Poisson process, sum of iid (not Poisson) random variables We are given a Poisson process $\{N_t\}_{t \ge 0}$ with intensity $\lambda$ and a sequence of iid random variables $\{Y_n\}_{n \in \mathbb{N}}$. Assume $\{Y_n\}_{n \in \mathbb{N}}$ and $\{N_t\}_{t \ge 0}$ are independent.
Could you help me find mean and covariance of $\{Z_t\}_{t \ge 0}$ with $Z_t= \sum_{k=1}^{N_t}Y_k$ (we assume we are given $\mathbb{E}Y_1$ and  $\mathbb{D}^2Y_1$).
I am not sure what to use here.
When it comes to expectation, I have that $$\mathbb{E}Z_t = \mathbb{E}\sum_{k=1}^{N_t}Y_k = \mathbb{E}N_t \mathbb{E}Y_1 \to \text{corrected}$$ 
For covariance I have $$cov(Z_t, Z_s) = cov(\sum_{k=1}^{N_t}Y_k, \sum_{n=1}^{N_s}Y_n) = \sum_{k=1}^{N_t}\sum_{n=1}^{N_s}(Y_k, Y_n)$$
What can I do now?
 A: 
$\text{cov}(Z_t,Z_s) = [...] = \sum_{k=1}^{N_t} \sum_{n=1}^{N_s} (Y_k,Y_n)$

Note that the covariance $\text{cov}(Z_t,Z_s)$ is a ("deterministic") real number. Since both $N_t$ and $N_s$ are random variables, this means that the identity doesn't make sense.
Since $(N_t)_{t \geq 0}$ is a Poisson process with intensity $\lambda$, we know that $N_t(\omega) \in \mathbb{N}_0$ for all $\omega$ and
$$\mathbb{P}(N_t = n) = e^{-\lambda t} \frac{(\lambda t)^n}{n!} \qquad \text{for all $n \geq 0$.} \tag{1}$$
Since
$$\sum_{k=1}^{N_t} Y_k = \sum_{n=1}^{\infty} \left( 1_{\{N_t = n\}} \sum_{k=1}^n Y_k \right)$$
we get
$$\mathbb{E} \left( \sum_{k=1}^{N_t} Y_k \right) = \sum_{n=1}^{\infty} \mathbb{E} \left( 1_{\{N_t = n\}} \sum_{k=1}^n Y_k \right).$$
Using the independence of $(N_t)_{t \geq 0}$ and $(Y_n)_{n \in \mathbb{N}}$, we obtain
$$\begin{align*} \mathbb{E} \left( \sum_{k=1}^{N_t} Y_k \right)=\sum_{n=1}^{\infty} \mathbb{P} \left( \sum_{k=1}^n Y_k \right)  &=\mathbb{E}(Y_1) \sum_{n=1}^{\infty} n \mathbb{P}(N_t = n) \\ &\stackrel{(1)}{=} \mathbb{E}(Y_1)e^{-\lambda t}  \sum_{n=1}^{\infty} \frac{(\lambda t)^n}{(n-1)!}. \end{align*}$$
Finally, as
$$ \sum_{n=1}^{\infty} \frac{(\lambda t)^n}{(n-1)!} = \lambda t \sum_{n=1}^{\infty} \frac{(\lambda t)^{n-1}}{(n-1)!} = \lambda t \sum_{n \geq 0} \frac{(\lambda t)^n}{n!} = \lambda t e^{\lambda t}$$
we conclude
$$\mathbb{E} \left( \sum_{k=1}^{N_t} Y_k \right) = \lambda t \mathbb{E}(Y_1).$$
Since the calculation of the covariance is very similar, I leave it to you.
Remark: The process $$X_t := \sum_{k=1}^{N_t} Z_k$$ is called compound Poisson process.
