance

In order to quantify the rate of Markov chain convergence, we need a way of quantifying the closeness of two probability distributions. This motivates the following definition:

Definition (total variation distance). The total variation distance between probability distributions $\rho$ and $\sigma$ on $\{1,\ldots,m\}$ is the maximum difference between the probability of an event $S$ under $\rho$ and under $\sigma$ :
$\norm{\rho - \sigma}_{\rm TV} = \max_{S \subseteq \{1,\ldots,m\}} |\rho(S) - \sigma(S)| = \frac{1}{2} \sum_{i=1}^m \left| \rho_i - \sigma_i \right|.$

The total variation distance is always between $0$ and $1$ . It is zero only when $\rho$ and $\sigma$ are the same distribution. It is one only when $\rho$ and $\sigma$ have disjoint supports—that is, there is no $i \in \{1,\ldots,m\}$ for which $\rho_i, \sigma_i > 0$ .

The total variation distance is a very strict way of comparing two probability distributions. Sinclair’s notes provide a vivid example. Suppose that $\rho$ denotes the uniform distribution on all possible ways of shuffling a deck of $N$ cards, and $\sigma$ denotes the uniform distribution on all ways of shuffling $N$ cards with the ace of spades at the top. Then the total variation distance between $\rho$ and $\sigma$ is $1 - 1/N$ . Thus, despite these distributions seeming quite similar to us, the total variation distance between $\rho$ and $\sigma$ is almost as far apart as possible. There are a number of alternative ways of measuring the closeness of probability distributions, some of which are less severe.

Couplings

Given a probability distribution $\rho$ , it can be helpful to work with random variables drawn from $\rho$ . Say a random variable $x$ is drawn from the distribution $\rho$ , written $x \sim \rho$ , if

$\prob \{x = i\} = \rho_i \quad \text{for $i=1,2,\ldots,m$}.$

To understand the total variation distance more, we shall need the following definition:

Definition (coupling). Given probability distributions $\rho,\sigma$ on $\{1,\ldots,m\}$ , a coupling $\gamma$ is a distribution on $\{1,\ldots,m\}^2$ such that if a pair of random variables $(x,y)\sim\gamma$ is drawn from $\gamma$ , then $x \sim \rho$ and $y \sim \sigma$ . Denote the set of all couplings of $\rho$ and $\sigma$ as $\operatorname{Couplings}(\rho,\sigma)$ .

More succinctly, a coupling of $\rho$ and $\sigma$ is a joint distribution with marginals $\rho$ and $\sigma$ .

Couplings are related to total variation distance by the following lemma:¹

Lemma (coupling lemma). Let $\rho$ and $\sigma$ be distributions on $\{1,\ldots,m\}$ . Then
$\norm{\rho - \sigma}_{\rm TV} = \min_{\gamma \in \operatorname{Couplings}(\rho,\sigma)} \prob_{(x,y) \sim \gamma} \{ x \ne y \}.$
Here, $\prob_{(x,y) \sim \gamma}$ represents the probability for variables $x,y$ drawn from joint distribution $\gamma$ .

To see a simple example, suppose $\rho = \sigma$ . Then the coupling lemma tells us that there is a coupling $\gamma$ of $\rho$ and itself such that $\prob \{ x \ne y \} = 0$ . Indeed, such a coupling can be obtained by drawing $x \sim \rho$ and setting $y \coloneqq x$ . This defines a joint distribution $\gamma$ under which $x = y$ with 100% probability.

To unpack the coupling lemma a little more, it really contains two statements:

For any coupling $\gamma$ between $\rho$ and $\sigma$ and $(x,y) \sim \gamma$ ,
$\norm{\rho - \sigma}_{\rm TV} \le \prob \{x \ne y \}.$
There exists a coupling $\gamma$ between $\rho$ and $\sigma$ such that when $(x,y) \sim \gamma$ , then
$\norm{\rho - \sigma}_{\rm TV} = \prob \{x \ne y\}.$

We will need both of these statements in our proof of the fundamental theorem.

Proof of the Fundamental Theorem

With these ingredients in place, we are now ready to prove the fundamental theorem of Markov chains. First, we will assume there exists a stationary distribution $\pi > 0$ . We will provide a proof of this fact at the end of this post.

Suppose we initialize the chain in distribution $\rho^{(0)}$ , and let $\rho^{(0)},\rho^{(1)},\rho^{(2)},\ldots$ denote the distributions of the chain at times $0,1,2,\ldots$ . Our goal will be to establish that $\norm{\rho^{(n)} - \pi}_{\rm TV} \to 0$ as $n\to\infty$ at an exponential rate.

Distance to Stationarity is Non-Increasing

First, let us establish the more modest claim that $\norm{\rho^{(n)} - \pi}_{\rm TV}$ is non-increasing

(1) $\norm{\rho^{(n+1)} - \pi}_{\rm TV} \le \norm{\rho^{(n)} - \pi}_{\rm TV} \quad \text{for every } n =0,1,2\ldots.$

We shall do this by means of the coupling lemma.

Consider two versions of the chain $x_0,x_1,x_2,\ldots$ and $y_0,y_1,y_2,\ldots$ , one initialized in $x_0 \sim \rho^{(0)}$ and the other initialized with $y_0 \sim \pi$ . We now apply the coupling lemma to the states $x_n$ and $y_n$ of the chains at time $n$ . By the coupling lemma, there exists a coupling of $x_n$ and $y_n$ such that

$\norm{\rho^{(n)} - \pi}_{\rm TV} = \prob \{x_n\ne y_n\}.$

Now construct a coupling of $x_{n+1}$ and $y_{n+1}$ according to the following rules:

If $x_n = y_n$ , then draw $x_{n+1}$ according to the transition matrix and set $y_{n+1} \coloneqq x_{n+1}$ .
If $x_n \ne y_n$ , then run the two chains independently to generate $x_{n+1}$ and $y_{n+1}$ .

By the way we’ve designed the coupling,

$\prob\{x_{n+1}\ne y_{n+1}\} \le \prob\{x_n \ne y_n\}.$

Thus, by the coupling lemma,

$\norm{\rho^{(n+1)} - \pi}_{\rm TV} \le \prob\{x_{n+1}\ne y_{n+1}\} \le \prob\{x_n \ne y_n\} = \norm{\rho^{(n)} - \pi}_{\rm TV}.$

We have established that the distance to stationarity is non-increasing.

This proof already contains the essence of the argument as to why Markov chains mix. We run two versions of the Markov chain, one initialized in an arbitrary distribution $\rho^{(0)}$ and the other initialized in the stationary distribution $\pi$ . While the states of the two chains are different, we run the chains independently. When the chains meet, we continue moving the chains together in synchrony. After running for long enough, the two chains are likely to meet, implying the chain has mixed.

The All-to-All Case

As another stepping stone to the complete proof, let us prove the fundamental theorem in the special case where there is a strictly positive probability of moving between any two states, i.e., assuming $P>0$ .

Consider the two chains $x_0,x_1,x_2,\ldots$ and $y_0,y_1,y_2,\ldots$ coupled as in the previous section. We compute the probability $\prob \{x_{n+1} \ne y_{n+1}\}$ more carefully. Write it as

(2) $\begin{align*}\prob \{x_{n+1} \ne y_{n+1}\} &= \prob \{x_{n+1} \ne y_{n+1} \mid x_n \ne y_n\}\prob \{x_n \ne y_n\} \\&= (1-\prob \{x_{n+1} = y_{n+1} \mid x_n \ne y_n\})\prob \{x_n \ne y_n\}. \end{align*}$

To compute $\prob \{x_{n+1} = y_{n+1} \mid x_n \ne y_n\})$ , break into cases for all possible values $i,j,k$ for $y_{n+1},x_n,y_n$ to take

$\begin{gather*}\prob \{x_{n+1} = y_{n+1} \mid x_n \ne y_n\} \\= \sum_{\substack{i,j,k\in \{1,\ldots,m\}\\ j\ne k}} \prob \{x_{n+1} =i \mid y_{n+1}=i,x_n=j,y_n=k\} \prob \{y_{n+1}=i,x_n=j,y_n=k \mid x_n \ne y_n\}.\end{gather*}$

We now are in a place to lower bound this probability. Let $p_{\rm min}$ be the minimum probability of moving between any two states

$p_{\rm min} \coloneqq \min_{1\le i,j\le m} P_{ij}.$

The probability of moving from, $j$ to $i$ is at least $p_{\rm min}$ . We conclude the lower bound

$\begin{align*}\prob \{x_{n+1} = y_{n+1} \mid x_n \ne y_n\} &\ge \sum_{\substack{i,j,k\in \{1,\ldots,m\}\\ j\ne k}} p_{\rm min} \prob \{y_{n+1}=i,x_n=j,y_n=k \mid x_n \ne y_n\} = p_{\rm min}.\end{align*}$

Substituting back in (2), we obtain

$\prob \{x_{n+1} \ne y_{n+1}\} \le (1 - p_{\rm min})\prob \{x_n \ne y_n\}.$

By the coupling lemma, we conclude

$\norm{\rho^{(n+1)}-\pi}_{\rm TV} \le \prob \{x_{n+1} \ne y_{n+1}\} \le (1 - p_{\rm min})\prob \{x_n \ne y_n\} = (1 - p_{\rm min}) \norm{\rho^{(n)} - \pi}_{\rm TV}.$

By iteration, we conclude

$\norm{\rho^{(n)} - \pi}_{\rm TV} \le (1 - p_{\rm min})^n \norm{\rho^{(0)} - \pi}_{\rm TV} \le (1 - p_{\rm min})^n.$

The chain converges to stationarity at an exponential rate, as claimed.

The General Case

We’ve now proved the fundamental theorem in the special case when $P > 0$ . Fortunately, together with our earlier observation that distance to stationarity is non-increasing, we can upgrade this proof into a proof for the general case.

We have assumed the Markov chain $x_0,x_1,x_2,\ldots$ is primitive, so there exists a time $n_0$ for which $P^{n_0} > 0$ . Construct an auxilliary Markov chain $z_0,z_1,z_2,\ldots$ such that one step of the auxilliary chain consists of running $n_0$ steps of the original chain:

$z_0 = x_0, \:z_1 = x_{n_0}, \:z_2 = x_{2n_0},\ldots.$

By the all-to-all case, we know that $z_0,z_1,z_2,\ldots$ converges to stationarity at an exponential rate. Thus, since the distribution of $z_k = x_{k\cdot n_0}$ is $\rho^{(k\cdot n_0)}$ , we have

$\norm{\rho^{(k\cdot n_0)} - \pi}_{\rm TV} \le (1-\delta)^k \norm{\rho^{(0)} - \pi}_{\rm TV} \le (1-\delta)^k \quad \text{for }k=0,1,2,\ldots,$

where $\delta \coloneqq \min_{1\le i,j\le m} (P^{n_0})_{ij} > 0$ . Thus, since distance to stationarity is non-increasing, we have

$\norm{\rho^{(n)} - \pi}_{\rm TV} \le \norm{\rho^{(n_0 \cdot \lfloor n/n_0\rfloor)} - \pi}_{\rm TV} \le (1-\delta)^{\lfloor n/n_0\rfloor} \norm{\rho^{(0)} - \pi}_{\rm TV} \le (1-\delta)^{\lfloor n/n_0\rfloor}.$

Thus, for any starting distribution $\rho^{(0)}$ , the distribution of the chain $\rho^{(n)}$ at time $n$ converges to stationarity at an exponential rate as $n\to\infty$ , proving the fundamental theorem.

Mixing Time

We’ve proven a quantiative version of the fundamental theorem of Markov chains, showing that the total variation distance to stationarity decreases exponentially as a function of time. For algorithmic applications of Markov chains, we also care about the rate of convergence, as it dictates how long we need to run the chain. To this end, we define the mixing time:

Definition (mixing time). The mixing time $\tau_{\rm mix}$ of a Markov chain is the number of steps required for the distance to stationarity to be at most $1/2e$ when started from a worst-case distribution:
$\tau_{\rm mix} \coloneqq \min \left\{ n \ge 1 : \max_{\rho^{(0)}} \norm{\rho^{(n)} - \pi}_{\rm TV} \le \frac{1}{2e} \right\}.$

The mixing time controls the rate of convergence for a Markov chain:

Theorem (mixing time as a convergence rate). For any starting distribution,
$\norm{\rho^{(n)} - \pi}_{\rm TV} \le e^{-\lfloor n / \tau_{\rm mix}\rfloor}.$

In particular, for $\rho^{(n)}$ to be within $\varepsilon$ total variation distance of $\pi$ , we only need to run the chain for $\tau_{\rm mix} \cdot \lceil \log(1/\varepsilon) \rceil$ steps:

Corollary (time to mix to $\varepsilon$ -stationarity). If $n\ge \tau_{\rm mix} \cdot \lceil \log(1/\epsilon)\rceil$ , then $\norm{\rho^{(n)} - \pi}_{\rm TV} \le \varepsilon$ .

These results can be proven using very similar techniques to the proof of the fundamental theorem from above. See Sinclair’s notes for more details.

Bonus: Existence of a Stationary Measure

To complete our probabilistic proof of the Markov chain convergence theorem, we must establish the existance of a stationary measure. We do this now.

Fix any state $i \in \{1,\ldots,m\}$ . Imagine starting the chain at $i$ and running it until it reaches $i$ again. Let $a_j$ be the expected number of times the chain hits $j$ in such a process, and set $a_i \coloneqq 1$ . Because the chain is primitive, all of the $a_j$ ‘s are well-defined, positive, and finite. Our claim will be that

$\pi_i = \frac{a_i}{\sum_{j=1}^m a_j}.$

is a stationary distribution for the chain. To prove this, it is sufficient to show that

(3) $a^\top P = a^\top.$

Let us prove this. Let $x_0 = i,x_1,x_2,\ldots$ denote the values of the chain and $n_{\rm ret}$ denote the time at which the chain returns to $i$ . By linearity of expectation, the expected number of hits $a_j$ is the sum over all times $n$ of the probability that the chain is at $j$ at time $n$ before hitting $i$ . That is,

$a_j = \sum_{n=1}^\infty \prob\{x_n = j, n_{\rm ret} > n\}.$

Break this sum into two pieces

$a_j = \prob\{x_1 = j\} + \sum_{n=2}^\infty \prob\{x_n = j, n_{\rm ret} > n\}.$

The first term is just the transition probability $P_{ij}$ . For the second term, break into cases depending on the value of the chain at time $n-1$ :

$\begin{align*}\prob\{x_n = j, n_{\rm ret} > n\} &= \sum_{k\ne i} \prob\{x_{n-1} = k, x_n = j, n_{\rm ret} > n-1 \} \\&= \sum_{k\ne i} \prob\{x_{n-1} = k, n_{\rm ret} > n-1 \} \prob\{x_n = j \mid x_{n-1} = k\} \\&= \sum_{k\ne i} \prob\{x_{n-1} = k, n_{\rm ret} > n-1 \} P_{kj}.\end{align*}$

Combining these two terms, we get

$a_j = P_{ij} + \sum_{n=2}^\infty \sum_{k\ne i} \prob\{x_{n-1} = k, n_{\rm ret} > n-1 \} P_{kj}.$

Relabel the outer sum to go from $n=1$ to $\infty$ and exchange the order of summation to obtain

$a_j = P_{ij} + \sum_{k\ne i} \left(\sum_{n=1}^\infty \prob\{x_n = k, n_{\rm ret} > n \}\right) P_{kj}.$

Recognize the term in the parentheses as $a_k$ . Thus, since $a_i = 1$ , we have

$a_j = a_iP_{ij} + \sum_{k\ne i} a_k P_{kj} = \sum_{k=1}^m a_k P_{kj},$

which is exactly the claim (3) we wanted to show.

Low-Rank Approximation Toolbox: Analysis of the Randomized SVD

June 22, 2023 by Ethan N. Epperly Leave a comment

In the previous post, we looked at the randomized SVD. In this post, we continue this discussion by looking at the analysis of the randomized SVD. Our approach is adapted from a new analysis of the randomized SVD by Joel A. Tropp and Robert J. Webber.

There are many types of analysis one can do for the randomized SVD. For instance, letting $B$ be a matrix and $X$ the rank- $k$ output of the randomized SVD, natural questions include:

What is the expected error $\mathbb{E} \norm{X-B}$ measured in some norm $\norm{\cdot}$ ? What about expected squared error $\mathbb{E} \left\|X-B\right\|^2$ ? How do these answers change with different norms?
How large can the randomized SVD error $\norm{X - B}$ get except for some small failure probability $\delta$ ?
How close is the randomized SVD truncated to rank $\rho$ compared to the best rank- $\rho$ approximation to $B$ ?
How close are the singular values and vectors of the randomized SVD approximation $X$ compared to those of $B$ ?

Implicitly and explicitly, the analysis of Tropp and Webber provides satisfactory answers to a number of these questions. In the interest of simplifying the presentation, we shall focus our presentation on just one of these questions, proving following result:

$\mathbb{E} \left\|B-X\right\|_{\rm F}^2 \le \left( 1 + \frac{k}{k-(r+1)}\right) \left\|B - \lowrank{B}_r \right\|_{\rm F}^2 \quad \text{for every $r \le k-2$}.$

Here, $\left\|\cdot\right\|_{\rm F}$ is the Frobenius norm. We encourage the interested reader to check out Tropp and Webber’s paper to see the methodology we summarize here used to prove numerous facts about the randomized SVD and its extensions using subspace iteration and block Krylov iteration.

Projection Formula

Let us recall the randomized SVD as we presented it in last post:

Collect information. Form $Y = B\Omega$ where $\Omega$ is an appropriately chosen $n\times k$ random matrix.
Orthogonalize. Compute an orthonormal basis $Q$ for the column space of $Y$ by, e.g., thin QR factorization.
Project. Form $C \coloneqq Q^*B$ , where ${}^*$ denotes the conjugate transpose.

The randomized SVD is the approximation

$B\approx X \coloneqq QC.$

It is easy to upgrade $X = QC$ into a compact SVD form $X = U\Sigma V^*$ , as we did as steps 4 and 5 in the previous post.

