Why is $\mathbb{P}(\tilde{X}=n)=\sum_{m\geq 0}\mathbb{P}(\tilde{X}=n \;\; \text{and}\;\; X-\tilde{X}=m)$?

Question

I try to understand the following proof: Let $X \sim P(\lambda)$ and let $(Y_n)$ be a family of independent, identically distributed random Bernoulli variables with parameter $p\in (0,1)$. We define further:$$\tilde{X}= \sum \limits_{n=0}^X Y_n$$ We wanna proof that $\tilde{X}$ is Bernoulli distributed with parameter $p\lambda$. So let $n,m \geq 0$, than for $m\geq 0$ we can sum up and get: $$\mathbb{P}(\tilde{X}=n)=\sum_{m\geq 0}\mathbb{P}(\tilde{X}=n \;\; \text{and}\;\; X-\tilde{X}=m)$$ Now I don't understand why we get that resultat. Can someone helps me? Thanks!

Answer 1

This result is an application of the law of total probability:

If events $B_0$, $B_1$, $B_2, \ldots,$ are mutually exclusive and exhaustive, then $$P(A) = \sum_{m\ge 0}P(A\cap B_m).$$

So if you have events $B_0$, $B_1$, $B_2, \ldots,$ that partition the sample space (i.e., the $B$'s are non-overlapping and exactly one of them must occur), then you can compute $P(A)$ by first carving up the event $A$ according to which one of the $B$'s occurred.

In your example, $A=\{\tilde X=n\}$ and $B_m = \{ X-\tilde X=m\}$. Notice that $X-\tilde X$ is non-negative, so the difference $X-\tilde X$ takes values $0, 1, 2,\ldots$. This is the reason why the summation ranges over $m\ge 0$.