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Random veriable $K$ has a uniform distribution on the interval $(1,5)$. Conditional probability distribution of $S_{N}=X_{1}+\dots+X_{N}$ given K has a compunded Poisson distribution. $N\approx Poiss(1)$ and $P(X>x|K=k)=e^{-kx},x>0$. Count $E(S_{N})$.

Well, I dont know how to solve it. I thought about sth like:

$E(S_{N})=EX\cdot EN=\dots$

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Since $(X_i \mid K=k)$ follows an exponential distribution with parameter $k$, we have $E[X_i \mid K=k] = \frac{1}{k}$. Then, $$E[X_i] = E[E[X_i \mid K]] = \frac{1}{4} \int_1^5 \frac{1}{k} \mathop{dk} = \frac{1}{4} \log 5.$$

Then, using your first step combined with $E[N]=1$, we have $$E[S_N] = E[E[S_N \mid N]]=E[X_1] E[N] = E[X_1] = \frac{1}{4} \log 5.$$

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  • $\begingroup$ Can you explain this step: $E[E[X_i \mid K]] = \frac{1}{4} \int_1^4 \frac{1}{k}$ $\endgroup$ Commented May 23, 2016 at 19:20
  • $\begingroup$ @WalterWhite $E[E[X_i \mid K]] = E[1/K] = \int_1^5 \frac{1}{k} p_K(k) \mathop{dk}$ where $p_K(k)=1/4$ is the pdf of the uniform distribution on $(1,5)$. $\endgroup$ Commented May 23, 2016 at 20:23
  • $\begingroup$ Thank you. It is clear now. $\endgroup$ Commented May 23, 2016 at 20:28
  • $\begingroup$ Since $E[S_{N}]=E[X]E[N]$ it true that $E[S_{N}|K]=E[X|K]E[N|K]$. If it is true. Could you tell me why? I was wondering how to solve it in another way. This is why I am asking this question. $\endgroup$ Commented May 31, 2016 at 12:31
  • $\begingroup$ @WalterWhite $$E[S_N \mid K] = E[E[S_N \mid N,K] \mid K] = E[N \cdot E[X_1 \mid N,K] \mid K] = E[X \mid K] E[N \mid K].$$ $\endgroup$ Commented May 31, 2016 at 16:27

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