For the analysis of the randomized SVD, it is helpful to notice that the approximation $X$ takes the form

$X = QC = QQ^*B = \Pi_{B\Omega} B.$

Here, $\Pi_{B\Omega}$ denotes the orthogonal projection onto the column space of the matrix $B\Omega$ . We call $X = \Pi_{B\Omega} B$ the projection formula for the randomized SVD.¹

Aside: Quadratic Unitarily Invariant Norms

To state our error bounds in the most general language, we can adopt the language of quadratic unitarily invariant norms. Recall that a square matrix $U$ is unitary if

$U^*U = UU^* = I.$

You may recall from my post on Nyström approximation that a norm $\left\|\cdot\right\|_{\rm UI}$ on matrices is unitarily invariant if

$\left\|UBV\right\|_{\rm UI} = \left\|B\right\|_{\rm UI} \quad \text{for all unitary matrices $U$, $V$ and any matrix $B$}.$

A unitarily invariant norm $\left\|\cdot\right\|_{\rm Q}$ is said to be quadratic if there exists another unitarily invariant norm $\left\|\cdot\right\|_{\rm UI}$ such that

(1) $\left\|B\right\|_{\rm Q}^2 = \left\|B^*B\right\|_{\rm UI} = \left\|BB^*\right\|_{\rm UI} \quad \text{for every matrix $B$.}$

Many examples of quadratic unitarily invariant norms are found among the Schatten $p$ -norms, defined as

$\left\|B\right\|_{S_p}^p \coloneqq \sum_i \sigma_i(B)^p.$

Here, $\sigma_1(B) \ge \sigma_2(B) \ge \cdots$ denote the decreasingly order singular values of $B$ . The Schatten $2$ -norm $\left\|\cdot\right\|_{S_2}$ is the Frobenius norm of a matrix. The spectral norm (i.e., operator 2-norm) is the Schatten $\infty$ -norm, defined to be

$\norm{B} = \left\|B\right\|_{S_\infty} \coloneqq \max_i \sigma_i(B) = \sigma_1(B).$

The Schatten $p$ -norms are unitarily invariant norms for every $1\le p \le \infty$ . However, the Schatten $p$ -norms are quadratic unitarily invariant norms only for $2\le p \le \infty$ since

$\left\|B\right\|_{S_p}^2 = \left\|B^*B\right\|_{S_{p/2}}.$

For the remainder of this post, we let $\left\|\cdot\right\|_{\rm Q}$ and $\left\|\cdot\right\|_{\rm UI}$ be a quadratic unitarily invariant norm pair satisfying (1).

Error Decomposition

The starting point of our analysis is the following decomposition of the error of the randomized SVD:

(2) $\left\|B - X\right\|_{\rm Q}^2 \le \left\|B - \lowrank{B\right\|_r}_{\rm Q}^2 + \left\|\lowrank{B\right\|_r - \Pi_{B\Omega} \lowrank{B}_r}_{\rm Q}^2.$

Recall that we have defined $\lowrank{B}_r$ to be an optimal rank- $r$ approximation to $B$ :

$\left\|B - \lowrank{B\right\|_r}_{\rm Q} \le \left\|B - C\right\|_{\rm Q} \quad \text{for any rank-$r$ matrix $C$}.$

Thus, the error decomposition says that the excess error is bounded by

$\left\|B - X\right\|_{\rm Q}^2 - \left\|B - \lowrank{B\right\|_r}_{\rm Q}^2 \le \left\|\lowrank{B\right\|_r - \Pi_{B\Omega} \lowrank{B}_r}_{\rm Q}^2.$

Using the projection formula, we can prove the error decomposition (2) directly:

$\begin{align*}\left\|B - X\right\|_{\rm Q}^2 &= \left\|B - \Pi_{B\Omega} B \right\|_{\rm Q}^2\\&= \norm{(I-\Pi_{B\Omega})BB^*(I-\Pi_{B\Omega})}_{\rm UI} \\&= \norm{(I-\Pi_{B\Omega})(BB^* - \lowrank{B}_r\lowrank{B}_r^*)(I-\Pi_{B\Omega})}_{\rm UI} \\&\qquad + \norm{(I-\Pi_{B\Omega})\lowrank{B}_r\lowrank{B}_r^*(I-\Pi_{B\Omega})}_{\rm UI}\\&\le \left\|BB^* - \lowrank{B}_r\lowrank{B}_r^*\right|_{\rm UI} + \left\|\lowrank{B}_r - \Pi_{B\Omega} \lowrank{B}_r\right\|_{\rm Q}^2 \\&\le \left\|B - \lowrank{B}_r \right\|_{\rm Q}^2 + \left\|\lowrank{B}_r- \Pi_{B\Omega} \lowrank{B}_r\right\|_{\rm Q}^2.\end{align*}$

The first line is the projection formula, the second is relation between $\norm{\cdot}_{\rm Q}$ and $\norm{\cdot}_{\rm UI}$ , and the third is the triangle inequality. For the fifth line, we use the fact that $I - \Pi_{B\Omega}$ is an orthoprojector and thus has unit spectral norm $\norm{I - \Pi_{B\Omega}}$ together with the fact that

$\left\|BCD\right\|_{\rm UI} \le \norm{B} \cdot \left\|C\right\|_{\rm UI}\cdot \norm{D} \quad \text{for all matrices $B,C,D$.}$

For the final inequality, we used commutation rule

$(B - \lowrank{B}_r)(B - \lowrank{B}_r)^* = BB^* - \lowrank{B}_r\lowrank{B}_r^*.$

Bounding the Error

In this section, we shall continue to refine the error decomposition (2). Consider a thin SVD of $B$ , partitioned as

$B = \onebytwo{U_1}{U_2} \twobytwo{\Sigma_1}{0}{0}{\Sigma_2}\onebytwo{V_1}{V_2}^*$

where

$\onebytwo{U_1}{U_2}$ and $\onebytwo{V_1}{V_2}$ both have orthonormal columns,
$\Sigma_1$ and $\Sigma_2$ are square diagonal matrices with nonnegative entries,
$U_1$ and $V_1$ have $r$ columns, and
$\Sigma_1$ is an $r\times r$ matrix.

Under this notation, the best rank- $r$ approximation is $\lowrank{B}_r = U_1\Sigma_1V_1^*$ . Define

$\twobyone{\Omega_1}{\Omega_2} \coloneqq \twobyone{V_1^*}{V_2^*} \Omega.$

We assume throughout that $\Omega_1$ is full-rank (i.e., $\rank \Omega_1 = r$ ).

Applying the error decomposition (2), we get

(3) $\begin{align*}\left\|B - X\right\|_{\rm Q}^2&\le \norm{ \onebytwo{U_1}{U_2} \twobytwo{0}{0}{0}{\Sigma_2}\onebytwo{V_1}{V_2}^*}_{\rm Q}^2 + \norm{(I-\Pi_{B\Omega})U_1\Sigma_1V_1^*}_{\rm Q}^2 \\&= \left\| \Sigma_2 \right\|_{\rm Q}^2 + \norm{(I-\Pi_{B\Omega})U_1\Sigma_1}_{\rm Q}^2. \end{align*}$

For the second line, we used the unitary invariance of $\left\|\cdot\right\|_{\rm Q}$ . Observe that the column space of $B\Omega$ is a subset of the column space of $B\Omega\Omega_1^\dagger$ so

(4) $\norm{(I-\Pi_{B\Omega})U_1\Sigma_1}_{\rm Q}^2 \le \norm{(I-\Pi_{B\Omega\Omega_1^\dagger})U_1\Sigma_1}_{\rm Q}^2 = \norm{\Sigma_1U_1^*(I-\Pi_{B\Omega\Omega_1^\dagger})U_1\Sigma_1}_{\rm UI}.$

Let’s work on simplifying the expression

$\Sigma_1U_1^*(I-\Pi_{B\Omega\Omega_1^\dagger})U_1\Sigma_1.$

First, observe that

$B\Omega\Omega_1^\dagger = B\onebytwo{V_1}{V_2}\twobyone{\Omega_1}{\Omega_2}\Omega_1^\dagger = \onebytwo{U_1}{U_2} \twobytwo{\Sigma_1}{0}{0}{\Sigma_2} \twobyone{I}{\Omega_2\Omega_1^\dagger} = U_1\Sigma_1 + U_2\Sigma_2\Omega_2\Omega_1^\dagger.$

Thus, the projector $\Pi_{B\Omega\Omega_1^\dagger}$ takes the form

$\begin{align*}\Pi_{B\Omega\Omega_1^\dagger} &= (B\Omega\Omega_1^\dagger) \left[(B\Omega\Omega_1^\dagger)^*(B\Omega\Omega_1^\dagger)\right]^\dagger (B\Omega\Omega_1^\dagger)^* \\&= (U_1\Sigma_1 + U_2\Sigma_2\Omega_2\Omega_1^\dagger) \left[\Sigma_1^2 + (\Sigma_2\Omega_2\Omega_1^\dagger)^*(\Sigma_2\Omega_2\Omega_1^\dagger)\right]^\dagger (U_1\Sigma_1 + U_2\Sigma_2\Omega_2\Omega_1^\dagger)^*.\end{align*}$

Here, ${}^\dagger$ denotes the Moore–Penrose pseudoinverse, which reduces to the ordinary matrix inverse for a square nonsingular matrix. Thus,

(5) $\Sigma_1U_1^*(I-\Pi_{B\Omega\Omega_1^\dagger})U_1\Sigma_1 = \Sigma_1^2 - \Sigma_1^2 \left[\Sigma_1^2 + (\Sigma_2\Omega_2\Omega_1^\dagger)^*(\Sigma_2\Omega_2\Omega_1^\dagger)\right]^\dagger\Sigma_1^2.$

Remarkably, this seemingly convoluted combination of matrices actually is a well-studied operation in matrix theory, the parallel sum. The parallel sum of positive semidefinite matrices $A$ and $H$ is defined as

(6) $A : H \coloneqq A - A(A+H)^\dagger A.$

We will have much more to say about the parallel sum soon. For now, we use this notation to rewrite (5) as

$\Sigma_1U_1^*(I-\Pi_{B\Omega\Omega_1^\dagger})U_1\Sigma_1 = \Sigma_1^2 : [(\Sigma_2\Omega_2\Omega_1^\dagger)^*(\Sigma_2\Omega_2\Omega_1^\dagger)].$

Plugging this back into (4) and then (3), we obtain the error bound

(7) $\left\|B - X\right\|_{\rm Q}^2\le \left\| \Sigma_2 \right\|_{\rm Q}^2 + \left\|\Sigma_1^2 : [(\Sigma_2\Omega_2\Omega_1^\dagger)^*(\Sigma_2\Omega_2\Omega_1^\dagger)]\right\|_{\rm UI}.$

Simplifying this expression further will require more knowledge about parallel sums, which we shall discuss now.

Aside: Parallel Sums

Let us put aside the randomized SVD for a moment and digress to the seemingly unrelated topic of electrical circuits. Suppose we have a battery of voltage $v$ and we connect the ends using a wire of resistance $a$ . The current $\mathrm{curr}_a$ is given by Ohm’s law

$v = a \cdot \mathrm{curr}_a.$

Similarly, if the wire is replaced by a different wire with resistance $h$ , the current $\mathrm{curr}_h$ is then

$v = h \cdot \mathrm{curr}_h.$

Now comes the interesting question. What if we connect the battery by both wires (resistances $a$ and $h$ ) simultaneously in parallel, so that current can flow through either wire? Since wires of resistances $a$ and $h$ still have voltage $v$ , Ohm’s law still applies and thus the total current is

$\mathrm{curr}_{\rm total} = \mathrm{curr}_a + \mathrm{curr}_h = \frac{v}{a} + \frac{v}{h} = v \left(a^{-1}+h^{-1}\right).$

The net effect of connecting resisting wires of resistances $a$ and $h$ is the same as a single resisting wire of resistance

(8) $a:h \coloneqq \frac{v}{\mathrm{curr}_{\rm total}} = \left( a^{-1}+h^{-1}\right)^{-1}.$

We call the the operation $a \mathbin{:} h$ the parallel sum of $a$ and $h$ because it describes how resistances combine when connected in parallel.

The parallel sum operation $:$ can be extended to all nonnegative numbers $a$ and $h$ by continuity:

$a:h \coloneqq \lim_{b\downarrow a, \: k\downarrow h} b:k = \begin{cases}\left( a^{-1}+h^{-1}\right)^{-1}, & a,h > 0 ,\\0, & \textrm{otherwise}.\end{cases}$

Electrically, this definition states that one obtains a short circuit (resistance $0$ ) if either of the wires carries zero resistance.

Parallel sums of matrices were introduced by William N. Anderson, Jr. and Richard J. Duffin as a generalization of the parallel sum operation from electrical circuits. There are several equivalent definitions. The natural definition (8) applies for positive definite matrices

$A:H = (A^{-1}+H^{-1})^{-1} \quad \text{for $A,H$ positive definite.}$

This definition can be extended to positive semidefinite matrices by continuity. It can be useful to have an explicit definition which applies even to positive semidefinite matrices. To do so, observe that the (scalar) parallel sum satisfies the identity

$a:h = \frac{1}{a^{-1}+h^{-1}} = \frac{ah}{a+h} = \frac{a(a+h)-a^2}{a+h} = a - a(a+h)^{-1}a.$

To extend to matrices, we capitalize the letters and use the Moore–Penrose pseudoinverse in place of the inverse:

$A : H = A - A(A+H)^\dagger A.$

This was the definition (6) of the parallel sum we gave above which we found naturally in our randomized SVD error bounds.

The parallel sum enjoys a number of beautiful properties, some of which we list here:

Symmetry. $A:H = H:A$ ,
Simplified formula. $A:H = (A^{-1}+H^{-1})^{-1}$ for positive definite matrices,
Bounds. $A:H \preceq A$ and $A:H \preceq H$ .
Monotone. $A\mapsto (A:H)$ is monotone in the Loewner order.
Concavity. The map $(A,H) \mapsto A:H$ is (jointly) concave (with respect to the Loewner order).
Conjugation. For any square $C$ , $C^*(A:H)C = (C^*AC):(C^*HC)$ .
Trace. $\tr(A:H) \le (\tr A):(\tr H)$ .²
Unitarily invariant norms. $\left\|A:H\right\|_{\rm UI} \le \left\|A\right\|_{\rm UI} : \left\|H\right\|_{\rm UI}$ .

For completionists, we include a proof of the last property for reference.

Proof of Property 8

Let $\left\|\cdot\right\|_{\rm UI}'$ be the unitarily invariant norm dual to $\left\|\cdot\right\|_{\rm UI}$ .³ By duality, we have

$\begin{align*}\left\|A:H\right\|_{\rm UI} &= \max_{M\succeq 0,\: \left\|M\right\|_{\rm UI}'\le 1} \tr((A:H)M) \\&= \max_{M\succeq 0,\: \left\|M\right\|_{\rm UI}'\le 1} \tr(M^{1/2}(A:H)M^{1/2})\\&= \max_{M\succeq 0,\: \left\|M\right\|_{\rm UI}'\le 1} \tr((M^{1/2}AM^{1/2}):(M^{1/2}HM^{1/2})) \\&\le \max_{M\succeq 0,\: \left\|M\right\|_{\rm UI}'\le 1} \left[\tr(M^{1/2}AM^{1/2}):\tr(M^{1/2}HM^{1/2}) \right] \\&\le \left[\max_{M\succeq 0,\: \left\|M\right\|_{\rm UI}'\le 1} \tr(M^{1/2}AM^{1/2})\right]:\left[\max_{M\succeq 0,\: \left\|M\right\|_{\rm UI}'\le 1} \tr(M^{1/2}HM^{1/2})\right]\\&= \left\|A\right\|_{\rm UI} : \left\|H\right\|_{\rm UI}.\end{align*}$

The first line is duality, the second holds because $M\succeq 0$ , the third line is property 6, the fourth line is property 7, the fifth line is property 4, and the sixth line is duality.

Nearing the Finish: Enter the Randomness

Equipped with knowledge about parallel sums, we are now prepared to return to bounding the error of the randomized SVD. Apply property 8 of the parallel sum to the randomized SVD bound (7) to obtain

$\begin{align*}\left\|B - X\right\|_{\rm Q}^2&\le \left\| \Sigma_2 \right\|_{\rm Q}^2 + \left\|\Sigma_1^2 : [(\Sigma_2\Omega_2\Omega_1^\dagger)^*(\Sigma_2\Omega_2\Omega_1^\dagger)]\right\|_{\rm UI} \\&\le \left\| \Sigma_2 \right\|_{\rm Q}^2 + \left\|\Sigma_1^2\right\|_{\rm UI} : \left\|[(\Sigma_2\Omega_2\Omega_1^\dagger)^*(\Sigma_2\Omega_2\Omega_1^\dagger)]\right\|_{\rm UI} \\&= \left\| \Sigma_2 \right\|_{\rm Q}^2 + \left\|\Sigma_1\right\|_{\rm Q}^2 : \left\|\Sigma_2\Omega_2\Omega_1^\dagger\right\|^2_{\rm Q}.\end{align*}$

This bound is totally deterministic: It holds for any choice of test matrix $\Omega$ , random or otherwise. We can use this bound to analyze the randomized SVD in many different ways for different kinds of random (or non-random) matrices $\Omega$ . Tropp and Webber provide a number of examples instantiating this general bound in different ways.

We shall present only the simplest error bound for randomized SVD. We make the following assumptions:

The test matrix $\Omega$ is standard Gaussian.⁴
The norm $\left\|\cdot\right\|_{\rm Q}$ is the Frobenius norm $\left\|\cdot\right\|_{\rm F}$ .
The randomized SVD rank $k$ is $\ge r-2$ .

Under these assumptions and using the property $a : h \le h$ , we obtain the following expression for the expected error of the randomized SVD

(9) $\mathbb{E} \left\|B - X\right\|_{\rm F}^2\le \left\| B - \lowrank{B\right\|_r }_{\rm F}^2 + \mathbb{E}\left\|\Sigma_2\Omega_2\Omega_1^\dagger\right\|^2_{\rm F}.$

All that is left to is compute the expectation of $\left\|\Sigma_2\Omega_2\Omega_1^\dagger\right\|_{\rm F}^2$ . Remarkably, this can be done in closed form.

