Tag: mathematics

Poisson approximations to the negative binomial distribution

This post is an introduction to the negative binomial distribution and a discussion of different ways of approximating the negative binomial distribution.

The negative binomial distribution describes the number of times a coin lands on tails before a certain number of heads are recorded. The distribution depends on two parameters p and r. The parameter p is the probability that the coin lands on heads and r is the number of heads. If X has the negative binomial distribution, then X = x means in the first x+r-1 tosses of the coin, there were r-1 heads and that toss number x+r was a head. This means that the probability that X=x is given by

\displaystyle{f(x) = \binom{x+r-1}{r-1}p^{r}\left(1-p\right)^x}

Here is a plot of the function f(x) for different values of r and p.

Poisson approximations

When the parameter r is large and p is close to one, the negative binomial distribution can be approximated by a Poisson distribution. More formally, suppose that r(1-p)=\lambda for some positive real number \lambda. If r is large then, the negative binomial random variable with parameters p and r, converges to a Poisson random variable with parameter \lambda. This is illustrated in the picture below where three negative binomial distributions with r(1-p)=5 approach the Poisson distribution with \lambda =5.

Total variation distance is a common way to measure the distance between two discrete probability distributions. The log-log plot below shows that the error from the Poisson approximation is on the order of 1/r and that the error is bigger if the limiting value of r(1-p) is larger.

It turns out that is is possible to get a more accurate approximation by using a different Poisson distribution. In the first approximation, we used a Poisson random variable with mean \lambda = r(1-p). However, the mean of the negative binomial distribution is r(1-p)/p. This suggests that we can get a better approximation by setting \lambda = r(1-p)/p.

The change from \lambda = r(1-p) to \lambda = r(1-p)/p is a small because p \approx 1. However, this small change gives a much better approximation, especially for larger values of r(1-p). The below plot shows that both approximations have errors on the order of 1/r, but the constant for the second approximation is much better.

Second order accurate approximation

It is possible to further improve the Poisson approximation by using a Gram–Charlier expansion. A Gram–Charlier approximation for the Poisson distribution is given in this paper.1 The approximation is

\displaystyle{f_{GC}(x) = P_\lambda(x) - \frac{1}{2}(1-p)\left((x-\lambda)P_\lambda(x)-(x-1-\lambda)P_\lambda(x-1)\right)},

where \lambda = \frac{k(1-p)}{p} as in the second Poisson approximation and P_\lambda(x) is the Poisson pmf evaluated at x.

The Gram–Charlier expansion is considerably more accurate than either Poisson approximation. The errors are on the order of 1/r^2. This higher accuracy means that the error curves for the Gram–Charlier expansion has a steeper slope.

  1. The approximation is given in equation (4) of the paper and is stated in terms of the CDF instead of the PMF. The equation also contains a small typo, it should say \frac{1}{2}q instead of \frac{1]{2}p. ↩︎

“Uniformly random”

The term “uniformly random” sounds like a contradiction. How can the word “uniform” be used to describe anything that’s random? Uniformly random actually has a precise meaning, and, in a sense, means “as random as possible.” I’ll explain this with an example about shuffling card.

Shuffling cards

Suppose I have a deck of ten cards labeled 1 through 10. Initially, the cards are face down and in perfect order. The card labeled 10 is on top of the deck. The card labeled 9 is second from the top, and so on down to the card labeled 1. The cards are definitely not random.

Next, I generate a random number between 1 and 10. I then find the card with the corresponding label and put it face down on top of the deck. The cards are now somewhat random. The number on top could anything, but the rest of the cards are still in order. The cards are random but they are not uniformly random.

Now suppose that I keep generating random numbers and moving cards to the top of the deck. Each time I do this, the cards get more random. Eventually (after about 30 moves1) the cards will be really jumbled up. Even if you knew the first few cards, it would be hard to predict the order of the remaining ones. Once the cards are really shuffled, they are uniformly random.

Uniformly random

A deck of cards is uniformly random if each of the possible arrangements of the cards are equally likely. After only moving one card, the deck of cards is not uniformly random. This is because there are only 10 possible arrangements of the deck. Once the deck is well-shuffled, all of the 3,628,800 possible arrangements are equally likely.

