# Conditioning and a sum of Poisson random variables

Posted by Jason Polak on 13. February 2018 · Write a comment · Categories: statistics · Tags:

Previously we talked about the Poisson distribution. The Poisson distribution with mean $\mu \gt 0$ is a distribution on the natural numbers whose density function is
$$f(n) = \frac{e^{-\mu}\mu^n}{n!}$$
We have already seen that the Poisson distribution essentially arises from the binomial distribution as a sort of "limiting case". In fact, the Poisson distribution is sometimes used as an approximation to the binomial distribution, even though it also arises in its own right in processes like observing the arrival of random particles from radioactive decay.

The Poisson distribution and the binomial distribution are related in another way, through conditioning on the sum of Poisson random variables.

Suppose we have two independent Poisson random variables $X_1$ and $X_2$ with means $E(X_i) = \mu_i$. Then the sum $X_1 + X_2$ also has a Poisson distribution with mean $\mu_1 + \mu_2$.

On the other hand, what is the conditional density $P(X_1 = n_1, X_2 = n_2~|~ X_1 + X_2 = n)$? Here, $n_1 + n_2 = n$. By definition, it is
$$\frac{P(X_1 = n_1, X_2 = n- n_1)}{P(X_1 + X_2 = n)}$$
This is
$$\binom{n}{n_1}p^{n_1}(1-p)^{n-n_1}$$
where $p = \mu_1/(\mu_1+\mu_2)$. So, the joint density of two Poisson random variables conditioning on their sum being $n$ is binomial!