Let $X_1, X_2, ..., X_N$~$_{iid}Exp(\beta)$ and N~$Poiss(\lambda)$ is independent from $X_1,...,X_N$.
Define $M_N=min\{X_1,...,X_N\}$. Find $Cov(N,M_N)$.
My approach:
$Cov(N,M_N)=E(N*min(X_1,...X_N))-EN*E(min(X_1,...,X_N))$.
We know that for $X_1, X_2, ..., X_n$~$_{iid}Exp(\beta)$ we have $min\{X_1,...,X_n\}$~$Exp(n\beta)$. Thus:
(1) $E(min(X_1,...,X_N))=E(E(min(X_1,...,X_N)|N))=E(\frac{1}{N \beta})=\frac{1}{\lambda \beta}$.
(2) $E(N*min(X_1,...,X_N))=E(E(N*min(X_1,...,X_N)|N))$. We are looking for a distribution of $A:=N*min(X_1,...,X_N)|N$:
$F_A(t)=P(N*min(X_1,...,X_N)<t|N)=P(min(X_1,...,X_N)<\frac{t}{N}|N)=1-e^{-\beta N \frac{t}{N}}=1-e^{-\beta t}$. Thus $A$~$Exp(\beta)$ and $E(E(N*min(X_1,...,X_N)|N))=E(\frac{1}{\beta})=\frac{1}{\beta}$.
But then
$Cov(N,M_N)=\frac{1}{\beta}-\lambda \frac{1}{\lambda \beta} = 0$. And it's wrong. Could you help me in this issue?
Edit:
I forgot to write that $M_0=0$.
And from below comment (thank you for that) I have:
$E(min(X_1,...,X_N))=\sum^{\infty}_{k=0}P(N=k)*E(min(X_1,...,X_N)|N=k)=0+\sum^{\infty}_{k=1}e^{-\lambda}\frac{\lambda^k}{k!}*\frac{1}{k\beta}=\frac{e^{-\lambda}}{\beta}\sum^{\infty}_{k=1}\frac{\lambda^k}{k!k}$
If it was $k+1$ instead of $k$ in denominator, it would be fine, but in that case I'm stuck. Any suggestions?
Applying the double expectation formula $$ \begin{align} EM_N &= E\left[E\left[M_N\right] |N\right] \\ &= E\int_0^\infty xN\beta e^{-\beta Nx}dx \\ &=E\frac{1}{N\beta} \\ &= \frac1\beta \sum_{n=1}^\infty \frac{\lambda^n}{nn!}e^{-\lambda} \\ \end{align} $$ The series above converges but does not have a close form, and $$ \begin{align} ENM_N &= E\left[NEM_N |N\right] \\ &= E\int_0^\infty xN^2\beta e^{-\beta Nx}dx \\ &=E\frac{1}{\beta} \\ &= \frac{1}{\beta} \end{align} $$ Hence, $$ cov(N, M_N) = \frac{1}{\beta} - \frac{1}{\beta} \sum_{n=1}^\infty \frac{\lambda^n}{nn!}e^{-\lambda} $$