Insurance claims Poisson problem derving expected value and variance If I have that claims arrive at an insurance company according to a Poisson process $\{N(t) : t \ge 0\}$ at a rate $\lambda > 0$ and $X_i$ denotes the claim size of the $ith$ claim.  I assume that $\{X_i, i=1,2,\ldots\}$ is independent and identically distributed sequence of positive values.  I also have that $\{X_i, i=1,2,\ldots\}$ is independent of $\{N(t) : t \ge 0\}$ and $S(t)$ is the aggregate loss or total amount of claims to the insurance company.  
If I know that $\mu=E[X_1]$ and $\sigma^2=\operatorname{Var}(X_1)$, I'm supposed to derive the formulas:
$$E[S(t)]=\lambda \mu t\text{ and }\operatorname{Var}(S(t))=\lambda (\sigma^2 + \mu^2)t$$
I'm confused as to how to go about this but this looks like a compound poisson problem.  My thinking is that I need to use conditional expectations to derive these formulas.  This is what I have tried:
$$E[S(t)]=E(E[S(t)\mid N(t)])=E[S(t)]E[N(t)]=E[S(t)] \times \mu$$ and
$$\operatorname{Var}(S(t))=\operatorname{E}[\operatorname{Var}(S(t)\mid N(t)] + \operatorname{Var}(E(S(t)\mid N(t)]$$
I'm stuck from here.  Any suggestions as to how to do this?
 A: You're intuition's dead on with using Conditional variance. 
What you're looking for is right here:
http://www.columbia.edu/~ww2040/3106F14/lec1023.pdf
A: If $$S(t) = X_1 + X_2 + \cdots + X_{N(t)}$$ is the aggregate claims and $N(t)$ is the claim count at time $t$, then $$\operatorname{E}[S(t)] = \operatorname{E}[\operatorname{E}[S(t) \mid N(t)]] = \operatorname{E}[N(t) \operatorname{E}[X_i]] = \operatorname{E}[X_i]\operatorname{E}[N(t)].$$  Since a single claim $X_i$ has expected size $\operatorname{E}[X_i] = \mu$, and the mean claim count at time $t$ is $\operatorname{E}[N(t)] = \lambda t$, the result immediately follows.  This is simply the law of total expectation.
To get the variance of the aggregate, we use the law of total variance:  $$\operatorname{Var}[S(t)] = \operatorname{Var}[\operatorname{E}[S(t) \mid N(t)]] + \operatorname{E}[\operatorname{Var}[S(t) \mid N(t)]].$$  The first expectation is what we found before:  $$\operatorname{E}[S(t) \mid N(t)] = N(t) \operatorname{E}[X_i] = N(t) \mu.$$  The conditional variance of the aggregate claim size given the number of claims up to time $t$ is $$\operatorname{Var}[S(t) \mid N(t)] = N(t) \operatorname{Var}[X_i] = N(t) \sigma^2,$$ because there are $N(t)$ iid claims $X_1, X_2, \ldots, X_{N(t)}$, so the variance of their sum is simply $N(t)$ times the variance of a single claim.  Then the total variance becomes $$\operatorname{Var}[S(t)] = \operatorname{Var}[\mu N(t)] + \operatorname{E}[\sigma^2 N(t)].$$  Since $N(t)$ is Poisson with mean equal to variance equal to $\lambda t$, we then get $$\operatorname{Var}[S(t)] = \mu^2 \operatorname{Var}[N(t)] + \sigma^2 \operatorname{E}[N(t)] = (\mu^2 + \sigma^2) \lambda t,$$ as claimed.