In general, something is uniformly random if each possibility is equally likely. So the outcome of rolling a fair 6-sided die is uniformly random, but rolling a loaded die is not. The word “uniform” refers to the chance of each possibility (1/6 for each side of the die). These chances are all the same and “uniform”.

This is why uniformly random can mean “as random as possible.” If you are using a fair die or a well-shuffled deck, there are no biases in the outcome. Mathematically, this means you can’t predict the outcome.

Communicating research

The inspiration for this post came from a conversation I had earlier in the week. I was telling someone about my research. As an example, I talked about how long it takes for a deck of cards to become uniformly random. They quickly stopped me and asked how the two words could ever go together. It was a good point! I use the words uniformly random all the time and had never realized this contradiction.2 It was a good reminder about the challenge of clear communication.

Footnotes

  1. The number of moves it takes for the deck to well-shuffled is actually random. But on average it takes around 30 moves. For the mathematical details, see Example 1 in Shuffling Cards and Stopping Times by David Aldous and Persi Diaconis. ↩︎
  2. Of the six posts I published last year, five contain the word uniform! ↩︎

Understanding the Ratio of Uniforms Distribution

The ratio of uniforms distribution is a useful distribution for rejection sampling. It gives a simple and fast way to sample from discrete distributions like the hypergeometric distribution1. To use the ratio of uniforms distribution in rejection sampling, we need to know the distributions density. This post summarizes some properties of the ratio of uniforms distribution and computes its density.

The ratio of uniforms distribution is the distribution of the ratio of two independent uniform random variables. Specifically, suppose U \in [-1,1] and V \in [0,1] are independent and uniformly distributed. Then R = U/V has the ratio of uniforms distribution. The plot below shows a histogram based on 10,000 samples from the ratio of uniforms distribution2.

The histogram has a flat section in the middle and then curves down on either side. This distinctive shape is called a “table mountain”. The density of R also has a table mountain shape.

And here is the density plotted on top of the histogram.

A formula for the density of R is

\displaystyle{h(R) = \begin{cases} \frac{1}{4} & \text{if } -1 \le R \le 1, \\\frac{1}{4R^2} & \text{if } R < -1 \text{ or } R > 1.\end{cases}}

The first case in the definition of h corresponds to the flat part of the table mountain. The second case corresponds to the sloping curves. The rest of this post use geometry to derive the above formula for h(R).

Calculating the density

The point (U,V) is uniformly distributed in the box B=[-1,1] \times [0,1]. The image below shows an example of a point (U,V) inside the box B.

We can compute the ratio R = U/V geometrically. First we draw a straight line that starts at (0,0) and goes through (U,V). This line will hit the horizontal line y=1. The x coordinate at this point is exactly R=U/V.

In the above picture, all of the points on the dashed line map to the same value of R. We can compute the density of R by computing an area. The probability that R is in a small interval [R,R+dR] is

\displaystyle{\frac{\text{Area}(\{(u,v) \in B : u/v \in [R, R+dR]\})}{\text{Area}(B)} = \frac{1}{2}\text{Area}(\{(u,v) \in B : u/v \in [R, R+dR]\}).}

If we can compute the above area, then we will know the density of R because by definition

\displaystyle{h(R) = \lim_{dR \to 0} \frac{1}{2dR}\text{Area}(\{(u,v) \in B : u/v \in [R, R+dR]\})}.

We will first work on the case when R is between -1 and 1. In this case, the set \{(u,v) \in B : u/v \in [R, R+dR]\} is a triangle. This triangle is drawn in blue below.

The horizontal edge of this triangle has length dR. The perpendicular height of the triangle from the horizontal edge is 1. This means that

\displaystyle{\text{Area}(\{(u,v) \in B : u/v \in [R, R+dR]\}) =\frac{1}{2}\times dR \times 1=\frac{dR}{2}}.

And so, when R \in [-1,1] we have

\displaystyle{h(R) = \lim_{dR\to 0} \frac{1}{2dR}\times \frac{dR}{2}=\frac{1}{4}}.

Now let’s work on the case when R is bigger than 1 or less than -1. In this case, the set \{(u,v) \in B : u/v \in [R, R+dR]\} is again triangle. But now the triangle has a vertical edge and is much skinnier. Below the triangle is drawn in red. Note that only points inside the box B are coloured in.