Aside: Expectation of Gaussian–Inverse Gaussian Matrix Product

The final ingredient in our randomized SVD analysis is the following computation involving Gaussian random matrices. Let $G$ and $\Gamma$ be independent standard Gaussian random matrices with $\Gamma$ of size $r\times k$ , $k\ge r-2$ . Then

$\mathbb{E} \left\|SG\Gamma^\dagger R\right\|_{\rm F}^2 = \frac{1}{k-(r+1)}\left\|S\right\|_{\rm F}^2\left\|R\right\|_{\rm F}^2.$

For the probability lovers, we will now go on to prove this. For those who are less probabilistically inclined, this result can be treated as a black box.

Proof of Random Matrix Formula

To prove this, first consider a simpler case where $\Gamma^\dagger$ does not appear:

$\mathbb{E} \left\|SGW\right\|_{\rm F}^2 = \mathbb{E} \sum_{ij} \left(\sum_{k\ell} s_{ik}g_{k\ell}w_{\ell j} \right)^2 = \mathbb{E} \sum_{ijk\ell} s_{ik}^2 w_{\ell j}^2 = \left\|S\right\|_{\rm F}^2\left\|W\right\|_{\rm F}^2.$

Here, we used the fact that the entries $g_{k\ell}$ are independent, mean-zero, and variance-one. Thus, applying this result conditionally on $\Gamma$ with $W = \Gamma^\dagger R$ , we get

(10) $\mathbb{E} \left\|SG\Gamma^\dagger R\right\|_{\rm F}^2 =\mathbb{E}\left[ \mathbb{E}\left[ \left\|SG\Gamma^\dagger R\right\|_{\rm F}^2 \,\middle|\, \Gamma\right]\right] =\left\|S\right\|_{\rm F}^2\cdot \mathbb{E} \left\|\Gamma^\dagger R\right\|_{\rm F}^2.$

To compute $\mathbb{E} \left\|\Gamma^\dagger R\right\|_{\rm F}^2$ , we rewrite using the trace

$\mathbb{E} \left\|\Gamma^\dagger R\right\|_{\rm F}^2 = \mathbb{E} \tr \left(\Gamma^\dagger RR^* \Gamma^{*\dagger} \right)= \tr \left(RR^* \mathbb{E}[\Gamma^{*\dagger}\Gamma^\dagger] \right) = \tr \left(RR^* \mathbb{E}[(\Gamma\Gamma^*)^{-1}] \right).$

The matrix $(\Gamma\Gamma^*)^{-1}$ is known as an inverse-Wishart matrix and is a well-studied random matrix model. In particular, its expectation is known to be $(k-(r+1))^{-1}I$ . Thus, we obtain

$\mathbb{E} \left\|\Gamma^\dagger R\right\|_{\rm F}^2 = \tr \left(RR^* \mathbb{E}[(\Gamma\Gamma^*)^{-1}] \right) = \frac{\tr \left(RR^* \right)}{k-(r+1)} = \frac{\left\|R\right\|_{\rm F}^2}{k-(r+1)}.$

Plugging into (10), we obtain the desired conclusion

$\mathbb{E} \left\|SG\Gamma^\dagger R\right\|_{\rm F}^2 =\left\|S\right\|_{\rm F}^2\cdot \mathbb{E} \left\|\Gamma^\dagger R\right\|_{\rm F}^2 = \frac{1}{k-(r+1)}\left\|S\right\|_{\rm F}^2\left\|R\right\|_{\rm F}^2.$

This completes the proof.

Wrapping it All Up

We’re now within striking distance. All that is left is to assemble the pieces we have been working with to obtain a final bound for the randomized SVD.

Recall we have chosen the random matrix $\Omega$ to be standard Gaussian. By the rotational invariance of the Gaussian distribution, $\Omega_1$ and $\Omega_2$ are independent and standard Gaussian as well. Plugging the matrix expectation bound to (9) then completes the analysis

$\begin{align*}\mathbb{E} \left\|B - X\right\|_{\rm F}^2&\le \left\| B - \lowrank{B\right\|_r }_{\rm F}^2 + \mathbb{E}\norm{\Sigma_2\Omega_2\Omega_1^\dagger I_{k\times k}}^2_{\rm F} \\&= \left\| B - \lowrank{B\right\|_r }_{\rm F}^2 + \frac{\left\|\Sigma_2\right\|_{\rm F}^2\norm{I_{k\times k}}_{\rm F}^2}{k-(r+1)} \\&= \left(1 + \frac{k}{k-(r+1)}\right) \left\| B - \lowrank{B\right\|_r }_{\rm F}^2.\end{align*}$

Low-Rank Approximation Toolbox: Randomized SVD

June 12, 2023 by Ethan N. Epperly 2 Comments

The randomized SVD and its relatives are workhorse algorithms for low-rank approximation. In this post, I hope to provide a fairly brief introduction to this useful method. In the following post, we’ll look into its analysis.

The randomized SVD is dead-simple to state: To approximate an $m\times n$ matrix $B$ by a rank- $k$ matrix, perform the following steps:

Collect information. Form $Y = B\Omega$ where $\Omega$ is an appropriately chosen $n\times k$ random matrix.
Orthogonalize. Compute an orthonormal basis $Q$ for the column space of $Y$ by, e.g., thin QR factorization.
Project. Form $C \coloneqq Q^*B$ . (Here, ${}^*$ denotes the conjugate transpose of a complex matrix, which reduces to the ordinary transpose if the matrix is real.)

If all one cares about is a low-rank approximation to the matrix $B$ , one can stop the randomized SVD here, having obtained the approximation

$B \approx X \coloneqq Q C.$

As the name “randomized SVD” suggests, one can easily “upgrade” the approximation $X = QC$ to a compact SVD format:

SVD. Compute a compact SVD of the matrix $C$ : $C = \hat{U}\Sigma V^*$ where $\hat{U}$ and $\Sigma$ and $k\times k$ and $V$ is $n\times k$ .
Clean up. Set $U \coloneqq Q\hat{U}$ .

We now have approximated $B$ by a rank- $k$ matrix $X$ in compact SVD form:

$B \approx X = QC = U\Sigma V^*.$

One can use the factors $U$ , $\Sigma$ , and $V$ of the approximation $X \approx B$ as estimates of the singular vectors and values of the matrix $B$ .

What Can You Do with the Randomized SVD?

The randomized SVD has many uses. Pretty much anywhere you would ordinarily use an SVD is a candidate for the randomized SVD. Applications include model reduction, fast direct solvers, computational biology, uncertainty quantification, among numerous others.

To speak in broad generalities, there are two ways to use the randomized SVD:

As an approximation. First, we could use $X$ as a compressed approximation to $B$ . Using the format $X = QC$ , $X$ requires just $(m+n)k$ numbers of storage, whereas $B$ requires a much larger $mn$ numbers of storage. As I detailed at length in my blog post on low-rank matrices, many operations are cheap for the low-rank matrix $X$ . For instance, we can compute a matrix–vector product $Xz$ in roughly $(m+n)k$ operations rather than the $mn$ operations to compute $Bz$ . For these use cases, we don’t need the “SVD” part of the randomized SVD.
As an approximate SVD. Second, we can actually use the $U$ , $\Sigma$ , and $V$ produced by the randomized SVD as approximations to the singular vectors and values of $B$ . In these applications, we should be careful. Just because $X$ is a good approximation to $B$ , it is not necessarily the case that $U$ , $\Sigma$ , and $V$ are close to the singular vectors and values of $B$ . To use the randomized SVD in this context, it is safest to use posterior diagnostics (such as the ones developed in this paper of mine) to ensure that the singular values/vectors of interest are computed to a high enough accuracy. A useful rule of thumb is the smallest two to five singular values and vectors computed by the randomized SVD are suspect and should be used in applications only with extreme caution. When the appropriate care is taken and for certain problems, the randomized SVD can provide accurate singular vectors far faster than direct methods.

Accuracy of the Randomized SVD

How accurate is the approximation $X$ produced by the randomized SVD? No rank- $k$ approximation can be more accurate than the best rank- $k$ approximation $\lowrank{B}_k$ to the matrix $B$ , furnished by an exact SVD of $B$ truncated to rank $k$ . Thus, we’re interested in how much larger $B - X$ is than $B - \lowrank{B}_k$ .

Frobenius Norm Error

A representative result describing the error of the randomized SVD is the following:

(1) $\mathbb{E} \left\|B - X\right\|_{\rm F}^2 \le \min_{r \le k-2} \left( 1 + \frac{r}{k-(r+1)} \right) \left\|B - \lowrank{B}_r\right\|^2_{\rm F}.$

This result states that the average squared Frobenius-norm error of the randomized SVD is comparable with the best rank- $r$ approximation error for any $r\le k-2$ . In particular, choosing $k = r+2$ , we see that the randomized SVD error is at most $(1+r)$ times the best rank- $r$ approximation

(2) $\mathbb{E} \left\|B - X\right\|_{\rm F}^2 \le (1+r) \left\|B - \lowrank{B}_r \right\|^2_{\rm F} \quad \text{for $k=r+2$}.$

Choosing $k = 2r+1$ , we see that the randomized SVD has at most twice the error of the best rank- $r$ approximation

(3) $\mathbb{E} \left\|B - X\right\|_{\rm F}^2 \le 2 \left\|B - \lowrank{B}_r\right\|^2_{\rm F} \quad \text{for $k=2r+1$}.$

In effect, these results tell us that if we want an approximation that is nearly as good as the best rank- $r$ approximation, using an approximation rank of, say, $k = r+2$ or $k = 2r$ for the randomized SVD suffices. These results hold even for worst-case matrices. For nice matrices with steadily decaying singular values, the randomized SVD can perform even better than equations (2)–(3) would suggest.

Spectral Norm Error

The spectral norm is often a better error metric for matrix computations than the Frobenius norm. Is the randomized SVD also comparable with the best rank- $k$ approximation when we use the spectral norm? In this setting, a representative error bound is

(4) $\mathbb{E} \left\|B - X\right\|^2 \le \min_{r \le k-2} \left( 1 + \frac{2r}{k-(r+1)} \right) \left(\left\|B - \lowrank{B}_r \right\|^2 + \frac{\mathrm{e}^2}{k-r} \left\|B - \lowrank{B}_r\right\|^2_{\rm F} \right).$

The spectral norm error of the randomized SVD depends on the Frobenius norm error of the best rank- $r$ approximation for $r \le k-2$ .

Recall that the spectral and Frobenius norm can be defined in terms of the singular values of the matrix $B$ :

$\norm{B} = \sigma_1(B),\quad \left\|B\right\|_{\rm F}^2 = \sum_i \sigma_i(B)^2.$

Rewriting (4) using these formulas, we get

$\mathbb{E} \left\|B - X\right\|^2 \le \min_{r \le k-2} \left( 1 + \frac{2r}{k-(r+1)} \right) \left(\sigma_{r+1}^2 + \frac{\mathrm{e}^2}{k-r} \sum_{i>r} \sigma_i^2 \right).$

Despite being interested in the largest singular value of the error matrix $B-X$ , this bound demonstrates that the randomized SVD incurs errors based on the entire tail of $B$ ‘s singular values $\sum_{i>r}\sigma_i^2$ . The randomized SVD is much worse than the best rank- $k$ approximation for a matrix $B$ with a long detail of slowly declining singular values.

Improving the Approximation: Subspace Iteration and Block Krylov Iteration

We saw that the randomized SVD produces nearly optimal low-rank approximations when we measure using the Frobenius norm. When we measure using the spectral norm, we have a split decision: If the singular values decay rapidly, the randomized SVD is near-optimal; if the singular values decay more slowly, the randomized SVD can suffer from large errors.

Fortunately, there are two fixes to improve the randomized SVD in the presence of slowly decaying singular values:

Subspace iteration: To improve the quality of the randomized SVD, we can combine it with the famous power method for eigenvalue computations. Instead of $Y = B\Omega$ , we use the powered expression $Y = B(B^*B)^q\Omega$ . Powering has the effect of amplifying the large singular values of $B$ and dimishing the influence of the tail.
Block Krylov iteration: Subspace iteration is effective, but fairly wasteful. To compute $B(B^*B)^q\Omega$ we compute the block Krylov sequence
$B\Omega, BB^*B\Omega, BB^*BB^*B\Omega,\ldots,B(B^*B)^q\Omega$
and throw away all but the last term. To make more economical use of resources, we can use the entire sequence as our information matrix $Y$ :
$Y = \begin{bmatrix} B\Omega & BB^*B\Omega & BB^*BB^*B\Omega& \cdots & B(B^*B)^q\Omega\end{bmatrix}.$
Storing the entire block Krylov sequence is more memory-intensive but makes more efficient use of matrix products than subspace iteration.

Both subspace iteration and block Krylov iteration should be carefully implemented to produce accurate results.

Both subspace iteration and block Krylov iteration diminish the effect of a slowly decaying tail of singular values on the accuracy of the randomized SVD. The more slowly the tail decays, the larger one must take the number of iterations $q$ to obtain accurate results.

The Hard Way to Prove Jensen’s Inequality

June 5, 2023 by Ethan N. Epperly Leave a comment

In this post, I want to discuss a very beautiful piece of mathematics I stumbled upon recently. As a warning, this post will be more mathematical than most, but I will still try and sand off the roughest mathematical edges. This post is adapted from a much more comprehensive post by Paata Ivanishvili. My goal is to distill the main idea to its essence, deferring the stochastic calculus until it cannot be avoided.

Jensen’s inequality is one of the most important results in probability.

Jensen’s inequality. Let $X$ be a (real) random variable and $f:\real\to\real$ a convex function such that both $\mathbb{E} X$ and $\mathbb{E} f(X)$ are defined. Then $f(\mathbb{E}X) \le \mathbb{E} [f(X)]$ .

Here is the standard proof. A convex function has supporting lines. That is, at a point $a \in \real$ , there exists a slope $m$ such that $m(x-a) + f(a) \le f(x)$ for all $x \in \real$ . Invoke this result at $a = \mathbb{E} X$ and $x = X$ and take expectations to conclude that

$\mathbb{E}[m(X - \mathbb{E}X) + f(\mathbb{E}X)] = f(\mathbb{E}X) \le \mathbb{E} [f(X)].$

In this post, I will outline a proof of Jensen’s inequality which is much longer and more complicated. Why do this? This more difficult proof illustrates an incredible powerful technique for proving inequalities, interpolation. The interpolation method can be used to prove a number of difficult and useful inequalities in probability theory and beyond. As an example, at the end of this post, we will see the Gaussian Jensen inequality, a striking generalization of Jensen’s inequality with many applications.

The idea of interpolation is as follows: Suppose I wish to prove $A_0 \le A_1$ for two numbers $A_0$ and $A_1$ . This may hard to do directly. With the interpolation method, I first construct a family of numbers $A_t$ , $0 \le t \le 1$ , such that $A_{t = 0} = A_0$ and $A_{t=1} = A_1$ and show that $(A_t : 0\le t\le 1)$ is (weakly) increasing in $t$ . This is typically accomplished by showing the derivative is nonnegative:

$\frac{d}{dt} A_t \ge 0.$

To prove Jensen’s inequality by interpolation, we shall begin with a special case. As often in probability, the simplest case is that of a Gaussian random variable.

Jensen’s inequality for a Gaussian. Let $X$ be a standard Gaussian random variable (i.e., mean-zero and variance $1$ ) and let $f:\real\to\real$ be a thrice-differentiable convex function satisfying a certain technical condition.¹ Then
$f(0) \le \mathbb{E} [f(X)].$

Note that the conclusion is exactly Jensen’s inequality, as we have assumed $X$ is mean-zero.

The difficulty with any proof by interpolation is to come up with the “right” $A_t$ . For us, the “right” answer will take the form

$A_t = \mathbb{E} [ f(X_t) ],$

where $X_0 = 0$ starts with no randomness and $X_1 = X$ is our standard Gaussian. To interpolate between these extremes, we increase the variance linearly from $0$ to $1$ . Thus, we define

$A_t = \mathbb{E} [ f(X_t)] \quad \text{where $X_t\sim\mathcal{N}(0,t)$}.$

Here, and throughout, $\mathcal{N}(0,v)$ denotes a Gaussian random variable with zero mean and variance $v$ .

Let’s compute the derivative of $A_t$ . To do this, let $\delta > 0$ denote a small parameter which we will later send to zero. For us, the key fact will be that a $\mathcal{N}(0,t+\delta)$ can be realized as a sum of independent $\mathcal{N}(0,t)$ and $\mathcal{N}(0,\delta)$ random variables. Therefore, we write

$X_{t+\delta} = X_t + \Delta \quad \text{where $\Delta \sim \mathcal{N}(0,\delta)$ is independent of $X_t$.}$

We now evaluate $f(X_t+\Delta)$ by using Taylor’s formula

(1) $f(X_t+\Delta) = f(X_t) + f'(X_t)\Delta + \frac{1}{2} f''(X_t) \Delta^2 + \frac{1}{6} f'''(\xi) \Delta^3,$

where $\xi$ lies between $X_t$ and $X_t+\Delta$ . Now, take expectations,

$\mathbb{E}[ f(X_t+\Delta)]=\mathbb{E}[f(X_t)] + \mathbb{E}[f'(X_t)\Delta] + \frac{1}{2} \mathbb{E}[f''(X_t)] \mathbb{E}[\Delta^2] + \underbrace{\frac{1}{6} \mathbb{E}[f'''(\xi) \Delta^3]}_{:=\mathrm{Rem}(\delta)}.$

The random variable $\Delta$ has mean zero and variance $\delta$ so this gives

$\mathbb{E} [f(X_t+\Delta)]=\mathbb{E}[f(X_t)] + \delta \frac{1}{2} \mathbb{E}[f''(X_t)] + \mathrm{Rem}(\delta).$

As we show below, the remainder term $\mathrm{Rem}(\delta)/\delta$ vanishes as $\delta\to 0$ . Thus, we can rearrange this expression to compute the derivative:

$\frac{d}{dt} A_t = \lim_{\delta \downarrow 0} \frac{\mathbb{E} f(X_t+\Delta)-\mathbb{E}[f(X_t)]}{\delta} = \lim_{\delta \downarrow 0} \frac{1}{2} \mathbb{E}[f''(X_t)] + \frac{\mathrm{Rem}(\delta)}{\delta} = \frac{1}{2} \mathbb{E}[f''(X_t)].$

The second derivative of a convex function is nonnegative: $f''(x) \ge 0$ for every $x$ . Therefore,

$\frac{d}{dt} A_t \ge 0 \quad \text{for all } t\in [0,1].$

Jensen’s inequality is proven! In fact, we’ve proven the stronger version of Jensen’s inequality:

$\mathbb{E} f(X) = f(0) + \frac{1}{2} \int_0^1 \mathbb{E} [f''(X_t)] \, dt.$

This strengthened version can yield improvements. For instance, if $f$ is $\beta$ -smooth

$f''(x) \le \beta \quad \text{for every } x \in \real,$

then we have

$f(0) \le \mathbb{E} f(X) \le f(0) + \frac{1}{2}\beta.$

This inequality isn’t too hard to prove directly, but it does show that we’ve obtained something more than the simple proof of Jensen’s inequality.

