# Conditional Expectation, given $N=n$

Expanding on this question: Conditional Probability Given N=n

Suppose that $$N$$ is a Poisson random variable with parameter $$\mu$$. Given $$N=n$$, random variables $$X_1, ... X_n$$ are independent with uniform (0,1) distribution. So there are a random number of $$X$$'s.

c) Let $$S_n=X_1+...+X_N$$ denote the sum of the random number of $$X$$'s (If $$N=0$$ then $$S_N=0$$). Find $$P(S_N=0$$).

I'm thinking: $$P(S_N=0)= P(S_N=0|N=0)\cdot P(N=0)$$ By condition $$(N=0) \to (S_N=0):$$ $$P(S_N=0|N=0)=1$$ so $$P(S_N=0)$$ is exponential distribution for $$0$$.

How can I carry this part forward to solve d) Find $$E(S_N)$$ $$E[E(S_N|N=n)]=E(pN)=pE(N)=p\mu$$

You have the law of total expectation: $$\mathbb{E}[S_N]=\mathbb{E}[\mathbb{E}[S_N|N]]$$ Now, $$\mathbb{E}[S_N|N] = \mathbb{E}[\sum_{n=1}^N X_n|N] = \sum_{n=1}^N \mathbb{E}[X_n] = \sum_{n=1}^N \frac{1}{2} = \frac{N}{2}$$ so $$\mathbb{E}[S_N]=\mathbb{E}\left[\frac{N}{2}\right]=\frac{\mu}{2}\,.$$
• We can simply use Wald's identity to compute $$\mathbb E[S_N] = \mathbb E[N]\mathbb E[X_1] = \frac\mu2.$$ – Math1000 Nov 24 '19 at 0:21
• @ElliottdeLaunay $X_n$ is just a r.v. uniform on $(0,1)$. Its expectation is $1/2$. – Clement C. Nov 24 '19 at 1:21