The vertical edge of the triangle has length \frac{1}{R} - \frac{1}{R+dR}= \frac{dR}{R(R+dR)}. The perpendicular height of the triangle from the vertical edge is 1. Putting this together

\displaystyle{\text{Area}(\{(u,v) \in B : u/v \in [R, R+dR]\}) =\frac{1}{2}\times \frac{dR}{R(R+dR)} \times 1=\frac{dR}{2R(R+dR)}}.

And so

\displaystyle{h(R) = \lim_{dR \to 0} \frac{1}{2dR} \times \frac{dR}{2 R(R+dR)} = \frac{1}{4R^2}}.

And so putting everything together

\displaystyle{h(R) = \begin{cases} \frac{1}{4} & \text{if } -1 \le R \le 1, \\\frac{1}{4R^2} & \text{if } R < -1 \text{ or } R > 1.\end{cases}}

Footnotes and references

  1. https://ieeexplore.ieee.org/document/718718 ↩︎
  2. For visual purposes, I restricted the sample to values of R between -8 and 8. This is because the ratio of uniform distribution has heavy tails. This meant that there were some very large values of R that made the plot hard to see. ↩︎

The discrete arcsine distribution

The discrete arcsine distribution is a probability distribution on \{0,1,\ldots,n\}. It is a u-shaped distribution. There are peaks at 0 and n and a dip in the middle. The figure below shows the probability distribution function for n=10,15, 20.

The probability distribution function of the arcsine distribution is given by

\displaystyle{p_n(k) = \frac{1}{2^{2n}}\binom{2k}{k}\binom{2n-2k)}{n-k}\text{ for } k \in \{0,1,\ldots,n\}}

The discrete arcsine distribution is related to simple random walks and to an interesting Markov chain called the Burnside process. The connection with simple random walks is explained in Chapter 3, Volume 1 of An Introduction to Probability and its applications by William Feller. The connection to the Burnside process was discovered by Persi Diaconis in Analysis of a Bose-Einstein Markov Chain.

The discrete arcsine distribution gets its name from the continuous arcsine distribution. Suppose X_n is distributed according to the discrete arcsine distribution with parameter n. Then the normalized random variables X_n/n converges in distribution to the continuous arcsine distribution on [0,1]. The continuous arcsine distribution has the probability density function

\displaystyle{f(x) = \frac{1}{\pi\sqrt{x(1-x)}} \text{ for } 0 \le x \le 1}

This means that continuous arcsine distribution is a beta distribution with \alpha=\beta=1/2. It is called the arcsine distribution because the cumulative distribution function involves the arcsine function

\displaystyle{F(x) = \int_0^x f(y)dy = \frac{2}{\pi}\arcsin(\sqrt{x}) \text{ for } 0 \le x \le 1}

There is another connection between the discrete and continuous arcsine distributions. The continuous arcsine distribution can be used to sample the discrete arcsine distribution. The two step procedure below produces a sample from the discrete arcsine distribution with parameter n:

  1. Sample p from the continuous arcsine distribution.
  2. Sample X from the binomial distribution with parameters n and p.

This means that the discrete arcsine distribution is actually the beta-binomial distribution with parameters \alpha = \beta =1/2. I was surprised when I was told this, and couldn’t find a reference. The rest of this blog post proves that the discrete arcsine distribution is an instance of the beta-binomial distribution.

As I showed in this post, the beta-binomial distribution has probability distribution function:

\displaystyle{q_{\alpha,\beta,n}(k) = \binom{n}{k}\frac{B(k+\alpha, n-k+\alpha)}{B(a,b)}},

where B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} is the Beta-function. To show that the discrete arc sine distribution is an instance of the beta-binomial distribution we need that p_n(k)=q_{1/2,1/2,n}(k). That is

\displaystyle{ \binom{n}{k}\frac{B(k+1/2, n-k+1/2)}{B(1/2,1/2)} = \frac{1}{2^{2n}}\binom{2k}{k}\binom{2n-2k}{n-k}},

for all k = 0,1,\ldots,n. To prove the above equation, we can first do some simplifying to q_{1/2,1/2,n}(k). By definition

\displaystyle{\frac{B(k+1/2, n-k+1/2)}{B(1/2,1/2)} = \frac{\frac{\Gamma(k+1/2)\Gamma(n-k+1/2)}{\Gamma(n+1)}}{\frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)}} = \frac{1}{n!}\frac{\Gamma(k+1/2)}{\Gamma(1/2)}\frac{\Gamma(n-k+1/2)}{\Gamma(1/2)}},

where I have used that \Gamma(m)=(m-1)! factorial if m is a natural number. The Gamma function \Gamma(x) also satisfies the property \Gamma(x+1)=x\Gamma(x). Using this repeatedly gives

\displaystyle{\Gamma(k+1/2) = (k-1/2) \times (k-3/2) \times \cdots \times \frac{3}{2}\times\frac{1}{2}\times\Gamma(1/2) }.

