How to find the expected value of the random sum of variables?

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$X$ is a discrete random variable with the following density function: $$ f_X(n) = \begin{cases} c e^{-2} \, \frac{2^n}{n!} & n \geq 0, \\ c 3^n & n < 0. \end{cases} $$ Define the variables $V$ as follows: $$ V = \begin{cases} 0 & X < 0 \\ X & X \geq 0. \end{cases} $$ Let $V_1, V_2, V_3, \dotsc$ be a sequence of independent and identically distributed random variables each having the same distribution as $V$.

Let $N$ be a poisson random variable with mean $5$. Assume $N$ is independent of the $V_i$’s. Define $Z = \sum_{i=1}^N V_i$. What is $E(Z)$?

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$C$ is a constant

Not a homework. The whole problem (but not very relevant) could be found at https://math.stackexchange.com/questions/2885578/find-the-expected-value-of-the-sum-of-random-variables


$\mathbb E[V]=\sum_{n\geq 0}nce^{-2}2^n/n!=2c$

The $\mathbb E[N]$ is given as $5$.

So the answer is $\mathbb E[V]\mathbb E[N]=10c$?

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Observe that: $$\mathbb E[Z\mid N=n]=\mathbb E\sum_{i=1}^nV_i=\sum_{i=1}^n\mathbb EV_i=n\mathbb EV$$Here the first equality is a consequence of independence of $N$ and the $V_i$.

Based on this we find $\mathbb E[Z\mid N]=N\mathbb EV$ and consequently:

$$\mathbb EZ=\mathbb E[\mathbb E[Z\mid N]]=\mathbb E[N\mathbb EV]=\mathbb EV\mathbb EN$$

I leave the rest (i.e. finding $\mathbb EV$ and $\mathbb EN$) to you.