Analyzing the Remainder Term

Let us quickly check that the remainder term vanishes $\mathrm{Rem}(\delta)/\delta$ as $\delta \to 0$ . Let’s do this. As an exercise, you can verify that our technical regularity condition implies $\mathbb{E} |f'''(\xi)|^p < +\infty$ . Thus, by Hölder’s inequality and setting $q$ to be $p$ ‘s Hölder conjugate ( $1/p = 1/q = 1$ ), we obtain

$\frac{|\mathrm{Rem}(\delta)|}{\delta} = \frac{|\mathbb{E}[f'''(\xi) \Delta^3]|}{6\delta} \le \frac{(|\mathbb{E} |f'''(\xi)|^p)^{1/p}| (\mathbb{E} |\Delta|^{3q})^{1/q}}{6\delta}.$

One can show that $(\mathbb{E} |\Delta|^{3q})^{1/q} \le C(q) \delta^{3/2}$ where $C(q)$ is a function of $q$ alone. Therefore, $|\mathrm{Rem}(\delta)|/\delta \le \mathrm{constant} \cdot \delta^{1/2} \to 0$ as $\delta \downarrow 0$ .

What’s Really Going On Here?

In our proof, we use a family of random variables $X_t \sim \mathcal{N}(0,t)$ , defined for each $0\le t \le 1$ . Rather than treating these quantities as independent, we can think of them as a collective, comprising a random function $t \mapsto X_t$ known as a Brownian motion.

The Brownian motion is a very natural way of interpolating between a constant $\mu$ and a Gaussian with mean $\mu$ .²

There is an entire subject known as stochastic calculus which allows us to perform computations with Brownian motion and other random processes. The rules of stochastic calculus can seem bizarre at first. For a function $f$ of a real number $x$ , we often write

$df = f'(x) \, dx$

For a function $f(X_t)$ of a Brownian motion, the analog is Itô’s formula

$df = f'(X_t) \, dX_t + \frac{1}{2} f''(X_t) \, dt.$

While this might seem odd at first, this formula may seem more sensible if we compare with (1) above. The idea, very roughly, is that for an increment of the Brownian motion $dX_t$ over a time interval $dt$ , $(dX_t)^2$ is a random variable with mean $dt$ , so we cannot drop the second term in the Taylor series, even up to first order in $dt$ . Fully diving into the subtleties of stochastic calculus is far beyond the scope of this short post. Hopefully, the rest of this post, which outlines some extensions of our proof of Jensen’s inequality that require more stochastic calculus, will serve as an enticement to learn more about this beautiful subject.

Proving Jensen by Interpolation

For the rest of this post, we will be less careful with mathematical technicalities. We can use the same idea that we used to prove Jensen’s inequality for a Gaussian random variable to prove Jensen’s inequality for any random variable $Y$ :

$f(\mathbb{E}Y) \le \mathbb{E}[f(Y)].$

Here is the idea of the proof.

First, realize that we can write any random variable $Y$ as a function of a standard Gaussian random variable $X$ . Indeed, letting $F_X$ and $F_Y$ denote the cumulative distribution functions of $X$ and $Y$ , one can show that

$g(X) := \inf \{ \alpha \in \real : F_Y(\alpha) \ge F_X(X) \}$

has the same distribution as $Y$ .

Now, as before, we can interpolate between $\mathbb{E} Y$ and $Y$ using a Brownian motion. As a first, idea, we might try

$A_t \stackrel{?}{=} \mathbb{E} [f(g(X_t))].$

Unfortunately, this choice of $A_t$ does not work! Indeed, $A_0 = \mathbb{E}[f(g(0))]$ does not even equal to $\mathbb{E} [f(Y)]$ ! Instead, we must define

$A_t = \mathbb{E} [f(\mathbb{E}[g(X_1) \mid X_t])].$

We define $A_t$ using the conditional expectation of the final value $g(X_1)$ conditional on the Brownian motion $X_t$ at an earlier time $t$ . Using a bit of elbow grease and stochastic calculus, one can show that

$\frac{d}{dt} A_t \ge 0 \quad \text{for all }t\in [0,1].$

This provides a proof of Jensen’s inequality in general by the method if interpolation.

Gaussian Jensen Inequality

Now, we’ve come to the real treat, the Gaussian Jensen inequality. In the last section, we saw the sketch of a proof of Jensen’s inequality using interpolation. While it is cool that this proof is possible, we learned anything new since we can prove Jensen’s inequality in other ways. The Gaussian Jensen inequality provides an application of this technique which is hard to prove other ways. This section, in particular, is cribbing quite heavily from Paata Ivanishvili‘s excellent post on the topic.

Here’s the big question:

If $Y_1,\ldots,Y_n$ are “somewhat dependent”, for which functions does the multivariate Jensen’s inequality
( $\star$ ) $f(\mathbb{E} Y_1,\ldots,\mathbb{E}Y_n) \le \mathbb{E} [f(Y_1,\ldots,Y_n)]$
hold?

Considering extreme cases, if $Y_1,\ldots,Y_n$ are entirely dependent, then we would only expect ( $\star$ ) to hold when $f$ is convex. But if $Y_1,\ldots,Y_n$ are independent, then we can apply Jensen’s inequality to each coordinate one at a time to deduce

$\text{($\star$) holds if $f$ is convex in each coordinate, separately.}$

We would like a result which interpolates between extremes {fully dependent, fully convex} and {independent, separately convex}. The Gaussian Jensen inequality provides exactly this tool.

As in the previous section, we can generate arbitrary random variables $Y_1,\ldots,Y_n$ as functions $g(X_1),\ldots,g(X_n)$ of Gaussian random variables $X_1,\ldots,X_n$ . We will use the covariance matrix $\Sigma$ of the Gaussian random variables $X_1,\ldots,X_n$ as our measure of the dependence of the random variables $Y_1,\ldots,Y_n$ . With this preparation in place, we have the following result:

Gaussian Jensen inequality. The conclusion of Jensen’s inequality
(2) $f(\mathbb{E}g_1(X_1),\ldots,\mathbb{E}g_n(X_n)) \le \mathbb{E} [f(g(X_1),\ldots,g(X_n))]$
holds for all test functions $g_1,\ldots,g_n$ if and only if
$\Sigma \circ \nabla^2 f(x) \text{ is positive semidefinite} \quad \text{for all $x \in \real^n$}.$
Here, $\nabla^2 f(x)$ is the Hessian matrix at $x$ and $\circ$ denotes the entrywise product of matrices.

This is a beautiful result with striking consequences (see Ivanishvili‘s post). The proof is essentially the same as the proof as Jensen’s inequality by interpolation with a little additional bookkeeping.

Let us confirm this result respects our extreme cases. In the case where $X_1=\cdots=X_n$ are equal (and variance one), $\Sigma$ is a matrix of all ones and $\Sigma \circ \nabla^2 f(x) = \nabla^2 f(x)$ for all $x$ . Thus, the Gaussian Jensen inequality states that (2) holds if and only if $\nabla^2 f(x)$ is positive semidefinite for every $x$ , which occurs precisely when $f$ is convex.

Next, suppose that $X_1,\ldots,X_n$ are independent and variance one, then $\Sigma$ is the identity matrix and

$\Sigma \circ \nabla^2 f(x) = \mathrm{diag} \left( \frac{\partial^2 f}{\partial x_i^2} : i=1,\ldots,n \right).$

A diagonal matrix is positive semidefinite if and only if its entries are nonnegative. Thus, (2) holds if and only if each of $f$ ‘s diagonal second derivatives are nonnegative $\partial_{x_i}^2 f \ge 0$ : this is precisely the condition for $f$ to be separately convex in each argument.

There’s much more to be said about the Gaussian Jensen inequality, and I encourage you to read Ivanishvili‘s post to see the proof and applications. What I find so compelling about this result—so compelling that I felt the need to write this post—is how interpolation and stochastic calculus can be used to prove inequalities which don’t feel like stochastic calculus problems. The Gaussian Jensen inequality is a statement about functions of dependent Gaussian random variables; there’s nothing dynamic happening. Yet, to prove this result, we inject dynamics into the problem, viewing the two sides of our inequality as endpoints of a random process connecting them. This is a such a beautiful idea that I couldn’t help but share it.

Big Ideas in Applied Math: Markov Chains

May 30, 2023 by Ethan N. Epperly 3 Comments

In this post, we’ll talk about Markov chains, a useful and general model of a random system evolving in time.

PageRank

To see how Markov chains can be useful in practice, we begin our discussion with the famous PageRank problem. The goal is assign a numerical ranking to each website on the internet measuring how important it is. To do this, we form a mathematical model of an internet user randomly surfing the web. The importance of each website will be measured by the amount of times this user visits each page.

The PageRank model of an internet user is as follows: Start the user at an arbitrary initial website $x_0$ . At each step, the user makes one of two choices:

With 85% probability, the user follows a random link on their current website.
With 15% probability, the user gets bored and jumps to a random website selected from the entire internet.

As with any mathematical model, this is a riduculously oversimplified description of how a person would surf the web. However, like any good mathematical model, it is useful. Because of the way the model is designed, the user will spend more time on websites with many incoming links. Thus, websites with many incoming links will be rated as important, which seems like a sensible choice.

An example of the PageRank distribution for a small internet is shown below. As one would expect, the surfer spends a large part of their time on website B, which has many incoming links. Interestingly, the user spends almost as much of their time on website C, whose only links are to and from B. Under the PageRank model, a website is important if it is linked to by an important website, even if that is the only website linking to it.

Markov Chains in General

Having seen one Markov chain, the PageRank internet surfer, let’s talk about Markov chains in general. A (time-homogeneous) Markov chain consists of two things: a set of states and probabilities for transitioning between states:

Set of states. For this discussion, we limit ourselves to Markov chains which can only exist in finitely many different states. To simplify our discussion, label the possible states using numbers $1,2,\ldots,m$ .
Transition probabilities. The definining property of a (time-homogeneous) Markov chain is that, at any point in time $n$ , if the state is $i$ , the probability of moving to state $j$ is a fixed number $P_{ij}$ . In particular, the probability $P_{ij}$ of moving from $i$ to $j$ does not depend on the time $n$ or the past history of the chain before time $n$ ; only the value of the chain at time $n$ matters.

Denote the state of the Markov chain at times $0,1,2,\ldots$ by $x_0,x_1,x_2,\ldots$ . Note that the states $x_0,x_1,x_2,\ldots$ are random quantities. We can write the Markov chain property using the language of conditional probability:

$\mathbb{P} \{ x_{n+1} = j \mid x_n = i,x_{n-1}=a_{n-1},\ldots,x_0=a_0\} = \mathbb{P}\{x_{n+1} = j \mid x_n = i\} = P_{ij}.$

This equation states that the probability that the system is in state $j$ at time $n+1$ given the entire history of the system depends only on the value $x_n = i$ of the chain at time $n$ . This probability is the transition probability $P_{ij}$ .

Let’s see how the PageRank internet surfer fits into this model:

Set of states. Here, the set of states are the websites, which we label $1,\ldots,m$ .
Transition probabilities. Consider two websites $i$ and $j$ . If $i$ does not have a link to $j$ , then the only way of going from $i$ to $j$ is if the surfer randomly gets bored (probability 15%) and picks website $j$ to visit at random (probability $1/m$ ). Thus,
( $i\not\to j$ ) $P_{ij} = \frac{0.15}{m}.$
Suppose instead that $i$ does link to $j$ and $i$ has $d_i$ outgoing links. Then, in addition to the $0.15/m$ probability computed before, user $i$ has an 85% percent chance of following a link and a $1/d_i$ chance of picking $j$ as that link. Thus,
( $i\to j$ ) $P_{ij} = \frac{0.85}{d_i} + \frac{0.15}{m}.$

Markov Chains and Linear Algebra

For a non-random process $y_0,y_1,y_2,\ldots$ , we can understand the processes evolution by determining its state $y_n$ at every point in time $n$ . Since Markov chains are random processes, it is not enough to track the state $x_n$ of the process at every time $n$ . Rather, we must understand the probability distribution of the state $x_n$ at every point in time $n$ .

It is customary in Markov chain theory to represent a probability distribution on the states $\{1,\ldots,n\}$ by a row vector $\rho^\top$ .¹ The $i$ th entry $\rho_i$ stores the probability that the system is in state $i$ . Naturally, as $\rho^\top$ is a probability distribution, its entries must be nonnegative ( $\rho_i \ge 0$ for every $i$ ) and add to one ( $\sum_{i=1}^n \rho_i = 1$ ).

Let $(\rho^{(0)})^\top, (\rho^{(1)})^\top,\ldots$ denote the probability distributions of the states $x_0,x_1,\ldots$ . It is natural to ask: How are the distributions $(\rho^{(0)})^\top, (\rho^{(1)})^\top,\ldots$ related to each other? Let’s answer this question.

The probability that $x_{n+1}$ is in state $j$ is the $j$ th entry of $\rho^{(n+1)}$ :

$\rho^{(n+1)}_j = \mathbb{P} \{x_{n+1} = j\}$

To compute this probability, we break into cases based on the value of the process at time $n$ : either $x_n = 1$ or $x_n = 2$ or … or $x_n = m$ ; only one of these cases can be true at once. When we have an “or” of random events and these events are mutually exclusive (only can be true at once), then the probabilities add:

$\rho^{(n+1)}_j = \mathbb{P} \{x_{n+1} = j\} = \sum_{i=1}^m \mathbb{P} \{x_{n+1} = j, x_n = i\}.$

Now we use some conditional probability. The probability that $x_{n+1} = j$ and $x_n = i$ is the probability that $x_n = i$ times the probability that $x_{n+1} = j$ conditional on $x_n = i$ . That is,

$\rho^{(n+1)}_j = \sum_{i=1}^m \mathbb{P} \{x_{n+1} = j, x_n = i\} = \sum_{i=1}^m \mathbb{P} \{x_n = i\} \mathbb{P}\{x_{n+1} = j \mid x_n = i\}.$

Now, we can simplify using our definitions. The probability that $x_n = i$ is just $\rho^{(n)}_i$ and the probability of moving from $i$ to $j$ is $P_{ij}$ . Thus, we conclude

$\rho_j^{(n+1)} = \sum_{i=1}^m \rho^{(n)}_i P_{ij} .$

Phrased in the language of linear algebra, we’ve shown

$\left(\rho^{(n+1)}\right)^\top = \left(\rho^{(n)}\right)^\top P \quad \text{for any } n = 0,1,2,\ldots.$

That is, if we view the transition probabilities $P_{ij}$ as comprising an $m\times m$ matrix $P$ , then the distribution at time $n+1$ is obtained by multiplying the distribution at time $n$ by transition matrix $P$ . In particular, if we iterate this result, we obtain that the distribution at time $n$ is given by

$\left(\rho^{(n)}\right)^\top = \left(\rho^{(n-1)}\right)^\top P = \left[\left(\rho^{(n-2)}\right)^\top P\right]P = \left(\rho^{(n-2)}\right)^\top P^2 = \cdots = \left(\rho^{(0)}\right)^\top P^n.$

Thus, the distribution at time $n$ is the distribution at time $0$ multiplied by the $n$ th power of the transition matrix $P$ .

Convergence to Stationarity

Let’s go back to our web surfer again. At time $0$ , we started our surfer at a particular website, say $1$ . As such, the probability distribution² $\rho^{(0)}$ at time $0$ is concentrated just on website $1$ , with no other website having any probability at all. In the first few steps, most of the probability will remain in the vacinity of website $1$ , in the websites linked to by $1$ and the websites linked to by the websites linked to by $1$ and so on. However, as we run the chain long enough, the surfer will have time to widely across the web and the probability distribution will become less and less influenced by the chain’s starting location. This motivates the following definition:

Definition. A Markov chain satisfies the mixing property if the probability distributions $\rho^{(0)}, \rho^{(1)}, \ldots$ converge to a single fixed probability distribution $\pi$ regardless of how the chain is initialized (i.e., independent of the starting distribution $\rho^{(0)}$ ).

The distribution $\pi$ for a mixing Markov chain is known as a stationary distribution because it does not change under the action of $P$ :

(St) $\pi^\top = \pi^\top P.$

To see this, recall the recurrence

$\left(\rho^{(n+1)}\right)^\top = \left(\rho^{(n)}\right)^\top P,$

take the limit as $n\to\infty$ , and observe that both $(\rho^{(n+1)})^\top$ and $(\rho^{(n)})^\top$ converge to $\pi^\top$ .

One of the basic questions in the theory of Markov chains is finding conditions under which the mixing property (or suitable weaker versions of it) hold. To answer this question, we will need the following definition:

A Markov chain is primitive if, after running the chain for some number $n$ steps, the chain has positive probability of moving between any two states. That is,
$\text{There exists $n$ such that, for any $i,j = 1,2,\ldots,m$, } \quad\mathbb{P}\{x_n = j \mid x_0 = i \} = (P^n)_{ij} > 0.$

The fundamental theorem of Markov chains is that primitive chains satisfy the mixing property.