This means that

\displaystyle{\frac{\Gamma(k+1/2)}{\Gamma(1/2)} = (k-1/2) \times (k-3/2) \times \cdots \times \frac{3}{2}\times\frac{1}{2} = \frac{(2k-1)\times(2k-3)\times \cdots \times 3 \times 1}{2^k}=\frac{(2k-1)!!}{2^k}},

where (2k-1)!!=(2k-1)\times (2k-3)\times\cdots \times 3 \times 1 is the double factorial. The same reasoning gives

\displaystyle{\frac{\Gamma(n-k+1/2)}{\Gamma(1/2)} =\frac{(2n-2k-1)!!}{2^{n-k}}}.

And so

\displaystyle{q_{1/2,1/2,n}(k) =\frac{1}{2^nk!(n-k)!}(2k-1)!!(2n-2k-1)!!}.

We’ll now show that p_n(k) is also equal to the above final expression. Recall

\displaystyle{p_n(k) = \frac{1}{2^{2n}} \binom{2k}{k}\binom{2(n-k)}{n-k} = \frac{1}{2^{2n}}\frac{(2k)!(2(n-k))!}{k!k!(n-k)!(n-k)!} = \frac{1}{2^nk!(n-k)!}\frac{(2k)!}{k!2^k}\frac{(2n-2k)!}{(n-k)!2^{n-k}}}.

And so it suffices to show \frac{(2k)!}{k!2^k} = (2k-1)!! (and hence \frac{(2n-2k)!}{(n-k)!2^{n-k}}=(2n-2k-1)!!). To see why this last claim holds, note that

\displaystyle{\frac{(2k)!}{k!2^k} = \frac{(2k)\times (2k-1)\times(2k-2)\times\cdots\times 3 \times 2 \times 1}{(2k)\times (2k-2)\times \cdots \times 2} = (2k-1)!!}

Showing that p_{n}(k)=q_{n,1/2,1/2}(k) as claimed.

The sample size required for importance sampling

My last post was about using importance sampling to estimate the volume of high-dimensional ball. The two figures below compare plain Monte Carlo to using importance sampling with a Gaussian proposal. Both plots use M=1,000 samples to estimate v_n, the volume of an n-dimensional ball

A friend of mine pointed out that the relative error does not seem to increase with the dimension n. He thought it was too good to be true. It turns out he was right and the relative error does increase with dimension but it increases very slowly. To estimate v_n the number of samples needs to grow on the order of \sqrt{n}.

To prove this, we will use the paper The sample size required for importance sampling by Chatterjee and Diaconis [1]. This paper shows that the sample size for importance sampling is determined by the Kullback-Liebler divergence. The relevant result from their paper is Theorem 1.3. This theorem is about the relative error in using importance sampling to estimate a probability.

In our setting the proposal distribution is Q=\mathcal{N}(0,\frac{1}{n}I_n). That is the distribution Q is an n-dimensional Gaussian vector with mean 0 and covariance \frac{1}{n}I_n. The conditional target distribution is P the uniform distribution on the n dimensional ball. Theorem 1.3 in [1] tells us how many samples are needed to estimate v_n. Informally, the required sample size is M = O(\exp(D(P \Vert Q))). Here D(P\Vert Q) is the Kullback-Liebler divergence between P and Q.

To use this theorem we need to compute D(P \Vert Q). Kullback-Liebler divergence is defined as integral. Specifically

\displaystyle{D(P\Vert Q) = \int_{\mathbb{R}^n} \log\frac{P(x)}{Q(x)}P(x)dx}

Computing the high-dimensional integral above looks challenging. Fortunately, it can reduced to a one-dimensional integral. This is because both the distributions P and Q are rotationally symmetric. To use this, define P_r,Q_r to be the distribution of the norm squared under P and Q. That is if X \sim P, then \Vert X \Vert_2^2 \sim P_R and likewise for Q_R. By the rotational symmetry of P and Q we have

D(P\Vert Q) = D(P_R \Vert Q_R).

