Moments of a random sum with bounds Poisson distributed?

Question

We have that $N$ and ${X_1,X_2,\dots}$ are all independent and that $f(x)=Cx^2(1-x)^2$.

Then, we have:

$$Z=\sum_{j=1}^{N+1}X_j$$

$N$~Poisson$\lambda$. Find the expectation and the variance of $Z$.

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The first thing I notice is that my function is actually ~beta$(3,3)$, meaning that the expectation is $1/2$ and the variance is $1/28$.

Okay, then I let $M=N+1$. Since $N$ ~Poisson$(\lambda)$, I have that $E[N]=\lambda$ and also that Var$[N]=\lambda$.

This implies that $E[M]=\lambda+1$ and also that Var$[M]=\lambda$.

But how do I put these pieces together to obtain $E[Z]$ and Var$[Z]$?

Answer 1

Edit: Maybe I should flesh out my answer a bit.

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Recall previous discussion on the matter.

\begin{align*} E[Z] &= \sum_{n=0}^\infty E[Z|N = n]P(N=n)\\ &=\sum_{n=0}^\infty E[X_1+\dotsb+X_{n+1}|N=n]P(N=n)\\ &=\sum_{n=0}^\infty (n+1)E[X_1|N_n]P(N=n)\\ &=\frac{1}{2}\left[\sum_{n=0}^\infty nP(N = n)+\sum_{n=0}^\infty P(N=n)\right]\\ &=\frac{1}{2}[E[N]+1]\\ &=\frac{1}{2}(1+\lambda) \end{align*}

Now, consider Michael's answer and previous answers regarding the variance and attempt to compute it for yourself.