Theorem (fundamental theorem of Markov chains). Every primitive Markov chain is mixing. In particular, there exists one and only probability distribution $\pi$ satisfying the stationary property (St) and the probability distributions $\rho^{(0)},\rho^{(1)},\ldots$ converge to $\pi$ when initialized in any probability distribution $\rho^{(0)}$ . Every entry of $\pi$ is strictly positive.

Let’s see an example of the fundamental theorem with the PageRank surfer. After $n=1$ step, there is at least a $0.15/m > 0$ chance of moving from any website $i$ to any other website $j$ . Thus, the chain is primitive. Consequently, there is a unique stationary distribution $\pi$ , and the surfer will converge to this stationary distribution regardless of which website they start at.

Going Backwards in Time

Often, it is helpful to consider what would happen if we ran a Markov chain backwards in time. To see why this is an interesting idea, suppose you run website $w$ and you’re interested in where your traffic is coming from. One way of achieving this would be to initialize the Markov chain at $w$ and run the chain backwards in time. Rather than asking, “given I’m at $w$ now, where would a user go next?”, you ask “given that I’m at $w$ now, where do I expect to have come from?”

Let’s formalize this notion a little bit. Consider a primitive Markov chain $x_0,x_1,x_2,\ldots$ with stationary distribution $\pi$ . We assume that we initialize this Markov chain in the stationary distribution. That is, we pick $\rho^{(0)} = \pi$ as our initial distribution for $x_0$ . The time-reversed Markov chain $y_0,y_1,\ldots$ is defined as follows: The probability $P^{\rm rev}_{ij}$ of moving from $i$ to $j$ in the time-reversed Markov chain is the probability that I was at state $j$ one step previously given that I’m at state $i$ now:

$\mathbb{P} \{y_1 = j \mid y_0 = i\} = P^{\rm rev}_{ij} = \mathbb{P} \{ x_0 = j \mid x_1 = i \}.$

To get a nice closed form expression for the reversed transition probabilities $P^{\rm rev}_{ij}$ , we can invoke Bayes’ theorem:

(Rev) $P^{\rm rev}_{ij} = \mathbb{P} \{ x_0 = j \mid x_1 = i \} = \frac{\mathbb{P} \{x_0 = j\} \mathbb{P} \{x_1 = i \mid x_0 = j\}}{\mathbb{P} \{x_1 = i\}} = \frac{ \pi_j P_{ji}}{\pi_i}.$

The time-reversed Markov chain can be a strange beast. For the reversed PageRank surfer, for instance, follows links “upstream” traveling from the linked site to the linking site. As such, our hypothetical website owner could get a good sense of where their traffic is coming from by initializing the reversed chain $y_0 = w$ at their website and following the chain one step back.

Reversible Markov Chains

We now have two different Markov chains: the original and its time-reversal. Call a Markov chain reversible if these processes are the same. That is, if the transition probabilities are the same:

$P^{\rm rev}_{ij} = P_{ij} \quad \text{for every } i,j=1,2,\ldots,m.$

Using our formula (Rev) for the reversed transition probability, the reversibility condition can be written more concisely as

$\pi_i P_{ij} = \pi_j P_{ji}.$

This condition is referred to as detailed balance.³ In words, it states that a Markov chain is reversible if, when initialized in the stationary distribution $\pi$ , the flow of probability mass from $i$ to $j$ (that is, $\pi_i P_{ij}$ ) is equal to the flow of probability mass from $j$ to $i$ (that is, $\pi_jP_{ji}$ ).

Many interesting Markov chains are reversible. One class of examples are Markov chain models of physical and chemical processes. Since physical laws like classical and quantum mechanics are reversible under time, so too should we expect Markov chain models built from theories to be reversible.

Not every interesting Markov chain is reversible, however. Indeed, except in special cases, the PageRank Markov chain is not reversible. If $i$ links to $j$ but. $j$ does not link to $i$ , then the flow of mass from $i$ to $j$ will be higher than the flow from $j$ to $i$ .

Before moving on, we note one useful fact about reversible Markov chains. Suppose a reversible, primitive Markov chain satisfies the detailed balance condition with a probability distribution $\sigma$ :

$\sigma_i P_{ij} = \sigma_j P_{ji}.$

Then $\sigma = \pi$ is the stationary distribution of this chain. To see why, we check the stationarity condition $\sigma^\top P = \sigma^\top$ . Indeed, for every $j$ ,

$(\sigma^\top P)_j = \sum_{i=1}^m \sigma_i P_{ij} = \sum_{i=1}^m \sigma_j P_{ji} = \sigma_j.$

The second equality is detailed balance and the third equality is just the condition that the sum of the transition probabilities from $j$ to each $i$ is one. Thus, $\sigma^\top P = \sigma^\top$ and $\sigma$ is a stationary distribution for $P$ . But a primitive chain has only one stationary distribution $\pi$ , so $\sigma = \pi$ .

Markov Chains as Algorithms

Markov chains are an amazingly flexible tool. One use of Markov chains is more scientific: Given a system in the real world, we can model it by a Markov chain. By simulating the chain or by studying its mathematical properties, we can hope to learn about the system we’ve modeled.

Another use of Markov chains is algorithmic. Rather than thinking of the Markov chain as modeling some real-world process, we instead design the Markov chain to serve a computationally useful end. The PageRank surfer is one example. We wanted to rank the importance of websites, so we designed a Markov chain to achieve this task.

One task we can use Markov chains to solve are sampling problems. Suppose we have a complicated probability distribution $\pi$ , and we want a random sample from $\pi$ —that is, a random quantity $y$ such that $\mathbb{P} \{ y = i \} = \pi_i$ for every $i$ . One way to achieve this goal is to design a primitive Markov chain with stationary distribution $\pi$ . Then, we run the chain for a large number of steps $n$ and use $x_n$ as an approximate sample from $\pi$ .

To design a Markov chain with stationary distribution $\pi$ , it is sufficient to generate transition probabilities $P$ such that $\pi$ and $P$ satisfy the detailed balance condition. Then, we are guaranteed that $\pi$ is a stationary distribution for the chain. (We also should check the primitiveness condition, but this is often straightforward.)

Here is an effective way of building a Markov chain to sample from a distribution $\pi$ . Suppose that the chain is in state $i$ at time $n$ , $x_n = i$ . To choose the next state, we begin by sampling $j$ from a proposal distribution $T$ . The proposal distribution $T$ can be almost anything we like, as long as it satisfies three conditions:

Probability distribution. For every $i$ , the transition probabilitie $T_{ij}$ add to one: $\sum_{j=1}^m T_{ij} = 1$ .
Bidirectional. If $T_{ij} > 0$ , then $T_{ji} > 0$ .
Primitive. The transition probabilities $T$ form a primitive Markov chain.

In order to sample from the correct distribution, we can’t just accept every proposal. Rather, given the proposal $i\to j$ , we accept with probability

$\min \left\{ 1 , \frac{\pi_j T_{ji}}{\pi_i T_{ij}} \right\}.$

If we accept the proposal, the next state of our chain is $x_{n+1} = j$ . Otherwise, we stay where we are $x_{n+1} = i$ . This Markov chain is known as a Metropolis–Hastings sampler.

For clarity, we list the steps of the Metropolis–Hastings sampler explicitly:

Initialize the chain in any state $x_0$ and set $n := 0$ .
Draw a proposal $x'$ with from the proposal distribution, $\mathbb{P} \{ x' = j \} = T_{x_nj}$ .
Compute the acceptance probability
$p_{\rm acc} := \min \left\{ 1 , \frac{\pi_j T_{ji}}{\pi_i T_{ij}} \right\}.$
With probability $p_{\rm acc}$ , set $x_{n+1} := x'$ . Otherwise, set $x_{n+1} := x_n$ .
Set $n := n+1$ and go back to step 2.

To check that $\pi$ is a stationary distribution of the Metropolis–Hastings distribution, all we need to do is check detailed balance. Note that the probability $P_{ij}$ of transitioning from $i$ to $j\ne i$ under the Metropolis–Hastings sampler is the proposal probability $T_{ij}$ times the acceptance probability:

$P_{ij} = T_{ij} \cdot \min \left\{ 1 , \frac{\pi_j T_{ji}}{\pi_i T_{ij}} \right\}.$

Detailed balance is confirmed by a short computation⁴

$\pi_i P_{ij} = \pi_i T_{ij} \cdot \min \left\{ 1 , \frac{\pi_j T_{ji}}{\pi_i T_{ij}} \right\} = \min \left\{ \pi_i T_{ij} , \pi_j T_{ji} \right\} = \pi_j P_{ji}.$

Thus the Metropolis–Hastings sampler has $\pi$ as stationary distribution.

Determinatal Point Processes: Diverse Items from a Collection

The uses of Markov chains in science, engineering, math, computer science, and machine learning are vast. I wanted to wrap up with one application that I find particularly neat.

Suppose I run a bakery and I sell $N$ different baked goods. I want to pick out $k$ special items for a display window to lure customers into my store. As a first approach, I might pick my top- $k$ selling items for the window. But I realize that there’s a problem. All of my top sellers are muffins, so all of the items in my display window are muffins. My display window is doing a good job luring in muffin-lovers, but a bad job of enticing lovers of other baked goods. In addition to rating the popularity of each item, I should also promote diversity in the items I select for my shop window.

Here’s a creative solution to my display case problems using linear algebra. Suppose that, rather than just looking at a list of the sales of each item, I define a matrix $A$ for my baked goods. In the $ii$ th entry $A_{ii}$ of my matrix, I write the number of sales for baked good $i$ . I populate the off-diagonal entries $A_{ij}$ of my matrix with a measure of similarity between items $i$ and $j$ .⁵ So if $i$ and $j$ are both muffins, $A_{ij}$ will be large. But if $i$ is a muffin and $j$ is a cookie, then $A_{ij}$ will be small. For mathematical reasons, we require $A$ to be symmetric and positive definite.

To populate my display case, I choose a random subset of $k$ items from my full menu of size $N$ according to the following strange probability distribution: The probability $\pi_S$ of picking items $S = \{s_1,\ldots,s_k\} \subseteq \{1,\ldots,N\}$ is proportional to the determinant of the submatrix $A(S,S)$ . More specifically,

( $k$ -DPP) $\pi_S = \frac{\det A(S,S)}{\sum_{\text{all subsets $T$ of size $k$}} \det A(T,T)}.$

Here, we let $A(S,S)$ denote the $k\times k$ submatrix of $A$ consisting of the entries appearing in rows and columns $s_1,\ldots,s_k$ . Such a random subset is known as a $k$ -determinantal point process ( $k$ -DPP). (See this survey for more about DPPs.)

To see why this makes any sense, let’s consider a simple example of $N = 3$ items and a display case of size $k = 2$ . Suppose I have three items: a pumpkin muffin, a chocolate chip muffin, and an oatmeal raisin cookies. Say the $A$ matrix looks like

$A = \begin{bmatrix} 10 & 9 & 0 \\ 9 & 10 & 0 \\ 0 & 0 & 5 \end{bmatrix}.$

We see that both muffins are equally popular $A_{11} = A_{22} = 10$ and much more popular than the cookie $A_{33} = 5$ . However, the two muffins are similar to each other and thus the corresponding submatrix has small determinant

$\det A(\{1,2\},\{1,2\}) = \det \twobytwo{10}{9}{9}{10} = 19.$

By contrast, if the cookie is disimilar to each muffin and the determinant is higher

$\det A(\{1,3\},\{1,3\}) = \det A(\{2,3\},\{2,3\}) = \det \twobytwo{10}{0}{0}{5} = 50.$

Thus, even though the muffins are more popular overall, choosing our display case from a $2$ -DPP, we have a $(50+50) / (50+50+19) \approx 84\%$ chance of choosing a muffin and a cookie for our display case. It is for this reason that we can say that a $k$ -DPP preferentially selects for diverse items.

Is sampling from a $k$ -DPP the best way of picking $k$ items for my display case? How does it compare to other possible methods?⁶ These are interesting questions for another time. For now, let us focus our attention on a different question: How would you sample from a $k$ -DPP?

Determinantal Point Process by Markov Chains

Sampling from a $k$ -DPP is a hard computational problem. Indeed, there are ${N \choose k}$ possible $k$ -element subspaces of a set of $N$ items. The number of possibilities gets large fast. If I have $N = 100$ items and want to pick $k = 10$ of them, there are already over 10 trillion possible combinations.

Markov chains offer one compelling way of sampling a $k$ -DPP. First, we need a proposal distribution. Let’s choose the simplest one we can think of:

Proposal for $k$ -DPP sampling. Suppose our current set of $k$ items is $S = \{s_1,\ldots,s_k\}$ . To generate a proposal, choose a uniformly random element $s_{\rm old}$ out of $S$ and a uniformly random element $s_{\rm new}$ out of $\{1,\ldots,N\}$ without $S$ . Propose $S'$ obtained from $S$ by replacing $s_{\rm old}$ with $s_{\rm new}$ (i.e., $S' = S \cup \{s_{\rm new}\} \setminus \{s_{\rm old}\}$ ).

Now, we need to compute the Metropolis–Hastings acceptance probability

$p_{\rm acc} = \min \left\{ 1 , \frac{\pi_{S'} T_{S'S}}{\pi_{S} T_{SS'}} \right\}.$

For $S$ and $S'$ which differ only by the addition of one element and the removal of another, the proposal probabilities $T_{S'S}$ and $T_{SS'}$ are both equal to $1/(kN)$ , $T_{S'S} = T_{SS'} = 1/(kN)$ . Using the formula for the probability $\pi_S$ of drawing $S$ from a $k$ -DPP, we compute that

$\frac{\pi_{S'}}{\pi_S} = \frac{\det A(S',S')}{\det A(S,S)}.$

Thus, the Metropolis–Hastings acceptance probability is just a ratio of determinants:

(Acc) $p_{\rm acc} = \min \left\{ 1 , \frac{\pi_{S'} T_{S'S}}{\pi_{S} T_{SS'}} \right\} = \min \left\{ 1, \frac{\det A(S',S')}{\det A(S,S)} \right\}.$

And we’re done. Let’s summarize our sampling algorithm:

Choose an initial set $S_0$ arbitrarily and set $n := 0$ .
Draw $s_{\rm old}$ uniformly at random from $S_n$ .
Draw $s_{\rm new}$ uniformly at random from $\{1,\ldots,N\} \setminus S_n$ .
Set $S' := S_n \cup \{s_{\rm new}\} \setminus \{s_{\rm old}\}$ .
With probability $p_{\rm acc}$ defined in (Acc), accept and set $S_{n+1} := S'$ . Otherwise, set $S_{n+1} := S_n$ .
Set $n := n+1$ and go to step 2.

This is a remarkably simple algorithm to sample from a complicated distribution. And its fairly efficient as well. Analysis by Anari, Oveis Gharan, and Rezaei shows that, when you pick a good enough initial set $S_0$ , this sampling algorithm produces approximate samples from a $k$ -DPP in roughly $Nk^2$ steps.⁷ Remarkably, if $k$ is much smaller than $N$ , this Markov chain-based algorithm samples from a $k$ -DPP without even looking at all $N^2$ entries of the matrix $A$ !

Upshot. Markov chains are a simple and general model for a state evolving randomly in time. Under mild conditions, Markov chains converge to a stationary distribution: In the limit of a large number of steps, the state of the system become randomly distributed in a way independent of how it was initialized. We can use Markov chains as algorithms to approximately sample from challenging distributions.

Note to Self: Norm of a Gaussian Random Vector

February 21, 2023 by Ethan N. Epperly Leave a comment

Let $g$ be a standard Gaussian vector—that is, a vector populated by independent standard normal random variables. What is the expected length $\mathbb{E} \|g\|$ of $g$ ? (Here, and throughout, $\|\cdot\|$ denotes the Euclidean norm of a vector.) The length of $g$ is the square root of the sum of $n$ independent standard normal random variables

$\|g\| = \sqrt{g_1^2 + \cdots + g_n^2},$

which is known as a $\chi$ random variable with $n$ degrees of freedom. (Not to be confused with a $\chi^\mathbf{2}$ random variable!) Its mean value is given by the rather unpleasant formula

$\mathbb{E} \|g\| = \sqrt{2} \frac{\Gamma((n+1)/2)}{\Gamma(n/2)},$

where $\Gamma(\cdot)$ is the gamma function. If you are familiar with the definition of the gamma function, the derivation of this formula is not too hard—it follows from a change of variables to $n$ -dimensional spherical coordinates.

This formula can be difficult to interpret and use. Fortunately, we have the rather nice bounds

(1) $\sqrt{n-1} < \frac{n}{\sqrt{n+1}} < \mathbb{E} \|g\| < \sqrt{n}.$

This result appears, for example, page 11 of this paper. These bounds show that, for large $n$ , $\mathbb{E} \|g\|$ is quite close to $\sqrt{n}$ . The authors of the paper remark that this inequality can be probed by induction. I had difficulty reproducing the inductive argument for myself. Fortunately, I found a different proof which I thought was very nice, so I thought I would share it here.

Our core tool will be Wendel’s inequality (see (7) in Wendel’s original paper): For $x > 0$ and $0 < s < 1$ , we have

(2) $\frac{x}{(x+s)^{1-s}} < \frac{\Gamma(x+s)}{\Gamma(x)} < x^s.$

Let us first use Wendel’s inequality to prove (1). Indeed, invoke Wendel’s inequality with $x = n/2$ and $s = 1/2$ and multiply by $\sqrt{2}$ to obtain

$\frac{\sqrt{2} \cdot n/2}{(n/2+1/2)^{1/2}} < \sqrt{2}\frac{\Gamma((n+1)/2)}{\Gamma(n/2)} = \mathbb{E}\|g\| < \sqrt{2}\cdot \sqrt{n/2},$

which simplifies directly to (1).