We can work out both P_R and Q_R. The distribution P is the uniform distribution on the n-dimensional ball. And so for X \sim P and any r \in [0,1]

\mathbb{P}(\Vert X \Vert_2^2 \le r) = \frac{v_n r^n}{v_n} = r^n.

Which implies that P_R has density P_R(r)=nr^{n-1}. This means that P_R is a Beta distribution with parameters \alpha = n, \beta = 1. The distribution Q is a multivariate Gaussian distribution with mean 0 and variance \frac{1}{n}I_n. This means that if X \sim Q, then \Vert X \Vert_2^2 = \sum_{i=1}^n X_i^2 is a scaled chi-squared variable. The shape parameter of Q_R is n and scale parameter is 1/n. The density for Q_R is therefor

Q_R(r) = \frac{n^{n/2}}{2^{n/2}\Gamma(n/2)}r^{n/2-1}e^{-nx/2}

The Kullback-Leibler divergence between P and Q is therefor

\displaystyle{D(P\Vert Q)=D(P_R\Vert Q_R) = \int_0^1 \log \frac{P_R(r)}{Q_R(r)} P_R(r)dr}

Getting Mathematica to do the above integral gives

D(P \Vert Q) = -\frac{1+2n}{2+2n} + \frac{n}{2}\log(2 e) - (1-\frac{n}{2})\log n + \log \Gamma(\frac{n}{2}).

Using the approximation \log \Gamma(z) \approx (z-\frac{1}{2})\log(z)-z+O(1) we get that for large n

D(P \Vert Q) = \frac{1}{2}\log n + O(1).

And so the required number of samples is O(\exp(D(P \Vert Q)) = O(\sqrt{n}).

[1] Chatterjee, Sourav, and Persi Diaconis. “THE SAMPLE SIZE REQUIRED IN IMPORTANCE SAMPLING.” The Annals of Applied Probability 28, no. 2 (2018): 1099–1135. https://www.jstor.org/stable/26542331. (Public preprint here https://arxiv.org/abs/1511.01437)

Monte Carlo integration in high dimensions

Monte Carlo integration

John Cook recently wrote a cautionary blog post about using Monte Carlo to estimate the volume of a high-dimensional ball. He points out that if \mathbf{X}=(X_1,\ldots,X_n) are independent and uniformly distributed on the interval [-1,1] then

\displaystyle{\mathbb{P}(X_1^2 + X_2^2 + \cdots + X_n^2 \le 1) = \frac{v_n}{2^n}},

where v_n is the volume of an n-dimensional ball with radius one. This observation means that we can use Monte Carlo to estimate v_n.

To do this we repeatedly sample vectors \mathbf{X}_m = (X_{m,1},\ldots,X_{m,n}) with X_{m,i} uniform on [-1,1] and m ranging from 1 to M. Next, we count the proportion of vectors \mathbf{X}_m with X_{1,m}^2 + \cdots + X_{n,m}^2 \le 1. Specifically, if Z_m is equal to one when X_{1,m}^2 + \cdots + X_{n,m}^2 \le 1 and equal to zero otherwise, then by the law of large numbers

\displaystyle{\frac{1}{M}\sum_{m=1}^M Z_m \approx \frac{v_n}{2^n}}

Which implies

v_n \approx 2^n \frac{1}{M}\sum_{m=1}^M Z_m

This method of approximating a volume or integral by sampling and counting is called Monte Carlo integration and is a powerful general tool in scientific computing.

The problem with Monte Carlo integration

As pointed out by John, Monte Carlo integration does not work very well in this example. The plot below shows a large difference between the true value of v_n with n ranging from 1 to 20 and the Monte Carlo approximation with M = 1,000.

The problem is that v_n is very small and the probability v_n/2^n is even smaller! For example when n = 10, v_n/2^n \approx 0.0025. This means that in our one thousand samples we only expect two or three occurrences of the event X_{m,1}^2 + \cdots + X_{m,n}^2 \le 1. As a result our estimate has a high variance.

