A kind of Poisson random variable?

Question

It is said that many are called but few are chosen. Suppose that the number called is $Poisson ( \lambda )$. Each person is called is chosen, independently, with probability $p$. Show that the distribution of number chosen is $Poisson ( p \lambda )$.

I need help in understanding the question. I tried relating it to Poisson Random Variable, but I'm not able to get it. I don't understand what $Poisson ( \lambda )$ is. I guess it is a function since it can't be a fixed number. But what is the domain of the function?

Edit 1: I don't want to know how to solve the question. I just need to understand it, so I can try it on my own.

Edit 2: I think I kind of understood the question. Here is my attempted solution:

Let $X \sim Poisson(\lambda)$. and $Y$: number of people chosen

Let $ q = 1-p$

$P[Y=m] = \sum_{k=0}^\infty P[X=k] \ p^m q^{k-m}$

$P[Y=m] = (\frac{p}{q})^m \sum_{k=0}^\infty \frac{(\lambda q)^k e^{-\lambda}}{k!}$

How do I proceed from here? I tried simplifying this but did not get anything useful.

Answer 1

Let's go through this methodically. Define the random number of people who are called as $X$, where $$X \sim \operatorname{Poisson}(\lambda), \quad \Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x = 0, 1, 2, \ldots.$$ Now, given that $X$ people are called, the random number actually chosen $Y$ is a binomial random variable with $n = X$ and probability of success $p$: $$Y \mid X \sim \operatorname{Binomial}(n = X, p), \quad \Pr[Y = y \mid X] = \binom{X}{y} p^y (1-p)^{X - y}, \quad y = 0, 1, \ldots, X.$$ Note this is conditionally binomial. So, what is the unconditional distribution of the random number chosen; i.e., what is $\Pr[Y = y]$? To find this, we simply recall the law of total probability and compute $$\begin{align*} \Pr[Y = y] &= \sum_{x=0}^\infty \Pr[Y = y \mid X = x]\Pr[X = x] \\ &= \sum_{x=y}^\infty \binom{x}{y} p^y (1-p)^{x-y} e^{-\lambda} \frac{\lambda^x}{x!} \\ &= e^{-\lambda} p^y \lambda^y \sum_{x=y}^\infty \frac{x!}{y!(x-y)!} \frac{((1-p)\lambda)^{x-y}}{x!} \\ &= e^{(1-p)\lambda} e^{-\lambda} \frac{(p\lambda)^y}{y!} \sum_{x=y}^\infty e^{-(1-p)\lambda} \frac{((1-p)\lambda)^{x-y}}{(x-y)!} \\ &= e^{-p\lambda} \frac{(p\lambda)^y}{y!} \sum_{m=0}^\infty e^{-(1-p)\lambda} \frac{((1-p)\lambda)^m}{m!}, \quad m = x-y \\ &= e^{-p\lambda} \frac{(p\lambda)^y}{y!}.\end{align*}$$ Note that in the final step we observed that the sum corresponds to the sum of a Poisson random variable with parameter $(1-p)\lambda$ over its support, and therefore equals $1$.

We could have done this more intuitively but I think it is instructive to look at the more computational approach.

Answer 2

Hint:

There are two parts to selection, first you must be called and then, if you are called, you must be selected.

If $n$ people are selected, that means that $k \ge n$ people were called and out of those $n$ of those were selected.

Informally, compute $\sum_{k=n}^\infty p [ k \ge n \text{ people called and } n \text{ of these selected}]$.

Since the choosing and selection are independent, this is tantamount to computing $\sum_{k=n}^\infty p [ k \ge n \text{ people called}] \cdot p'[n \text{ of } k \text{ chosen}]$.

Now substitute the relevant probabilities and simplify. Remember to expand $\binom{k}{n} = { k! \over n! (k-n)!}$.