Now, let’s prove Wendel’s inequality (2). The key property for us will be the strict log-convexity of the gamma function: For real numbers $x,y > 0$ and $0 < s < 1$ ,

(3) $\Gamma((1-s)x + sy) < \Gamma(x)^{1-s} \Gamma(y)^s.$

We take this property as established and use it to prove Wendel’s inequality. First, use the log-convexity property (3) with $y = x+1$ to obtain

$\Gamma(x+s) = \Gamma((1-s)x + s(x+1)) < \Gamma(x)^{1-s} \Gamma(x+1)^s.$

Divide by $\Gamma(x)$ and use the property that $\Gamma(x+1)/\Gamma(x) = x$ to conclude

(4) $\frac{\Gamma(x+s)}{\Gamma(x)} < \left( \frac{\Gamma(x+1)}{\Gamma(x)} \right)^s = x^s.$

This proves the upper bound in Wendel’s inequality (2). To prove the lower bound, invoke the upper bound (4) with $x+s$ in place of $x$ and $1-s$ in place of $s$ to obtain

$\frac{\Gamma(x+1)}{\Gamma(x+s)} < (x+s)^{1-s}.$

Multiplying by $\Gamma(x+s)$ , dividing by $(x+s)^{1-s}\Gamma(x)$ , and using $\Gamma(x+1)/\Gamma(x) = x$ again yields

$\frac{\Gamma(x+s)}{\Gamma(x)} > \frac{\Gamma(x+1)}{\Gamma(x)} \cdot \frac{1}{(x+s)^{1-s}} = \frac{x}{(x+s)^{1-s}},$

finishing the proof of Wendel’s inequality.

Notes. The upper bound in (1) can be proven directly by Lyapunov’s inequality: $\mathbb{E} \|g\| \le (\mathbb{E} \|g\|^2)^{1/2} = n^{1/2}$ , where we use the fact that $\|g\|^2 = g_1^2 + \cdots + g_n^2$ is the sum of $n$ random variables with mean one. The weaker lower bound $\mathbb{E} \|g\| \ge \sqrt{n-1}$ follows from a weaker version of Wendel’s inequality, Gautschi’s inequality.

After the initial publication of this post, Sam Buchanan mentioned another proof of the lower bound $\mathbb{E} \|g\| \ge \sqrt{n-1}$ using the Gaussian Poincaré inequality. This inequality states that, for a function $f : \real^n \to \real$ ,

$\Var(f(g)) \le \mathbb{E} \| \nabla f(g)\|^2.$

To prove the lower bound, set $f(g) := \|g\|$ which has gradient $\nabla f(g) = g/\|g\|$ . Thus,

$\mathbb{E} \| \nabla f(g)\|^2 = 1 \ge \Var(f(g)) = \mathbb{E} \|g\|^2 - (\mathbb{E} \|g\|)^2 = n - (\mathbb{E} \|g\|)^2.$

Rearrange to obtain $\mathbb{E} \|g\| \ge \sqrt{n-1}$ .

Stochastic Trace Estimation

January 26, 2023 by Ethan N. Epperly 4 Comments

I am delighted to share that me, Joel A. Tropp, and Robert J. Webber‘s paper XTrace: Making the Most of Every Sample in Stochastic Trace Estimation has recently been released as a preprint on arXiv. In it, we consider the implicit trace estimation problem:

Implicit trace estimation problem: Given access to a square matrix $A$ via the matrix–vector product operation $\omega \mapsto A\omega$ , estimate its trace $\tr A = \sum_{i=1}^n A_{ii}$ .

Algorithms for this task have many uses such as log-determinant computations in machine learning, partition function calculations in statistical physics, and generalized cross validation for smoothing splines. I described another application to counting triangles in a large network in a previous blog post.

Our paper presents new trace estimators XTrace and XNysTrace which are highly efficient, producing accurate trace approximations using a small budget of matrix–vector products. In addition, these algorithms are fast to run and are supported by theoretical results which explain their excellent performance. I really hope that you will check out the paper to learn more about these estimators!

For the rest of this post, I’m going to talk about the most basic stochastic trace estimation algorithm, the Girard–Hutchinson estimator. This seemingly simple algorithm exhibits a number of nuances and forms the backbone for more sophisticated trace estimates such as Hutch++, Nyström++, XTrace, and XNysTrace. Toward the end, this blog post will be fairly mathematical, but I hope that the beginning will be fairly accessible to all.

Girard–Hutchinson Estimator: The Basics

The Girard–Hutchinson estimator for the trace of a square matrix $A$ is

(1) $\hat{\tr} = \frac{1}{m} \sum_{i=1}^m \omega_i^* A \omega_i.$

Here, $\omega_1,\ldots,\omega_m$ are random vectors, usually chosen to be statistically independent, and ${}^*$ denotes the conjugate transpose of a vector or matrix. The Girard–Hutchinson estimator only depends on the matrix $A$ through the matrix–vector products $A\omega_1,\ldots,A\omega_m$ .

Unbiasedness

Provided the random vectors are isotropic

(2) $\mathbb{E} [\omega_i\omega_i^*] = I,$

the Girard–Hutchinson estimator is unbiased:

(3) $\mathbb{E} [\hat{\tr}] = \tr A.$

Let us confirm this claim in some detail. First, we use linearity of expectation to evaluate

(4) $\mathbb{E} [\hat{\tr}] = \mathbb{E} \left[ \frac{1}{m} \sum_{i=1}^m \omega_i^*A\omega_i \right] = \frac{1}{m} \sum_{i=1}^m \mathbb{E} \left[ \omega_i^* A \omega_i\right].$

Therefore, to prove that $\mathbb{E} [\hat{\tr}] = \tr A$ , it is sufficient to prove that $\mathbb{E} \left[\omega_i^*A\omega_i\right] = \tr A$ for each $i$ .

When working with traces, there are two tricks that solve 90% of derivations. The first trick is that, if we view a number as a $1\times 1$ matrix, then a number equals its trace, $x = \tr x$ . The second trick is the cyclic property: For a $k\times p$ matrix $B$ and a $p\times k$ matrix $C$ , we have $\tr (BC) = \tr (CB)$ . The cyclic property should be handled with care when one works with a product of three or more matrices. For instance, we have

$\tr[BCD] = \tr[(BC)D] = \tr[D(BC)] = \tr[DBC].$

However,

$\tr [BCD] \ne \tr[CBD] \quad \text{in general}.$

One should think of the matrix product $BCD$ as beads on a closed loop of string. One can move the last bead $D$ to the front of the other two, $\tr [BCD] = \tr[DBC]$ , but not interchange two beads, $\tr[BCD] \ne \tr[CBD]$ .

With this trick in hand, let’s return to proving that $\mathbb{E} \left[\omega_i^*A\omega_i\right] = \tr A$ for every $i$ . Apply our two tricks:

$\mathbb{E} \left[\omega_i^*A\omega_i\right] = \mathbb{E} \tr \left[\omega_i^*A\omega_i\right] = \mathbb{E} \tr \left[A\omega_i\omega_i^*\right].$

The expectation is a linear operation and the matrix $A$ is non-random, so we can bring the expectation into the trace as

$\mathbb{E} \left[\omega_i^*A\omega_i\right] = \mathbb{E} \tr \left[A\omega_i\omega_i^*\right] = \tr(A \mathbb{E}[\omega_i\omega_i^*] ).$

Invoke the isotropy condition (2) and conclude:

$\mathbb{E} \left[\omega_i^*A\omega_i\right] = \tr(A \mathbb{E}[\omega_i\omega_i^*] ) = \tr(A\cdot I) = \tr A.$

Plugging this into (4) confirms the unbiasedness claim (3).

Variance

Continue to assume that the $\omega_i$ ‘s are isotropic (3) and now assume that $\omega_1,\ldots,\omega_m$ are independent. By independence, the variance can be written as

$\Var(\hat{\tr}) = \frac{1}{m^2} \sum_{i=1}^m \Var(\omega_i^*A\omega_i).$

Assuming that $\omega_1,\ldots,\omega_m$ are identically distributed $\omega_1,\ldots,\omega_m \sim \omega$ , we then get

$\Var(\hat{\tr}) = \frac{1}{m} \Var(\omega^*A\omega).$

The variance decreases like $1/m$ , which is characteristic of Monte Carlo-type algorithms. Since $\hat{\tr}$ is unbiased (i.e, (3)), this means that the mean square error decays like $1/m$ so the average error (more precisely root-mean-square error) decays like

$\left| \hat{\tr} - \tr A \right| \lessapprox \frac{\mathrm{const}}{\sqrt{m}}.$

This type of convergence is very slow. If I want to decrease the error by a factor of $10$ , I must do $100\times$ the work!

Variance-reduced trace estimators like Hutch++ and our new trace estimator XTrace improve the rate of convergence substantially. Even in the worst case, Hutch++ and XTrace reduce the variance at a rate $1/m^2$ and (root-mean-square) error at rates $1/m$ :

$\Var(\hat{\tr}_{\text{H++ or X}}) \le \frac{\mathrm{const}}{m^2},\quad \left| \hat{\tr}_{\text{H++ or X}} - \tr A \right| \lessapprox \frac{\mathrm{const}}{m}.$

For matrices with rapidly decreasing singular values, the variance and error can decrease much faster than this.

Variance Formulas

As the rate of convergence for the Girard–Hutchinson estimator is so slow, it is imperative to pick a distribution on test vectors $\omega$ that makes the variance of the single–sample estimate $\omega^*A\omega$ as low as possible. In this section, we will provide several explicit formulas for the variance of the Girard–Hutchinson estimator. Derivations of these formulas will appear at the end of this post. These variance formulas help illuminate the benefits and drawbacks of different test vector distributions.

To express the formulas, we will need some notation. For a complex number $z = a + bi$ we use $\Re(z) = a$ and $\Im(z) = b$ to denote the real and imaginary parts. The variance of a random complex number $z$ is

$\Var(z) := \mathbb{E} |z - \mathbb{E} z|^2 = \Var(\Re z) + \Var(\Im z).$

The Frobenius norm of a matrix $A$ is

$\left\|A\right\|_{\rm F}^2 = \sum_{i,j} |A_{ij}|^2.$

If $A$ is real symmetric or complex Hermitian with (real) eigenvalues $\lambda_1,\ldots,\lambda_n$ , we have

(5) $\left\|A\right\|_{\rm F}^2 = \sum_{i=1}^n \lambda_i^2.$

$A^\top$ denotes the ordinary transpose of $A$ and $A^*$ denotes the conjugate transpose of $A$ .

Real-Valued Test Vectors

We first focus on real-valued test vectors $\omega$ . Since $\omega$ is real, we can use the ordinary transpose ${}^\top$ rather than the conjugate transpose ${}^*$ . Since $\omega^\top A\omega$ is a number, it is equal to its own transpose:

$\omega^\top A \omega = (\omega^\top A \omega)^\top = \omega^\top A^\top \omega.$

Therefore,

$\omega^\top A\omega = \frac{\omega^\top A \omega + \omega^\top A^\top \omega}{2} = \omega^\top \left( \frac{A + A^\top}{2} \right)\omega.$

The Girard–Hutchinson trace estimator applied to $A$ is the same as the Girard–Hutchinson estimator applied to the symmetric part of $A$ , $(A+A^\top)/2$ .

For the following results, assume $A$ is symmetric, $A = A^\top$ .

Real Gaussian: $\omega_1,\ldots,\omega_m$ are independent standard normal random vectors.
$\Var(\omega^\top A\omega) = 2 \left\|A\right\|_{\rm F}^2.$
Uniform signs (Rademachers): $\omega_1,\ldots,\omega_m$ are independent random vectors with uniform $\pm 1$ coordinates.
$\Var(\omega^\top A \omega) = 2\sum_{i\ne j} |A_{ij}|^2.$
Real sphere: Assume $\omega_1,\ldots,\omega_n$ are uniformly distributed on the real sphere of radius $\sqrt{n}$ : $\omega \sim \text{Uniform} \{x\in \mathbb{R}^n : x^\top x = n\}$ .
$\Var(\omega^\top A\omega) = \frac{2n}{n+2} \left( \left\|A\right\|_{\rm F}^2 - \frac{1}{n} |\tr A|^2 \right).$

These formulas continue to hold for nonsymmetric $A$ by replacing $A$ by its symmetric part $(A+A^\top)/2$ on the right-hand sides of these variance formulas.

Complex-Valued Test Vectors

We now move our focus to complex-valued test vectors $\omega$ . As a rule of thumb, one should typically expect that the variance for complex-valued test vectors applied to a real symmetric matrix $A$ is about half the natural real counterpart—e.g., for complex Gaussians, you get about half the variance than with real Gaussians.

A square complex matrix has a Cartesian decomposition

$A = A^{\rm H} + i A^{\rm SH}$

where

$A^{\rm H} = \frac{A+A^*}{2} ,\quad A^{\rm SH} = \frac{A - A^*}{2i}$

denote the Hermitian and skew-Hermitian parts of $A$ . Similar to how the imaginary part of a complex number is real, the skew-Hermitian part of a complex matrix is Hermitian (and $i A^{\rm SH}$ is skew-Hermitian). Since $A^{\rm H}$ and $A^{\rm SH}$ are both Hermitian, we have

$\Re(\omega^* A\omega) = \omega^* A^{\rm H} \omega, \quad \Im (\omega^* A \omega) = \omega^* A^{\rm SH} \omega.$

Consequently, the variance of $\omega^*A \omega$ can be broken into Hermitian and skew-Hermitian parts:

$\Var(\omega^* A\omega) = \Var(\omega^* A^{\rm H}\omega) + \Var(\omega^* A^{\rm SH}\omega).$

For this reason, we will state the variance formulas only for Hermitian $A$ , with the formula for general $A$ following from the Cartesian decomposition.

For the following results, assume $A$ is Hermitian, $A = A^*$ .

Complex Gaussian: $\omega_1,\ldots,\omega_m$ are independent standard complex random vectors, i.e., each $\omega_i$ has iid entries distributed as $(g_1+ig_2)/\sqrt{2}$ for $g_1,g_2$ standard normal random variables.
$\Var(\omega^* A\omega) = \left\|A\right\|_{\rm F}^2.$
Uniform phases (Steinhauses): $\omega_1,\ldots,\omega_m$ are independent random vectors whose entries are uniform on the complex unit circle $\{ z \in \complex : |z| \}$ .
$\Var(\omega^* A \omega) = \sum_{i\ne j} |A_{ij}|^2.$
Complex sphere: Assume $\omega_1,\ldots,\omega_n$ are uniformly distributed on the complex sphere of radius $\sqrt{n}$ : $\omega \sim \text{Uniform} \{x\in \complex^n : x^* x = n\}$ .
$\Var(\omega^* A\omega) = \frac{n}{n+1} \left( \left\|A\right\|_{\rm F}^2 - \frac{1}{n} |\tr A|^2 \right).$

Optimality Properties

Let us finally address the question of what the best choice of test vectors is for the Girard–Hutchinson estimator. We will state two results with different restrictions on $\omega_1,\ldots,\omega_m$ .

Our first result, due to Hutchinson, is valid for real symmetric matrices with real test vectors.

Optimality (independent test vectors with independent coordinates). If the test vectors $\omega_1,\ldots,\omega_m \in \mathbb{R}^n$ are isotropic (2), independent from each other, and have independent entries, then for any fixed real symmetric matrix $A$ , the minimum variance for $\hat{\tr}$ is obtained when $\omega_1,\ldots,\omega_m$ are populated with random signs $(\omega_i)_j \sim \textnormal{Uniform} \{\pm 1\}$ .

The next optimality results will have real and complex versions. To present the results for $\mathbb{R}$ -valued and an $\complex$ -valued test vectors on unified footing, let $\field$ denote either $\mathbb{R}$ or $\complex$ . We let a $\field$ -Hermitian matrix be either a real symmetric matrix (if $\field = \mathbb{R}$ ) or a complex Hermitian matrix (if $\field = \complex$ ). Let a $\field$ -unitary matrix be either a real orthogonal matrix (if $\field = \mathbb{R}$ ) or a complex unitary matrix (if $\field = \complex$ ).

The condition that the vectors $\omega_1,\ldots,\omega_m$ have independent entries is often too restrictive in practice. It rules out, for instance, the case of uniform vectors on the sphere. If we relax this condition, we get a different optimal distribution:

Optimality (independent test vectors). Consider any set $\mathscr{A}$ of $\field$ -Hermitian matrices which is invariant under $\field$ -unitary similary transformations:
$\text{If $A \in \mathscr{A}$ and $U$ is $\field$-unitary, then $U^*AU \in \mathscr{A}$.}$
Assume that the test vectors $\omega_1,\ldots,\omega_m$ are independent and isotropic (2). The worst-case variance $\sup_{A \in \mathscr{A}} \Var(\hat{\tr})$ is minimized by choosing $\omega_1,\ldots,\omega_m$ uniformly on the $\field$ -sphere: $\omega_1,\ldots,\omega_m \sim \text{Uniform} \{ x \in \field^n : x^*x =n \}$ .

More simply, if you wants your stochastic trace estimator to be effective for a class of inputs $\mathscr{A}$ (closed under $\field$ -unitary similarity transformations) rather than a single input matrix $A$ , then the best distribution are test vectors drawn uniformly from the sphere. Examples of classes of matrices $\mathscr{A}$ include:

Fixed eigenvalues. For fixed real eigenvalues $\lambda_1,\ldots,\lambda_n \in \mathbb{R}$ , the set of all $\field$ -Hermitian matrices with these eigenvalues.
Density matrices. The class of all trace-one psd matrices.
Frobenius norm ball. The class of all $\field$ -Hermitian matrices of Frobenius norm at most 1.

Derivation of Formulas

In this section, we provide derivations of the variance formulas. I have chosen to focus on derivations which are shorter but use more advanced techniques rather than derivations which are longer but use fewer tricks.

Real Gaussians

First assume $A$ is real. Since $A$ is real symmetric, $A$ has an eigenvalue decomposition $A = Q\Lambda Q^\top$ , where $Q$ is orthogonal and $\Lambda$ is a diagonal matrix reporting $A$ ‘s eigenvalues. Since the real Gaussian distribution is invariant under orthogonal transformations, $\omega^\top A\omega = (Q^\top \omega)^\top \Lambda (Q^\top\omega)$ has the same distribution as $\omega^\top \Lambda \omega$ . Therefore,

$\Var(\omega^\top A \omega) = \Var(\omega^\top \Lambda \omega) = \Var \left( \sum_{i=1}^n \lambda_i \omega_i^2 \right) = \sum_{i=1}^n \lambda_i^2 \Var(\omega_i^2) = 2\sum_{i=1}^n \lambda_i^2 = 2\left\|A\right\|_{\rm F}^2.$

Here, we used that the variance of a squared standard normal random variable is two.