The results get even worse as n increases. The probability v_n/2^n does to zero faster than exponentially. Even with a large value of M, our estimate will be zero. Since v_n \neq 0, the relative error in the approximation is 100%.

Importance sampling

Monte Carlo can still be used to approximate v_n. Instead of using plain Monte Carlo, we can use a variance reduction technique called importance sampling (IS). Instead of sampling the points X_{m,1},\ldots, X_{m,n} from the uniform distribution on [-1,1], we can instead sample the from some other distribution called a proposal distribution. The proposal distribution should be chosen so that that the event X_{m,1}^2 +\cdots +X_{m,n}^2 \le 1 becomes more likely.

In importance sampling, we need to correct for the fact that we are using a new distribution instead of the uniform distribution. Instead of the counting the number of times X_{m,1}^2+\cdots+X_{m,n}^2 \le 1 , we give weights to each of samples and then add up the weights.

If p is the density of the uniform distribution on [-1,1] (the target distribution) and q is the density of the proposal distribution, then the IS Monte Carlo estimate of v_n is

\displaystyle{\frac{1}{M}\sum_{m=1}^M Z_m \prod_{i=1}^n \frac{p(X_{m,i})}{q(X_{m,i})}},

where as before Z_m is one if X_{m,1}^2+\cdots +X_{m,n}^2 \le 1 and Z_m is zero otherwise. As long as q(x)=0 implies p(x)=0, the IS Monte Carlo estimate will be an unbiased estimate of v_n. More importantly, a good choice of the proposal distribution q can drastically reduce the variance of the IS estimate compared to the plain Monte Carlo estimate.

In this example, a good choice of proposal distribution is the normal distribution with mean 0 and variance 1/n. Under this proposal distribution, the expected value of X_{m,1}^2 +\cdots+ X_{m,n}^2 is one and so the event X_{m,1}^2 + \cdots + X_{m,n}^2 \le 1 is much more likely.

Here are the IS Monte Carlo estimates with again M = 1,000 and n ranging from 1 to 20. The results speak for themselves.

The relative error is typically less than 10%. A big improvement over the 100% relative error of plain Monte Carlo.

The next plot shows a close agreement between v_n and the IS Monte Carlo approximation on the log scale with n going all the way up to 100.

Notes

  1. There are exact formulas for v_n (available on Wikipedia). I used these to compare the approximations and compute relative errors. There are related problems where no formulas exist and Monte Carlo integration is one of the only ways to get an approximate answer.
  2. The post by John Cook also talks about why the central limit theorem can’t be used to approximate v_n. I initially thought a technique called large deviations could be used to approximate v_n but again this does not directly apply. I was happy to discover that importance sampling worked so well!

More sampling posts

Uniformly sampling orthogonal matrices

An n \times n matrix M \in \mathbb{R}^{n \times n} is orthogonal if M^\top M = I. The set of all n \times n orthogonal matrices is a compact group written as O_n. The uniform distribution on O_n is called Haar measure. There are many ways to generate random matrices for Haar measure. One of which is based on the Gram-Schmidt algorithm.

Proposition. Let X be a n \times n matrix such that each entry X_{ij} is an independent standard normal random variable. Let x_1,x_2,\ldots,x_n \in \mathbb{R}^n be the columns of X and let q_1,q_2,\ldots,q_n \in \mathbb{R}^n be the result of applying Gram-Schmidt to x_1,x_2,\ldots,x_n. Then the matrix M=[q_1,q_2,\ldots,q_n] \in O_n is distributed according to Haar measure.

Another way of stating the above proposition is that if X has i.i.d. standard normal entries, and Q,R is a QR-factorization of X calculated using Gram-Schmidt, then Q \in O_n is distributed according to Haar measure. However, most numerical libraries do not use Gram-Schmidt to calculate the QR-factorization of a matrix. This means that if you generate a random X and then use your computer’s QR-factorization function, then the result might not be Haar distributed as the plot below shows.

The plot shows an estimate of the distribution of \sqrt{n}M_{11}, the top-left entry of a matrix M. When M is correctly sampled, the distribution is symmetric about 0. When M is incorrectly sampled the distribution is clearly biased towards negative numbers.

