$N \sim \text{Po}(\lambda)$ and $X_1,X_2,....,X_N$ are iid and independent of $N$, what is distribution of $Z_N = \max \{X_i\}_{i=1}^{N}$ I think the title covers most of my concerns. The distribution of the $X_i$ does not really matter, I am just experiencing difficulties in finding an expression for
$$\text{Pr}(Z_N \leq x) = F(x)^N$$
which is not "random", i.e. not exponent $N$. The $X_i \sim \text{GPD}$ with distribution
$$F(x) = 1-\left(1+ \xi \dfrac{x}{\sigma}\right)^{-1/\xi}$$
Any hints are very appreciated.
Edit: My "belief" is that the expression would be something like
$$\text{Pr}(M_N \leq x) = \text{Pr}(N=n)\text{Pr}(M_n \leq x)$$
due to being independent.
 A: I assume that you take $Z$ to be the supremum of the set $\{X_1,X_2,\dots,X_N\}$ since this is also reasonably defined (as negative infinity) when the set is empty (in the event that $N=0$).
Conditional on $N$, $Z$ has cdf
\begin{align}
F_{Z|N}(z)&=P(\sup\{X_i\}_{i=1}^N\le z|N)
\\&=P(X_1\le z \cap \dots \cap X_N\le z|N)
\\&=F_X(z)^N.
\end{align}
This holds also for $N=0$, in which case
$F_{Z|N=0}(z)=P(Z\le z|N=0)=1$, that is, the supremum of the empty set ($-\infty$) is smaller than any $z$ with probability 1.
Using the law of total probability,
\begin{align}
F_Z(z)&=\sum_{n=0}^\infty P(N=n)F_X(z)^n
\\&=E(F_X(z)^N)
\\&=G_N(F_X(z))
\end{align}
where $G_N$ is the probability generating function of $N$ given by
$$
G_N(s)=e^{-\lambda(1-s)}
$$
when $N\sim\operatorname{Poisson}(\lambda)$.
Hence,
\begin{align}
F_Z(z)&=e^{-\lambda(1-F_X(z))}.
\end{align}
Note that $F_Z(z) \rightarrow e^{-\lambda}$ as $z\rightarrow -\infty$ in agreement with the "point mass" at $-\infty$.
For the standard generalized Pareto case
$$
F_X(x) = \begin{cases}
1-(1+ \xi x)^{-1/\xi}, & x\ge 0 \\
0, & x<0
\end{cases}
$$
the cdf of $Z$ thus becomes
$$
F_Z(z) = \begin{cases}
e^{-\lambda(1+ \xi z)^{-1/\xi})}, & z\ge 0 \\
e^{-\lambda}, & z<0.
\end{cases}
$$
