Covariance of a poisson distribution and a condition sum of variables 
Let ${N(t),t ≥ 0}$ be a Poisson process with rate $λ$ that is
  independent of the sequence $X_1,X_2,\ldots$ of independent and identically
  distributed random variables with mean $μ$ and variance $σ^2$. Find
  $$
\operatorname{Cov} \left(N(t),\sum_{i=1}^{N(t)} X_i \right)
$$

I think the way to go is using the Covariance formula 
\begin{align}
\operatorname{Cov}(X,Y) = \operatorname{E}[XY] - \operatorname{E}(X)\operatorname{E}(Y)
\end{align}
I can find the second term relatively easily
\begin{align}
\operatorname{E}[N(t)]  & = \lambda t \\[10pt]
 \operatorname{E} \left[ \sum_{i=1}^{N(t)}(X_i)\right] & = \operatorname{E} \left[\operatorname{E}\sum_{i=1}^{N(t)}(X_i)\mid N(t) = N\right] \\[10pt]
 & = \operatorname{E}[N(t) \mu] \\[10pt]
 & = \lambda t \mu 
\end{align}
 Hence the second term gives 
\begin{align}
\operatorname{E}(X)\operatorname{E}(Y) =  (\lambda t)^2 \mu
\end{align}
However, for the first term, I am not sure how to find the expectation of the product of the two distributions.  Any advice on the route would be much appreciated.
 A: The term you are missing is calculated the same way you did with the second one:
$$E\left[ N(t)\sum_{i=1}^{N(t)}(X_i) \right]  = E \left [E\left( N(t) \sum_{i=1}^{N(t)} X_i  \mid N(t)\right)\right]$$
$$= E\left[N(t)E\left(\sum_{i=1}^{N(t)} X_i  \mid N(t)\right)\right] = E[N^2(t)\mu] = \mu[\lambda t+(\lambda t)^2].$$
A: You could make use of the tower property as follows (similar to your own calculations)
$$
E \left[ N(t) \sum_{i=1}^{N(t)} X_i \right]
= E \left\{ E \left[ N(t) \sum_{i=1}^{N(t)} X_i \mid N(t) \right] \right\}
= E[(N(t))^2 EX_1]
= (VN(t) + (EN(t))^2) \cdot EX_1.
$$
Using this and your result, we have 
$$
\text{Cov}(N(t), \sum_{i=1}^{N(t)} X_i)
= EX \cdot VN(t).
$$
This is a general solution, where $N(t)$ need not be Poisson distributed.
A: $\newcommand{\c}{\operatorname{cov}}\newcommand{\E}{\operatorname{E}}$One can use the identity $$\c(X,Y) = \c(\E(X\mid Z),\E(Y\mid Z)) + \E(\c(X,Y\mid Z)).$$
Thus we have:
\begin{align}
\c\left( N ,\sum_{i=1}^N X_i \right) & = \c\left( \E(N\mid N), \E\left( \sum_{i=1}^N X_i \mid N \right) \right) + \E\left( \c\left( N,\sum_{i=1}^N X_i \mid N \right) \right) \\[10pt]
& = \c\left( N,N\mu \right) + 0 \\
& \qquad\qquad \text{(assuming the sequence $X_1,X_2,\ldots$ is independent of $N$)} \\[10pt]
& = \mu\c(N,N) = \mu\lambda t.
\end{align}