Fortunately there is a fix described in [1]! We can first use any QR-factorization algorithm to produce Q,R with X=QR. We then compute a diagonal matrix L with L_{ii} = \mathrm{Sign}(R_{ii}). Then, the matrix M=QL is Haar distributed. The following python code thus samples a Haar distributed matrix in O_n.

import numpy as np

def sample_M(n):
    M = np.random.randn(n, n)
    Q, R = np.linalg.qr(M)
    L = np.sign(np.diag(R))
    return Q*L[None,:]

References

[1] Mezzadri, Francesco. “How to generate random matrices from the classical compact groups.” arXiv preprint math-ph/0609050 (2006). https://arxiv.org/abs/math-ph/0609050

I recently gave a talk at the Stanford student probability seminar on Haar distributed random matrices. My notes and many further references are available here.

Auxiliary variables sampler

The auxiliary variables sampler is a general Markov chain Monte Carlo (MCMC) technique for sampling from probability distributions with unknown normalizing constants [1, Section 3.1]. Specifically, suppose we have n functions f_i : \mathcal{X} \to (0,\infty) and we want to sample from the probability distribution P(x) \propto \prod_{i=1}^n f_i(x). That is

\displaystyle{ P(x) = \frac{1}{Z}\prod_{i=1}^n f_i(x)},

where Z = \sum_{x \in \mathcal{X}} \prod_{i=1}^n f_i(x) is a normalizing constant. If the set \mathcal{X} is very large, then it may be difficult to compute Z or sample from P(x). To approximately sample from P(x) we can run an ergodic Markov chain with P(x) as a stationary distribution. Adding auxiliary variables is one way to create such a Markov chain. For each i \in \{1,\ldots,n\}, we add a auxiliary variable U_i \ge 0 such that

\displaystyle{P(U \mid X) = \prod_{i=1}^n \mathrm{Unif}[0,f_i(X)]}.

That is, conditional on X, the auxiliary variables U_1,\ldots,U_n are independent and U_i is uniformly distributed on the interval [0,f_i(X)]. If X is distributed according to P and U_1,\ldots,U_n have the above auxiliary variable distribution, then

\displaystyle{P(X,U) =P(X)P(U\mid X)\propto \prod_{i=1}^n f_i(X) \frac{1}{f_i(X)} I[U_i \le f(X_i)] = \prod_{i=1}^n I[U_i \le f(X_i)]}.

This means that the joint distribution of (X,U) is uniform on the set

\widetilde{\mathcal{X}} = \{(x,u) \in \mathcal{X} \times (0,\infty)^n : u_i \le f(x_i) \text{ for all } i\}.

Put another way, suppose we could jointly sample (X,U) from the uniform distribution on \widetilde{\mathcal{X}}. Then, the above calculation shows that if we discard U and only keep X, then X will be sampled from our target distribution P.

The auxiliary variables sampler approximately samples from the uniform distribution on \widetilde{\mathcal{X}} is by using a Gibbs sampler. The Gibbs samplers alternates between sampling U' from P(U \mid X) and then sampling X' from P(X \mid U'). Since the joint distribution of (X,U) is uniform on \widetilde{\mathcal{X}}, the conditional distributions P(X \mid U) and P(U \mid X) are also uniform. The auxiliary variables sampler thus transitions from X to X' according to the two step process

  1. Independently sample U_i uniformly from [0,f_i(X)].
  2. Sample X' uniformly from the set \{x \in \mathcal{X} : f_i(x) \ge u_i \text{ for all } i\}.

Since the auxiliary variables sampler is based on a Gibbs sampler, we know that the joint distribution of (X,U) will converge to the uniform distribution on \mathcal{X}. So when we discard U, the distribution of X will converge to the target distribution P!

Auxiliary variables in practice

To perform step 2 of the auxiliary variables sampler we have to be able to sample uniformly from the sets

\mathcal{X}_u = \{x \in \mathcal{X}:f_i(x) \ge u_i \text{ for all } i\}.

Depending on the nature of the set \mathcal{X} and the functions f_i, it might be difficult to do this. Fortunately, there are some notable examples where this step has been worked out. The very first example of auxiliary variables is the Swendsen-Wang algorithm for sampling from the Ising model [2]. In this model it is possible to sample uniformly from \mathcal{X}_u. Another setting where we can sample exactly is when \mathcal{X} is the real numbers \mathcal{R} and each f_i is an increasing function of x. This is explored in [3] where they apply auxiliary variables to sampling from Bayesian posteriors.