For $A$ non-real matrix, we can break the matrix $A$ into its entrywise real and imaginary parts $A = \mathfrak{R}(A) + i \, \mathfrak{I}(A)$ . Thus,

$\Var(\omega^\top A \omega) = \Var(\omega^\top \mathfrak{R}(A) \omega) + \Var(\omega^\top \mathfrak{I}(A) \omega) = 2\left\|\mathfrak{R}(A)\right\|_{\rm F}^2 + 2\left\|\mathfrak{I}(A)\right\|_{\rm F}^2 = 2\left\|A\right\|_{\rm F}^2.$

Uniform Signs

First, compute

$\omega^\top A \omega - \mathbb{E}[\omega^\top A \omega] = \sum_{i,j=1}^n A_{ij} \omega_i\omega_j - \sum_{i=1}^n A_{ii} = \sum_{i\ne j} A_{ij} \omega_i\omega_j + \sum_{i=1}^n A_{ii}(\omega_i^2-1).$

For a vector $\omega$ of uniform random signs, we have $\omega_i^2 = 1$ for every $i$ , so the second sum vanishes. Note that we have assumed $A$ symmetric, so the sum over $i\ne j$ can be replaced by two times the sum over $i < j$ :

$\omega^\top A \omega - \mathbb{E}[\omega^\top A \omega] = 2\sum_{i< j} A_{ij} \omega_i\omega_j.$

Note that $\{ \omega_i \omega_j : i < j\}$ are pairwise independent. As a simple exercise, one can verify that the identity

$\Var(a_1 X_1+\cdots+a_kX_k) = |a_1|^2 \Var(X_1) + \cdots + |a_k|^2 \Var(X_k)$

holds for any pairwise independent family of random variances $X_1,\ldots,X_k$ and numbers $a_1,\ldots,a_k$ . Ergo,

$\begin{align*}\Var(\omega^\top A\omega) &= \Var(\omega^\top A \omega - \mathbb{E}[\omega^\top A \omega]) \\&= \Var\left(\sum_{i< j} 2A_{ij} \omega_i\omega_j\right) \\&= \sum_{i<j} 4 |A_{ij}|^2 \Var(\omega_i\omega_j) \\&= \sum_{i<j} 4 |A_{ij}|^2 \\&= 2 \sum_{i\ne j} |A_{ij}|^2.\end{align*}$

In the second-to-last line, we use the fact that $\omega_i\omega_j$ is a uniform random sign, which has variance $1$ . The final line is a consequence of the symmetry of $A$ .

Uniform on the Real Sphere

The simplest proof is I know is by the “camel principle”. Here’s the story (a lightly edited quotation from MathOverflow):

A father left 17 camels to his three sons and, according to the will, the eldest son was to be given a half of the camels, the middle son one-third, and the youngest son the one-ninth. The sons did not know what to do since 17 is not evenly divisible into either two, three, or nine parts, but a wise man helped the sons: he added his own camel, the oldest son took $18/2=9$ camels, the second son took $18/3=6$ camels, the third son $18/9=2$ camels and the wise man took his own camel and went away.

We are interested in a vector $\omega$ which is uniform on the sphere of radius $\sqrt{n}$ . Performing averages on the sphere is hard, so we add a camel to the problem by “upgrading” $\omega$ to a spherically symmetric vector $g$ which has a random length. We want to pick a distribution for which the computation $\Var(g^\top A g)$ is easy. Fortunately, we already know such a distribution, the Gaussian distribution, for which we already calculated $\Var(g^\top A g) = 2\left\|A\right\|_{\rm F}^2$ .

The Gaussian vector $g$ and the uniform vector $\omega$ on the sphere are related by

$g = \sqrt{\frac{a}{n}} \omega,$

where $a$ is the squared length of the Gaussian vector $g$ . In particular, $a$ has the distribution of the sum of $n$ squared Gaussian random variables, which is known as a $\chi^2$ random variable with $n$ degrees of freedom.

Now, we take the camel back. Compute the variance of $g^\top A g$ using the chain rule for variance:

$\Var(g^\top A g) = \mathbb{E}[\Var(g^\top A g \mid a)] + \Var(\mathbb{E}[g^\top A g \mid a]).$

Here, $\Var(\cdot \mid a)$ and $\mathbb{E}[ \cdot \mid a]$ denote the conditional variance and conditional expectation with respect to the random variable $a$ . The quick and dirty ways of working with these are to treat the random variable $a$ “like a constant” with respect to the conditional variance and expectation.

Plugging in the formula $g = \sqrt{a/n} \cdot \omega$ and treating $a$ “like a constant”, we obtain

$\begin{align*}\Var(g^\top A g) &= \mathbb{E}[\Var(a/n \cdot \omega^\top A \omega \mid a)] + \Var(\mathbb{E}[a/n \cdot \omega^\top A \omega \mid a]) \\&=\mathbb{E}[(a/n)^2\Var(\omega^\top A \omega)] + \Var(a/n \cdot \mathbb{E}[\omega^\top A \omega]) \\&= \frac{1}{n^2} \mathbb{E}[a^2] \cdot \Var(\omega^\top A \omega) + \frac{1}{n^2} \Var(a) |\mathbb{E} [\omega^\top A \omega]|^2.\end{align*}$

As we mentioned, $a$ is a $\chi^2$ random variable with $n$ degrees of freedom and $\mathbb{E}[a^2]$ and $\Var(a)$ are known quantities that can be looked up:

$\mathbb{E}[a^2] = n(n+2), \quad \Var(a) = 2n.$

We know $\Var(g^\top A g) = 2\left\|A\right\|_{\rm F}^2$ and $\mathbb{E} [\omega^\top A \omega] = \tr A$ . Plugging these all in, we get

$2\left\|A\right\|_{\rm F}^2 = \frac{n+2}{n} \Var(\omega^\top A\omega) + \frac{2}{n} |\tr A|^2.$

Rearranging, we obtain

$\Var(\omega^\top A\omega) = \frac{2n}{n+2} \left( \left\|A\right\|_{\rm F}^2 - \frac{1}{n}|\tr A|^2\right).$

Complex Gaussians

The trick is the same as for real Gaussians. By invariance of complex Gaussian random vectors under unitary transformations, we can reduce to the case where $A$ is a diagonal matrix populated with eigenvalues $\lambda_1,\ldots,\lambda_n$ . Then

$\Var(\omega^*A \omega) = \Var \left( \sum_{i=1}^n \lambda_i |\omega_i|^2 \right) = \sum_{i=1}^n \Var(|\omega_i|^2) \lambda_i^2 = \sum_{i=1}^n \lambda_i^2 = \left\|A\right\|_{\rm F}^2.$

Here, we use the fact that $2|\omega_i|^2$ is a $\chi^2$ random variable with two degrees of freedom, which has variance four.

Random Phases

The trick is the same as for uniform signs. A short calculation (remembering that $A$ is Hermitian and thus $\overline{A_{ij}} = A_{ji}$ ) reveals that

$\Var\left( \omega^* A \omega \right) = \Var \left( \sum_{i<j} 2 \Re(A_{ij} \overline{\omega_i} \omega_j) \right).$

The random variables $\{\overline{\omega_i} \omega_j : i < j\}$ are pairwise independent so we have

$\Var\left( \omega^* A \omega \right) = \Var \left( \sum_{i<j} 2 \Re(A_{ij} \overline{\omega_i} \omega_j) \right) = 4\sum_{i<j} \Var \left( \Re(A_{ij} \overline{\omega_i} \omega_j) \right).$

Since $\overline{\omega}_i \omega_j$ is uniformly distributed on the complex unit circle, we can assume without loss of generality that $A_{ij} = |A_{ij}|$ . Thus, letting $\phi$ be uniform on the complex unit circle,

$\Var\left( \omega^* A \omega \right) = 4\sum_{i<j} \Var \left( |A_{ij}|\Re(\phi)) \right) = 4\Var\left( \Re(\phi) \right)\sum_{i<j}|A_{ij}|^2.$

The real and imaginary parts of $\phi$ have the same distribution so

$1 = \Var(\phi) = \Var(\Re \phi) + \Var(\Im \phi) = 2 \Var(\Re \phi)$

so $\Var(\Re \phi) = 1/2$ . Thus

$\Var\left( \omega^* A \omega \right) = 2 \sum_{i<j}|A_{ij}|^2 = \sum_{i\ne j} |A_{ij}|^2.$

Uniform on the Complex Sphere: Derivation 1 by Reduction to Real Case

There are at least three simple ways of deriving this result: the camel trick, reduction to the real case, and Haar integration. Each of these techniques illustrates a trick that is useful in its own right beyond the context of trace estimation. Since we have already seen an example of the camel trick for the real sphere, I will present the other two derivations.

Let us begin with the reduction to the real case. Let $\mathfrak{R}(\cdot)$ and $\mathfrak{I}(\cdot)$ denote the real and imaginary parts of a vector or matrix, taken entrywise. The key insight is that if $\omega$ is a uniform random vector on the complex sphere of radius $\sqrt{n}$ , then

$\mathscr{R}(\omega) := \twobyone{\mathfrak{R}(\omega)}{\mathfrak{I}(\omega)}\in\real^{2n} \quad \text{is a uniform random vector on the real sphere of radius $\sqrt{n}$}.$

We’ve converted the complex vector $\omega$ into a real vector $\mathscr{R}(\omega)$ .

Now, we need to convert the complex matrix $A$ into a real matrix $\mathscr{R}(A)$ . To do this, recall that one way of representing complex numbers is by $2\times 2$ matrices:

$a + bi \iff \twobytwo{a}{-b}{b}{a}.$

Using this correspondence addition and multiplication of complex numbers can be carried by addition and multiplication of the corresponding matrices.

To convert complex matrices to real matrices, we use a matrix-version of the same representation:

$\mathscr{R}(A) = \twobytwo{\mathfrak{R}(A)}{-\mathfrak{I}(A)}{\mathfrak{I}(A)}{\mathfrak{R}(A)}.$

One can check that addition and multiplication of complex matrices can be carried out by addition and multiplication of the corresponding “realified” matrices, i.e.,

$\mathscr{R}(A + B) = \mathscr{R}(A) + \mathscr{R}(B), \quad \mathscr{R}(A\cdot B) = \mathscr{R}(A) \cdot \mathscr{R}(B)$

holds for all complex matrices $A$ and $B$ .

We’ve now converted complex matrix $A$ and vector $\omega$ into real matrix $\mathscr{R}(A)$ and vector $\mathscr{R}(\omega)$ . Let’s compare $\omega^*A\omega$ to $\mathscr{R}(\omega)^\top\mathscr{R}(A)\mathscr{R}(\omega)$ . A short calculation reveals

$\omega^*A\omega = \mathscr{R}(\omega)^\top \mathscr{R}(A)\mathscr{R}(\omega) .$

Since $\mathscr{R}(\omega)$ is a uniform random vector on the sphere of radius $\sqrt{n}$ , $\sqrt{2}\cdot \mathscr{R}(\omega)$ is a uniform random vector on the sphere of radius $\sqrt{2n}$ . Thus, by the variance formula for the real sphere, we get

$\Var(\omega^*A\omega) = \Var[(\sqrt{2}\mathscr{R}(\omega))^\top (\mathscr{R}(A)/2)(\sqrt{2}\mathscr{R}(\omega) )] = \frac{4n}{2n+2} \left[ \|\mathscr{R}(A)/2\|_{\rm F}^2 - \frac{1}{8n}(\tr\mathscr{R}(A))^2 \right].$

A short calculation verifies that $\tr \mathscr{R}(A) = 2\tr A$ and $\norm{\mathscr{R}(A)}_{\rm F}^2 = 2\|A\|_{\rm F}^2$ . Plugging this in, we obtain

$\Var(\omega^*A\omega)= \frac{n}{n+1} \left[ \|A\|_{\rm F}^2 - \frac{1}{n}(\tr A)^2 \right].$

Uniform on the Complex Sphere: Derivation 2 by Haar Integration

The proof by reduction to the real case requires some cumbersome calculations and requires that we have already computed the variance in the real case by some other means. The method of Haar integration is more slick, but it requires some pretty high-power machinery. Haar integration may be a little bit overkill for this problem, but this technique is worth learning as it can handle some truly nasty expected value computations that appear, for example, in quantum information.

We seek to compute

$\mathbb{E} [(\omega^*A \omega)^2].$

The first trick will be to write this expession using a single matrix trace using the tensor (Kronecker) product $\otimes$ . For those unfamiliar with the tensor product, the main properties we will be using are

(6) $(A\otimes B) (C\otimes D) = (AB) \otimes (CD), \quad \tr(A\otimes B) = \tr A \cdot \tr B.$

We saw in the proof of unbiasedness that

$\omega^* A \omega = \tr (\omega^*A\omega) = \tr (A \omega\omega^*).$

Therefore, by (6),

$(\omega^*A\omega)^2 = (\tr [A \omega\omega^*])^2 = \tr [A\omega\omega^* \otimes A\omega\omega^*] = \tr [(A\otimes A) (\omega\omega^* \otimes \omega\omega^*)].$

Thus, to evaluate $\mathbb{E}[(\omega^*A\omega)^2]$ , it will be sufficient to evaluate $\mathbb{E}[\omega\omega^* \otimes \omega\omega^*]$ . Forunately, there is a useful formula for these expectation provided by a field of mathematics known as representation theory (see Lemma 1 in this paper):

$\mathbb{E}[ \omega\omega^* \otimes \omega\omega^*] = \frac{2n}{n+1} \operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}.$

Here, $\operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}$ is the orthogonal projection onto the space of symmetric two-tensors $\operatorname{Sym}^2(\complex^n) = \operatorname{span} \{ v \otimes v : v \in \complex^n \}$ . Therefore, we have that

$\mathbb{E}[(\omega^*A\omega)^2] = \tr [(A\otimes A) \mathbb{E}(\omega\omega^* \otimes \omega\omega^*)] = \frac{2n}{n+1} \tr [(A\otimes A) \operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}].$

To evalute the trace on the right-hand side of this equation, there is another formula (see Lemma 6 in this paper):

$\tr \left[(A\otimes B) \operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}\right] = \frac{1}{2} \left( \tr(AB) + \tr A \cdot \tr B \right).$

Therefore, we conclude

$\begin{align*}\Var(\omega^* A \omega) &= \mathbb{E}[(\omega^*A\omega)^2] - (\mathbb{E}[\omega^*A\omega])^2 \\&= \frac{2n}{n+1}\tr [(A\otimes A) \operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}] - (\tr A)^2 \\&= \frac{n}{n+1}\left[ \tr A^2 + (\tr A)^2 \right] - (\tr A)^2 \\&= \frac{n}{n+1}\left[ \left\|A\right\|_{\rm F}^2 - \frac{1}{n} (\tr A)^2 \right].\end{align*}$

Proof of Optimality Properties

In this section, we provide proofs of the two optimality properties.

Optimality: Independent Vectors with Independent Coordinates

Assume $A$ is real and symmetric and suppose that $\omega$ is isotropic (2) with independent coordinates. The isotropy condition

$\mathbb{E}[\omega\omega^\top] = I$

implies that $\mathbb{E}[\omega_i\omega_j] = \delta_{ij}$ , where $\delta$ is the Kronecker symbol. Using this fact, we compute the second moment:

$\begin{align*}\mathbb{E}[ (\omega^*A \omega)^2] &= \mathbb{E}\left[ \left( \sum_{i=1}^n A_{ii} \omega_i^2 +2 \sum_{i<j} A_{ij}\omega_i\omega_j) \right)^2\right] \\&= \sum_{i=1}^n A_{ii}^2 \mathbb{E}[\omega_i^4] + \sum_{i<j} (2A_{ii}A_{jj}+4A_{ij}^2) \mathbb{E}[\omega_i^2]\mathbb{E}[\omega_j^2] \\&= \sum_{i=1}^n A_{ii}^2 \mathbb{E}[\omega_i^4] + \sum_{i<j} (2A_{ii}A_{jj}+4A_{ij}^2) .\end{align*}$

Thus

$\Var(\omega^*A\omega) = \mathbb{E}[ (\omega^*A \omega)^2] - (\mathbb{E}[\omega^* A \omega])^2 = \sum_{i=1}^n A_{ii}^2 (\mathbb{E}[|\omega_i|^4]-1) + 4\sum_{i<j} A_{ij}^2.$

The variance is minimized by choosing $\omega$ with $\mathbb{E} \omega_i^4$ as small as possible. Since $\mathbb{E} \omega_i^2 = 1$ , the smallest possible value for $\mathbb{E} \omega_i^4$ is $\mathbb{E} \omega_i^4 = 1$ , which is obtained by populating $\omega$ with random signs.

Optimality: Independent Vectors

This result appears to have first been proven by Richard Kueng in unpublished work. We use an argument suggested to me by Robert J. Webber.

