## Introduction

I've seen some really interesting probability & numerical solutions using a strategy called Poissonization, but Googling for it revealed very few resources (just some references in some textbooks that I don't have quick access to). Below are my notes and repository for PoissonizationAfter we introduce the theory, we'll do some examples.

The technique relies on the following theorem:

Theorem: Let $$N \sim \text{Poi}(\lambda)$$ and suppose $$N=n, (X_1, X_2, ... X_k) \sim \text{Multi}(n, p_1, p_2, ..., p_k)$$. Then, marginally, $$X_1, X_2, ..., X_k$$ are are independent Poisson, with $$X_i \sim \text{Poi}(p_i \lambda)$$. [1]

The proof of this theorem is as follows. First, by the law of total probability:

\begin{align*} P(X_1=x_1, ..., X_k = x_k\;|\; n, \mathbf{p}) & = \sum_{n=0}^{\infty} P(X_1=x_1, ..., X_k = x_k \;|\; N=n) \frac{e^{-\lambda}\lambda^n}{n!} \\ & = \sum_{n=0}^{\infty} \frac{(x_1 + ... + x_k)!}{x_1!x_2! \cdots x_k!}p_1^{x_1}\cdots p_k^{x_k}\frac{e^{-\lambda}\lambda^n}{n!}I_{x_1 + ... + x_k = n} \\ & = e^{-\lambda}\lambda^{x_1} \cdots \lambda^{x_k}p_1^{x_1} \cdots p_k^{x_k}\frac{1}{x_1!x_2! \cdots x_k!}\\ & = e^{-\lambda}(\lambda p_1)^{x_1} \cdots (\lambda p_k)^{x_k}\frac{1}{x_1!x_2! \cdots x_k!}\\ & = \prod_{i=1}^k \frac{e^{-\lambda p_i}(\lambda p_i)^{x_i}}{x_i!}\;\;\;\;\blacksquare \end{align*}

Let's see how we can use this to solve probability problems. Let $$(Y_1, Y_2, ...Y_k)$$ be $$\text{Multi}(n, p_1, p_2, ..., p_k)$$, and we are interested in when the realized values fall into some set $$A$$. Let $$X_i$$ be defined as above, that is, $$X_i \sim \text{Poi}(p_i \lambda)$$. Then, from above, we can write:

\begin{align*} P((X_1, ... X_k) \in A) & = \sum_{i=0}^\infty \frac{e^{-\lambda}\lambda^n}{n!} P((Y_1, ... Y_k) \in A)\\ & = e^{-\lambda} \sum_{i=0}^\infty \frac{\lambda^n}{n!} P((Y_1, ... Y_k) \in A)\\ & e^{\lambda}P((X_1, ... X_k) \in A) = \sum_{i=0}^\infty \frac{\lambda^n}{n!} P((Y_1, ... Y_k) \in A) \;\;\;(★) \end{align*}

Both sides of the last line are infinite series of $$\lambda^n$$, and thus their coefficients must be equal. Thus, as $$N=n$$, $$P((Y_1, ... Y_k) \in A)$$ is equal to  $$n! c(n)$$ where $$c(n)$$ is the coefficient of $$\lambda^n$$ in the expansion of $$e^{\lambda}P((X_1, ... X_k) \in A)$$.

## Example 1

Let's try an example:

n defects are randomly distributed amongst k integrated-circuit chips produced by a factory (any number of defects may be found on a chip and each defect is independent of the other defects). If n = 4, k=7, find the probability that there is a chip with at least 3 defects.

This is a classic case where it's easier to compute the complementary probability: what is the probability that all chips have less than 3 defects (and then we subtract this from 1 to get the original probability).