There is an alternative to sampling exactly from the uniform distribution on \mathcal{X}_u. Instead of sampling X' uniformly from \mathcal{X}_u, we can run a Markov chain from the old X that has the uniform distribution as a stationary distribution. This approach leads to another special case of auxiliary variables which is called “slice sampling” [4].

References

[1] Andersen HC, Diaconis P. Hit and run as a unifying device. Journal de la societe francaise de statistique & revue de statistique appliquee. 2007;148(4):5-28. http://www.numdam.org/item/JSFS_2007__148_4_5_0/
[2] Swendsen RH, Wang JS. Nonuniversal critical dynamics in Monte Carlo simulations. Physical review letters. 1987 Jan 12;58(2):86. https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.58.86
[3] Damlen P, Wakefield J, Walker S. Gibbs sampling for Bayesian non‐conjugate and hierarchical models by using auxiliary variables. Journal of the Royal Statistical Society: Series B (Statistical Methodology). 1999 Apr;61(2):331-44. https://rss.onlinelibrary.wiley.com/doi/abs/10.1111/1467-9868.00179
[4] Neal RM. Slice sampling. The annals of statistics. 2003 Jun;31(3):705-67. https://projecteuclid.org/journals/annals-of-statistics/volume-31/issue-3/Slice-sampling/10.1214/aos/1056562461.full

Doing Calvin’s homework

Growing up, my siblings and I would read a lot of Bill Watterson’s Calvin and Hobbes. I have fond memories of spending hours reading and re-reading our grandparents collection during the school holidays. The comic strip is witty, heartwarming and beautifully drawn.

Recently, I came across this strip online

Originally published Feb 5th, 1990 by Bill Watterson. All creative credit goes to Bill. You can read the rest of the story arc here: https://www.gocomics.com/calvinandhobbes/1990/02/05

Seeing Calvin take this test took me back to those school holidays. I remember grabbing paper and a pencil to work out the answer (I also remember getting teased by those previously mentioned siblings). Reading the strip years later, I realized there’s a neat way to work out the answer without having to write much down. Here’s the question again,

Jack and Joe leave their homes at the same time and drive toward each other. Jack drives at 60 mph, while Joe drives at 30 mph. They pass each other in 10 minutes.

How far apart were Jack and Joe when they started?

It can be hard to think about Jack and Joe traveling at different speeds. But we can simplify the problem by thinking from Joe’s perspective or frame of reference. From Joe’s frame of reference, he is staying perfectly still but Jack is zooming towards him at 30+60=90 mph. Jack reaches Joe in 10 minutes which is one sixth of an hour. This means that the initial distance between them must have been 90 \times 1/6 = 15 miles.

It’s not as creative as Calvin’s private eye approach but at least Susie and I got the same answer.

Originally published Feb 10th, 1990 by Bill Watterson. Again all creative credit goes to Bill. https://www.gocomics.com/calvinandhobbes/1990/02/10

Solving a system of equations vs inverting a matrix

I used to have trouble understanding why inverting an n \times n matrix required the same order of operations as solving an n \times n system of linear equations. Specifically, if A is an n\times n invertible matrix and b is a length n vector, then computing A^{-1} and solving the equation Ax = b can both be done in O(n^3) floating point operations (flops). This surprised me because naively computing the columns of A^{-1} requires solving the n systems of equations

Ax = e_1, Ax=e_2,\ldots, Ax = e_n,

where e_1,e_2,\ldots, e_n are the standard basis vectors. I thought this would mean that calculating A^{-1} would require roughly n times as many flops as solving a single system of equations. It was only in a convex optimization lecture that I realized what was going on.

To solve a single system of equations Ax = b, we first compute a factorization of A such as the LU factorization. Computing this factorization takes O(n^3) flops but once we have it, we can use it solve any new system with O(n^2) flops. Now to solve Ax=e_1,Ax=e_2,\ldots,Ax=e_n, we can simply factor A once and the perform n solves using the factorization each of which requires an addition O(n^2) flops. The total flop count is then O(n^3)+nO(n^2)=O(n^3). Inverting the matrix requires the same order of flops as a single solve!

Of course, as John D Cook points out: you shouldn’t ever invert a matrix. Even if inverting and solving take the same order of flops, inverting is still more computational expensive and requires more memory.

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