A kind of Poisson random variable? 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.
 A: 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.
A: 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
$\binom{k}{n} = { k! \over n! (k-n)!}$.
