$Y_1,Y_2,...$ are independent random variables with a distribution identical to that of $Y$. $N(t)$ is a poisson process with parameter $\lambda$.
$$X(t)=\sum\limits_{n=1}^{N(t)}Y_n$$
Find the expected value $\mu_X(t)$ and the covariance $K_X(t,t')$. Is $X(t)$ stationary?
For the expected value I've found the following:
$\mu_X(t)=E[X(t)]=E[E[X(t)\mid N]]$
since $E[X(t)\mid N]=E[\sum\limits_{n=1}^N Y_n]=N\mu_Y$,
$E[E[X(t)\mid N]]=E[N\mu_Y]=t\lambda\mu_Y=\mu_X(t)$
But I'm stuck with the covariance $K_X(t,t')$. I tried this: $K_X(t,t')=E[X(t)X(t')]-\mu_X(t)\mu_X(t')$ and I suppose I would also have to find $E[X(t)X(t')]$ by computing $E[E[X(t)X(t')\mid N,N']]$.
Then $E[X(t)X(t')\mid N,N']=E\left[\sum\limits_{n=1}^N Y_n \sum\limits_{{n'}=1}^{N'} Y_{n'}\right].$ But I can't seem to work this out. Any tips on this?
If $t\leqslant s$, then $N(t)\leqslant N(s)$ hence $$X(t)X(s)=\sum_{n,k}Y_nY_k\mathbf 1_{n\leqslant N(t),k\leqslant N(s)}=\sum_{n}Y_n^2\mathbf 1_{n\leqslant N(t)}+\sum_{n\ne k}Y_nY_k\mathbf 1_{n\leqslant N(t),k\leqslant N(s)},$$ which implies that $E(X(t)X(s))=E(Y^2)U+E(Y)^2(V-U)$ with $$U=\sum_{n}P(n\leqslant N(t))=E(N(t)),$$ and $$ V=\sum_{n,k}P(n\leqslant N(t),k\leqslant N(s))=E(N(t)N(s)).$$ The value of $E(N(t))$ should be known. On the other hand, using $N(s)=N(t)+(N(s)-N(t))$ where $N(s)-N(t)$ is independent of $N(t)$ and distributed as $N(s-t)$, one gets $$V=E(N(t)^2)+E(N(t))E(N(s-t)),$$ from which the value of $V$, hence of $E(X(t)X(s))$, should arise...