Since each defect is randomly distributed, $$p_i = p = \frac{1}{7}$$. Below, we invoke Poissonization by letting $$n$$ be randomly distributed as $$\text{Poi}(\lambda)$$ (later we will condition on its actual value of 4). That means that we can assume the number of defects on a chip is a Poisson random variable with rate $$\lambda p = \frac{\lambda}{7}$$.

\begin{align*} P(\text{All chips have \le 2 defects}) & = \Big(P(\text{Chip has 0 defect}) + P(\text{Chip has 1 defect}) + P(\text{Chip has 2 defect})\Big)^7 \\ & = \Big(e^{-\lambda/7} + \frac{e^{-\lambda/7}(\lambda/7)^1}{1!} + \frac{e^{-\lambda/7}(\lambda/7)^2}{2!}\Big)^7\\ & = e^{-\lambda}\Big(1 + \frac{\lambda}{7} + \frac{\lambda^2}{2!7^2}\Big)^7\\ \end{align*}

When we multiply the above by $$e^{\lambda}$$, and imagine the right-hand side is expanded, we get a form that is familiar to equation (). This means that we are interested in the coefficient of $$\lambda^4$$. We could expand this by hand, but instead let's expand this computationally. For this, it's useful to know a bit about multinomial expansions. But basically you are looking for solutions to

$$k_0 + k_1 + k_2 = 7\\ 0 k_0 + 1 k_2 + 2 k_2 = 4$$

The first equation comes from the constraint for the powers of a multinomial expansion to sum to the outside exponent. The second equation comes from our need to find powers of $$\lambda$$ that sum up to 4 (the 0,1,2 come from the existing powers of $$\lambda$$). Two equations, three unknowns, solving for the values:

\begin{align*} k_2 &= k_0 - 3 \\ k_1 &= 10 - 2k_0\\ 3 &\le k_0 \le 5 \end{align*}

Computationally, we can write this in Python:

We get the answer 0.92711, so the complement to this is 0.07289 and is the probability of the original question.

## Example 2

This comes from FiveThiryEight's Riddler series: Santa Needs Some Help With Math. I'll quote the problem here:

In Santa’s workshop, elves make toys during a shift each day. On the overhead radio, Christmas music plays, with a program randomly selecting songs from a large playlist.

During any given shift, the elves hear 100 songs. A cranky elf named Cranky has taken to throwing snowballs at everyone if he hears the same song twice. This has happened during about half of the shifts. One day, a mathematically inclined elf named Mathy tires of Cranky’s sodden outbursts. So Mathy decides to use what he knows to figure out how large Santa’s playlist actually is.

Help Mathy out: How large is Santa’s playlist?

Okay, so let's translate this to our set up. We have $$k$$ songs to choose from, and we sample 100 of them. Turns out that P(any song is played more than once) = 0.5. So in this case, we know the probability, but we don't know the $$k$$. So, similarly to how we started above:

\begin{align*} P(\text{songs have 1 or less plays}) & = \Big(P(\text{songs has 0 plays}) + P(\text{songs has 1 plays})\Big)^k \\ & = \Big(e^{-\lambda/k} + \frac{e^{-\lambda/k}(\lambda/k)^1}{1!}\Big)^k\\ & = e^{-\lambda}\Big(1 + \frac{\lambda}{k}\Big)^k\\ \end{align*}

Yay we have a binomial! They are much easier to work with. Now we need to answer:

What is the coefficient of $$λ^{100}$$?

Call this coefficient $$c_{100}(k)$$. Since we have a binomial, expanding the above is easy:

$$c_{100}(k) = \frac{C(k, 100)}{k^{100}}$$

Using the other piece of information above, that the probability of an "event" is 0.5, we can write an equation down:

\begin{align*} 0.5 & = 100! c_{100}(k) \\ & = 100! \frac{C(k, 100)}{k^{100}} \\ & = \frac{k!}{(k-100)! k^{100}} \\ \end{align*}

$$2\frac{k!}{(k-100)!} - k^{100} = 0$$

We can use Python (!!) to numerically solve this equation for $$k$$.

Because the sign flips at 7175, we conclude that either 7174 or 7175 is the closest answer.

Looking for more examples?

[1] DasGupta, A. Probability for Statistics and Machine Learning. http://link.springer.com/book/10.1007%2F978-1-4419-9634-3