Tag: math

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.

Why is the fundamental theorem of arithmetic a theorem?

The fundamental theorem of arithmetic states that every natural number can be factorized uniquely as a product of prime numbers. The word “uniquely” here means unique up to rearranging. The theorem means that if you and I take the same number n and I write n = p_1p_2\ldots p_k and you write n = q_1q_2\ldots q_l where each p_i and q_i is a prime number, then in fact k=l and we wrote the same prime numbers (but maybe in a different order).

Most people happily accept this theorem as self evident and believe it without proof. Indeed some people take it to be so self evident they feel it doesn’t really deserve the name “theorem” – hence the title of this blog post. In this post I want to highlight two situations where an analogous theorem fails.

Situation One: The Even Numbers

Imagine a world where everything comes in twos. In this world nobody knows of the number one or indeed any odd number. Their counting numbers are the even numbers \mathbb{E} = \{2,4,6,8,\ldots\}. People in this world can add numbers and multiply numbers just like we can. They can even talk about divisibility, for example 2 divides 8 since 8 = 4\cdot 2. Note that things are already getting a bit strange in this world. Since there is no number one, numbers in this world do not divide themselves.

Once people can talk about divisibility, they can talk about prime numbers. A number is prime in this world if it is not divisible by any other number. For example 2 is prime but as we saw 8 is not prime. Surprisingly the number 6 is also prime in this world. This is because there are no two even numbers that multiply together to make 6.

If a number is not prime in this world, we can reduce it to a product of primes. This is because if n is not prime, then there are two number a and b such that n = ab. Since a and b are both smaller than n, we can apply the same argument and recursively write n as a product of primes.

Now we can ask whether or not the fundamental theorem of arthimetic holds in this world. Namely we want to know if their is a unique way to factorize each number in this world. To get an idea we can start with some small even numbers.

  • 2 is prime.
  • 4 = 2 \cdot 2 can be factorized uniquely.
  • 6 is prime.
  • 8 = 2\cdot 2 \cdot 2 can be factorized uniquely.
  • 10 is prime.
  • 12 = 2 \cdot 6 can be factorized uniquely.
  • 14 is prime.
  • 16 = 2\cdot 2 \cdot 2 \cdot 2 can be factorized uniquely.
  • 18 is prime.
  • 20 = 2 \cdot 10 can be factorized uniquely.

Thus it seems as though there might be some hope for this theorem. It at least holds for the first handful of numbers. Unfortunately we eventually get to 36 and we have:

36 = 2 \cdot 18 and 36 = 6 \cdot 6.

Thus there are two distinct ways of writing 36 as a product of primes in this world and thus the fundamental theorem of arithmetic does not hold.

Situtation Two: A Number Ring

While the first example is fun and interesting, it is somewhat artificial. We are unlikely to encounter a situation where we only have the even numbers. It is however common and natural for mathematicians to be lead into certain worlds called number rings. We will see one example here and see what an effect the fundamental theorem of arithmetic can have.

Consider wanting to solve the equation x^2+19=y^3 where x and y are both integers. One way to try to solve this is by rewriting the equation as (x+\sqrt{-19})(x-\sqrt{-19}) = y^3. With this rewriting we have left the familiar world of the whole numbers and entered the number ring \mathbb{Z}[\sqrt{-19}].

In \mathbb{Z}[\sqrt{-19}] all numbers have the form a + b \sqrt{-19}, where a and b are integers. Addition of two such numbers is defined like so

(a+b\sqrt{-19}) + (c + d \sqrt{-19}) = (a+c) + (b+d)\sqrt{-19}.

Multiplication is define by using the distributive law and the fact that \sqrt{-19}^2 = -19. Thus

(a+b\sqrt{-19})(c+d\sqrt{-19}) = (ac-19bd) + (ad+bc)\sqrt{-19}.

Since we have multiplication we can talk about when a number in \mathbb{Z}[\sqrt{-19}] divides another and hence define primes in \mathbb{Z}[\sqrt{-19}]. One can show that if x^2 + 19 = y^3, then x+\sqrt{-19} and x-\sqrt{-19} are coprime in \mathbb{Z}[\sqrt{-19}] (see the references at the end of this post).

This means that there are no primes in \mathbb{Z}[\sqrt{-19}] that divides both x+\sqrt{-19} and x-\sqrt{-19}. If we assume that the fundamental theorem of arthimetic holds in \mathbb{Z}[\sqrt{-19}], then this implies that x+\sqrt{-19} must itself be a cube. This is because (x+\sqrt{-19})(x-\sqrt{-19})=y^3 is a cube and if two coprime numbers multiply to be a cube, then both of those coprime numbers must be cubes.

Thus we can conclude that there are integers a and b such that x+\sqrt{-19} = (a+b\sqrt{-19})^3 . If we expand out this cube we can conclude that

x+\sqrt{-19} = (a^3-57ab^2)+(3a^2b-19b^3)\sqrt{-19}.

Thus in particular we have 1=3a^2b-19b^3=(3a^2-19b^2)b. This implies that b = \pm 1 and 3a^2-19b^2=\pm 1. Hence b^2=1 and 3a^2-19 = \pm 1. Now if 3a^2 -19 =-1, then a^2=6 – a contradiction. Similarly if 3a^2-19=1, then 3a^2=20 – another contradiction. Thus we can conclude there are no integer solutions to the equation x^2+19=y^3!

Unfortunately however, a bit of searching reveals that 18^2+19=343=7^3. Thus simply assuming that that the ring \mathbb{Z}[\sqrt{-19}] has unique factorization led us to incorrectly conclude that an equation had no solutions. The question of unique factorization in number rings such as \mathbb{Z}[\sqrt{-19}] is a subtle and important one. Some of the flawed proofs of Fermat’s Last Theorem incorrectly assume that certain number rings have unique factorization – like we did above.

References

The lecturer David Smyth showed us that the even integers do not have unique factorization during a lecture of the great course MATH2222.

The example of \mathbb{Z}[\sqrt{-19}] failing to have unique factorization and the consequences of this was shown in a lecture for a course on algebraic number theory by James Borger. In this class we followed the (freely available) textbook “Number Rings” by P. Stevenhagen. Problem 1.4 on page 8 is the example I used in this post. By viewing the textbook you can see a complete solution to the problem.