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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 $E(S_N)$ $$E[E(S_N|N=n)]=E(pN)=pE(N)=p\mu$$

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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 $$ \mathbb{E}[S_N]=\mathbb{E}\left[\frac{N}{2}\right]=\frac{\mu}{2}\,. $$

Note: this is a special case of compound Poisson distribution.

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  • $\begingroup$ We can simply use Wald's identity to compute $$ \mathbb E[S_N] = \mathbb E[N]\mathbb E[X_1] = \frac\mu2. $$ $\endgroup$ – Math1000 Nov 24 '19 at 0:21
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    $\begingroup$ Yes. But I didn't want to assume knowledge of Wald's identity here. It's useful to derive the result at least once by oneself... $\endgroup$ – Clement C. Nov 24 '19 at 0:22
  • $\begingroup$ I suppose the law of total expectation is more elementary. $\endgroup$ – Math1000 Nov 24 '19 at 0:23
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    $\begingroup$ Well, I'd say so. Without it, conditional expectations become quite unmanageable... $\endgroup$ – Clement C. Nov 24 '19 at 0:24
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    $\begingroup$ @ElliottdeLaunay $X_n$ is just a r.v. uniform on $(0,1)$. Its expectation is $1/2$. $\endgroup$ – Clement C. Nov 24 '19 at 1:21

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