Assume $\mathscr{A}$ is a class of $\field$ -Hermitian matrices closed under $\field$ -unitary similarity transformations and that $\omega$ is an isotropic random vector (2). Decompose the test vector as

$\omega = a \cdot s \quad \text{for} \quad a \in [0,+\infty), \: s \in\{x\in \field^n : x^*x = n \}.$

First, we shall show that the variance is reduced by replacing $s$ with a vector $t$ drawn uniformly from the sphere

(7) $\sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega}) \le \sup_{A\in\mathscr{A}} \Var(\omega^*A\omega$

where

(8) $\tilde{\omega} = a\cdot t \quad \text{and}\quad t\sim \text{Uniform} \{ x \in \field^n :x^*x = n \} \quad \text{is independent of $a$}.$

Note that such a $t$ can be generated as $t = Qs$ for a uniformly random $\field$ -unitary matrix $Q$ . Therefore, we have

$\begin{align*}\sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega})&= \sup_{A\in\mathscr{A}} \left[\mathbb{E}[(\tilde{\omega}^*A\tilde{\omega})^2] - (\tr A)^2\right]\\&= \sup_{A\in\mathscr{A}} \left[\mathbb{E}[a^2 \cdot s^*(Q^*AQ)s] - (\tr (Q^*AQ))^2\right].\end{align*}$

Now apply Jensen’s inequality only over the randomness in $Q$ to obtain

$\begin{align*}\sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega})&= \sup_{A\in\mathscr{A}} \left[\mathbb{E}[a^2 \cdot s^*(Q^*AQ)s] - (\tr (Q^*AQ))^2\right] \\&\le \mathbb{E}_Q \sup_{A\in\mathscr{A}} \left[\mathbb{E}_{a,s}[a^2 \cdot s^*(Q^*AQ)s] - (\tr (Q^*AQ))^2\right].\end{align*}$

Finally, note that since $\mathscr{A}$ is closed under $\field$ -unitary similarity transformations, the supremum over $Q^*AQ$ for $A \in \mathscr{A}$ is the same as the supremum of $A \in \mathscr{A}$ , so we obtain

$\begin{align*}\sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega})&\le \mathbb{E}_Q \sup_{A\in\mathscr{A}} \left[\mathbb{E}_{a,s}[a^2 \cdot s^*(Q^*AQ)s] - (\tr (Q^*AQ))^2\right] \\&= \mathbb{E}_Q \sup_{A\in\mathscr{A}} \left[\mathbb{E}_{a,s}[a^2 \cdot s^*As] - (\tr A)^2\right] \\&= \sup_{A\in\mathscr{A}} \Var(\omega^*A\omega).\end{align*}$

We have successfully proven (7). This argument is a specialized version of a far more general result which appears as Proposition 4.1 in this paper.

Next, we shall prove

(9) $\sup_{A\in\mathscr{A}} \Var(t^*At) \le \sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega}),$

where $t$ is still defined as in (8). Indeed, using the chain rule for variance, we obtain

$\begin{align*}\Var(\tilde{\omega}^*A\tilde{\omega})&= \Var(a^2\cdot t^*At) \\&= \mathbb{E}[\Var(a^2\cdot t^* A t \mid a)] + \Var(\mathbb{E}[a^2\cdot t^* A t \mid a]) \\&= \mathbb{E}[a^4]\Var(t^* A t )+ (\tr A)^2\Var(a^2) \\&\ge \mathbb{E}[a^4]\Var(t^* A t ).\end{align*}$

Here, we have used that $t$ is uniform on the sphere and thus $\mathbb{E}[t^*At] = \tr A$ . By definition, $a$ is the length of $\omega$ divided by $\sqrt{n}$ . Therefore,

$\mathbb{E}[a^2] = \frac{1}{n}\mathbb{E}[\omega^*\omega] = \frac{1}{n} \mathbb{E}[\tr (\omega\omega^*)] = \frac{1}{n} \tr (\mathbb{E}[\omega\omega^*]) = \frac{\tr I}{n} = 1.$

Therefore, by Jensen’s inequality,

$\mathbb{E}[a^4] = \mathbb{E}[(a^2)^2] \ge (\mathbb{E}[a^2])^2 = 1.$

Thus

$\Var(\tilde{\omega}^*A\tilde{\omega}) \ge \mathbb{E}[a^4]\Var(t^* A t ) \ge \Var(t^*At) \quad \text{for every }A,$

which proves (9).

Low-Rank Approximation Toolbox: Nyström, Cholesky, and Schur

October 24, 2022 by Ethan N. Epperly Leave a comment

In the last post, we looked at the Nyström approximation to an $N\times N$ positive semidefinite (psd) matrix $A$ . A special case was the column Nyström approximation, defined to be¹

(Nys) $\hat{A} = A(:,S) \, A(S,S)^{-1} \, A(S,:),$

where $S = \{s_1,\ldots,s_k\} \subseteq \{1,2,\ldots,N\}$ identifies a subset of $k$ columns of $A$ . Provided $k\ll N$ , this allowed us to approximate all $N^2$ entries of the matrix $A$ using only the $kN$ entries in columns $s_1,\ldots,s_k$ of $A$ , a huge savings of computational effort!

With the column Nyström approximation presented just as such, many questions remain:

Why this formula?
Where did it come from?
How do we pick the columns $s_1,\ldots,s_k$ ?
What is the residual $A - \hat{A}$ of the approximation?

In this post, we will answer all of these questions by drawing a connection between low-rank approximation by Nyström approximation and solving linear systems of equations by Gaussian elimination. The connection between these two seemingly unrelated areas of matrix computations will pay dividends, leading to effective algorithms to compute Nyström approximations by the (partial) Cholesky factorization of a positive (semi)definite matrix and an elegant description of the residual of the Nyström approximation as the Schur complement.

Cholesky: Solving Linear Systems

Suppose we want solve the system of linear equations $Ax = b$ , where $A$ is a real $N\times N$ invertible matrix and $b$ is a vector of length $N$ . The standard way of doing this in modern practice (at least for non-huge matrices $A$ ) is by means of Gaussian elimination/LU factorization. We factor the matrix $A$ as a product $A = LU$ of a lower triangular matrix $L$ and an upper triangular matrix $U$ .² The system $Ax = b$ is solved by first solving $Ly = b$ for $y$ and then $Ux = y$ for $x$ ; the triangularity of $L$ and $U$ make solving the associated systems of linear equations easy.

For real symmetric positive definite matrix $A$ , a simplification is possible. In this case, one can compute an LU factorization where the matrices $L$ and $U$ are transposes of each other, $U = L^\top$ . This factorization $A = LL^\top$ is known as a Cholesky factorization of the matrix $A$ .

The Cholesky factorization can be easily generalized to allow the matrix $A$ to be complex-valued. For a complex-valued positive definite matrix $A$ , its Cholesky decomposition takes the form $A = LL^*$ , where $L$ is again a lower triangular matrix. All that has changed is that the transpose ${}^\top$ has been replaced by the conjugate transpose ${}^*$ . We shall work with the more general complex case going forward, though the reader is free to imagine all matrices as real and interpret the operation ${}^*$ as ordinary transposition if they so choose.

Schur: Computing the Cholesky Factorization

Here’s one way of computing the Cholesky factorization using recursion. Write the matrix $A$ in block form as

$A = \twobytwo{A_{11}}{A_{12}}{A_{21}}{A_{22}}.$

Our first step will be “block Cholesky factorize” the matrix $A$ , factoring $A$ as a product of matrices which are only block triangular. Then, we’ll “upgrade” this block factorization into a full Cholesky factorization.

The core idea of Gaussian elimination is to combine rows of a matrix to introduce zero entries. For our case, observe that multiplying the first block row of $A$ by $A_{21}A_{11}^{-1}$ and subtracting this from the second block row introduces a matrix of zeros into the bottom left block of $A$ . (The matrix $A_{11}$ is a principal submatrix of $A$ and is therefore guaranteed to be positive definite and thus invertible.³) In matrix language,

$\twobytwo{I}{0}{-A_{21}A_{11}^{-1}}{I}\twobytwo{A_{11}}{A_{12}}{A_{21}}{A_{22}} = \twobytwo{A_{11}}{A_{12}}{0}{A_{22} - A_{21}A_{11}^{-1}A_{12}}.$

Isolating $A$ on the left-hand side of this equation by multiplying by

$\twobytwo{I}{0}{-A_{21}A_{11}^{-1}}{I}^{-1} = \twobytwo{I}{0}{A_{21}A_{11}^{-1}}{I}$

yields the block triangular factorization

$A = \twobytwo{A_{11}}{A_{12}}{A_{21}}{A_{22}} = \twobytwo{I}{0}{A_{21}A_{11}^{-1}}{I} \twobytwo{A_{11}}{A_{12}}{0}{A_{22} - A_{21}A_{11}^{-1}A_{12}}.$

We’ve factored $A$ into block triangular pieces, but these pieces are not (conjugate) transposes of each other. Thus, to make this equation more symmetrical, we can further decompose

(1) $A = \twobytwo{A_{11}}{A_{12}}{A_{21}}{A_{22}} = \twobytwo{I}{0}{A_{21}A_{11}^{-1}}{I} \twobytwo{A_{11}}{0}{0}{A_{22} - A_{21}A_{11}^{-1}A_{12}} \twobytwo{I}{0}{A_{21}A_{11}^{-1}}{I}^*.$

This is a block version of the Cholesky decomposition of the matrix $A$ taking the form $A = LDL^*$ , where $L$ is a block lower triangular matrix and $D$ is a block diagonal matrix.

We’ve met the second of our main characters, the Schur complement

(Sch) $S = A_{22} - A_{21}A_{11}^{-1}A_{12}.$

This seemingly innocuous combination of matrices has a tendency to show up in surprising places when one works with matrices.⁴ It’s appearance in any one place is unremarkable, but the shear ubiquity of $A_{22} - A_{21}A_{11}^{-1}A_{12}$ in matrix theory makes it deserving of its special name, the Schur complement. To us for now, the Schur complement is just the matrix appearing in the bottom right corner of our block Cholesky factorization.

The Schur complement enjoys the following property:⁵

Positivity of the Schur complement: If $A=\twobytwo{A_{11}}{A_{12}}{A_{21}}{A_{22}}$ is positive (semi)definite, then the Schur complement $S = A_{22} - A_{21}A_{11}^{-1}A_{12}$ is positive (semi)definite.

As a consequence of this property, we conclude that both $A_{11}$ and $S$ are positive definite.

With the positive definiteness of the Schur complement in hand, we now return to our Cholesky factorization algorithm. Continue by recursively⁶ computing Cholesky factorizations of the diagonal blocks

$A_{11} = L_{11}^{\vphantom{*}}L_{11}^*, \quad S = L_{22}^{\vphantom{*}}L_{22}^*.$

Inserting these into the block $LDL^*$ factorization (1) and simplifying gives a Cholesky factorization, as desired:

$A = \twobytwo{L_{11}}{0}{A_{21}^{\vphantom{*}}(L_{11}^{*})^{-1}}{L_{22}}\twobytwo{L_{11}}{0}{A_{21}^{\vphantom{*}}(L_{11}^{*})^{-1}}{L_{22}}^* =: LL^*.$

Voilà, we have obtained a Cholesky factorization of a positive definite matrix $A$ !

By unwinding the recursions (and always choosing the top left block $A_{11}$ to be of size $1\times 1$ ), our recursive Cholesky algorithm becomes the following iterative algorithm: Initialize $L$ to be the zero matrix. For $j = 1,2,3,\ldots,N$ , perform the following steps:

Update $L$ . Set the $j$ th column of $L$ :
$L(j:N,j) \leftarrow A(j:N,j)/\sqrt{a_{jj}}.$
Update $A$ . Update the bottom right portion of $A$ to be the Schur complement:
$A(j+1:N,j+1:N)\leftarrow A(j+1:N,j+1:N) - \frac{A(j+1:N,j)A(j,j+1:N)}{a_{jj}}.$

This iterative algorithm is how Cholesky factorization is typically presented in textbooks.

Nyström: Using Cholesky Factorization for Low-Rank Approximation

Our motivating interest in studying the Cholesky factorization was the solution of linear systems of equations $Ax = b$ for a positive definite matrix $A$ . We can also use the Cholesky factorization for a very different task, low-rank approximation.

Let’s first look at things through the lense of the recursive form of the Cholesky factorization. The first step of the factorization was to form the block Cholesky factorization

$A = \twobytwo{I}{0}{A_{21}A_{11}^{-1}}{I} \twobytwo{A_{11}}{0}{0}{A_{22} - A_{21}A_{11}^{-1}A_{12}} \twobytwo{I}{0}{A_{21}A_{11}^{-1}}{I}^*.$

Suppose that we choose the top left $A_{11}$ block to be of size $k\times k$ , where $k$ is much smaller than $N$ . The most expensive part of the Cholesky factorization will be the recursive factorization of the Schur complement $A_{22} - A_{21}A_{11}^{-1} A_{12}$ , which is a large matrix of size $(N-k)\times (N-k)$ .

To reduce computational cost, we ask the provocative question: What if we simply didn’t factorize the Schur complement? Observe that we can write the block Cholesky factorization as a sum of two terms

(2) $A = \twobyone{I}{A_{21}{A_{11}^{-1}}} A_{11} \twobyone{I}{A_{22}{A_{11}^{-1}}}^* + \twobytwo{0}{0}{0}{A_{22}-A_{21}A_{11}^{-1}A_{12}}.$

We can use the first term of this sum as a rank- $k$ approximation to the matrix $A$ . The low-rank approximation, which we can write out more conveniently as

$\hat{A} = \twobyone{A_{11}}{A_{21}} A_{11}^{-1} \onebytwo{A_{11}}{A_{12}} = \twobytwo{A_{11}}{A_{12}}{A_{21}}{A_{21}A_{11}^{-1}A_{12}},$

is the column Nyström approximation (Nys) to $A$ associated with the index set $S = \{1,2,3,\ldots,k\}$ and is the final of our three titular characters. The residual of the Nyström approximation is the second term in (2), which is none other than the Schur complement (Sch), padded by rows and columns of zeros:

$A - \hat{A} = \twobytwo{0}{0}{0}{A_{22}-A_{21}A_{11}^{-1}A_{12}}.$

Observe that the approximation $\hat{A}$ is obtained from the process of terminating a Cholesky factorization midway through algorithm execution, so we say that the Nyström approximation results from a partial Cholesky factorization of the matrix $A$ .

Summing things up:

If we perform a partial Cholesky factorization on a positive (semi)definite matrix, we obtain a low-rank approximation known as the column Nyström approximation. The residual of this approximation is the Schur complement, padded by rows and columns of zeros.

The idea of obtaining a low-rank approximation from a partial matrix factorization is very common in matrix computations. Indeed, the optimal low-rank approximation to a real symmetric matrix is obtained by truncating its eigenvalue decomposition and the optimal low-rank approximation to a general matrix is obtained by truncating its singular value decomposition. While the column Nyström approximation is not the optimal rank- $k$ approximation to $A$ (though it does satisfy a weaker notion of optimality, as discussed in this previous post), it has a killer feature not possessed by the optimal approximation:

The column Nyström approximation is formed from only $k$ columns from the matrix $A$ . A column Nyström approximation approximates an $N\times N$ matrix by only reading a fraction of its entries!

Unfortunately there’s not a free lunch here. The column Nyström is only a good low-rank approximation if the Schur complement has small entries. In general, this need not be the case. Fortunately, we can improve our situation by means of pivoting.

Our iterative Cholesky algorithm first performs elimination using the entry in position $(1,1)$ followed by position $(2,2)$ and so on. There’s no need to insist on this exact ordering of elimination steps. Indeed, at each step of the Cholesky algorithm, we can choose whichever diagonal position $(j,j)$ that we want to perform elimination. The entry we choose to perform elimination with is called a pivot.

Obtaining good column Nyström approximations requires identifying the a good choice for the pivots to reduce the size of the entries of the Schur complement at each step of the algorithm. With general pivot selection, an iterative algorithm for computing a column Nyström approximation by partial Cholesky factorization proceeds as follows: Initialize an $N\times k$ matrix $F$ to store the column Nyström approximation $\hat{A} = FF^*$ , in factored form. For $j = 1,2,\ldots,k$ , perform the following steps:

Select pivot. Choose a pivot $s_j$ .
Update the approximation. $F(:,j) \leftarrow A(:,s_j) / \sqrt{a_{s_js_j}}$ .
Update the residual. $A \leftarrow A - \frac{A(:,s_j)A(s_j,:)}{a_{s_js_j}}$ .

This procedure results in the Nyström approximation (Nys) with column set $S = \{s_1,\ldots,s_k\}$ :

$\hat{A} = FF^* = A(:,S) \, A(S,S)^{-1} \, A(S,:).$

The pivoted Cholesky steps 1–3 requires updating the entire matrix $A$ at every step. With a little more cleverness, we can optimize this procedure to only update the entries of the matrix $A$ we need to form the approximation $\hat{A} = FF^*$ . See Algorithm 2 in this preprint of my coauthors and I for details.

How should we choose the pivots? Two simple strategies immediately suggest themselves:

Uniformly random. At each step $j$ , select $s_j$ uniformly at random from among the un-selected pivot indices.
Greedy.⁷ At each step $j$ , select $s_j$ according to the largest diagonal entry of the current residual $A$ :
$s_j = \argmax_{1\le k\le N} a_{kk}.$

The greedy strategy often (but not always) works well, and the uniformly random approach can work surprisingly well if the matrix $A$ is “incoherent”, with all rows and columns of the matrix possessing “similar importance”.

Despite often working fairly well, both the uniform and greedy schemes can fail significantly, producing very low-quality approximations. My research (joint with Yifan Chen, Joel A. Tropp, and Robert J. Webber) has investigated a third strategy striking a balance between these two approaches:

Diagonally weighted random. At each step $j$ , select $s_j$ at random according to the probability weights based on the current diagonal of the matrix
$\mathbb{P} \{ s_j = k \} = \frac{a_{kk}}{\operatorname{tr} A}.$

Our paper provides theoretical analysis and empirical evidence showing that this diagonally-weighted random pivot selection (which we call randomly pivoted Cholesky aka RPCholesky) performs well at approximating all positive semidefinite matrices $A$ , even those for which uniform random and greedy pivot selection fail. The success of this approach can be seen in the following examples (from Figure 1 in the paper), which shows RPCholesky can produce much smaller errors than the greedy and uniform methods.

Conclusions

In this post, we’ve seen that a column Nyström approximation can be obtained from a partial Cholesky decomposition. The residual of the approximation is the Schur complement. I hope you agree that this is a very nice connection between these three ideas. But beyond its mathematical beauty, why do we care about this connection? Here are my reasons for caring:

Analysis. Cholesky factorization and the Schur complement are very well-studied in matrix theory. We can use known facts about Cholesky factorization and Schur complements to prove things about the Nyström approximation, as we did when we invoked the positivity of the Schur complement.
Algorithms. Cholesky factorization-based algorithms like randomly pivoted Cholesky are effective in practice at producing high-quality column Nyström approximations.

On a broader level, our tale of Nyström, Cholesky, and Schur demonstrates that there are rich connections between low-rank approximation and (partial versions of) classical matrix factorizations like LU (with partial pivoting), Cholesky, QR, eigendecomposition, and SVD for to full-rank matrices. These connections can be vital to analyzing low-rank approximation algorithms and developing improvements.