How do I find the cumulative distribution function of the maximum of a random number of I.I.D variables? The question is as follows. 
Let $Y = \text{ max } \{ X_{1}, X_{2}, X_{3} \ldots X_{N} \}$ 
Where $X_{i} \sim U(0,1)$ and $ N \sim Po(\lambda)$ 
Determine $F_{Y}(y) \text{ and } f_{Y}(y) $
So far I have done the following. 
$F_{Y}(y) = 0 \text{ if } y <0 \text{ and } 1 \text { if } y> 1$
For $0 < y < 1$,  $F_{Y}(y) = P[Y < y] = P[X_{1}+X_{2}+X_{3} + \ldots 
+ X_{n} < ny]$
$ \psi_{X_{1}+X_{2}+X_{3}+\ldots + X_{n}}(t) = \psi_{N}(\ln(\psi_{X_{i}}(t))$ 
$\psi_{N} = e^{\lambda(e^{t}-1)} \; \psi_{X_{i}} = \frac{e^{1}-1}{t} \text{ so } \psi_N(\ln(\psi_{X_{i}})) = e^{\lambda(\frac{e^t-1}{t}-1)}$
But from this point I am stuck. 
 A: Use the Law of Total Probability, mutual independence of $N$ and all $X_\star$, and identical distribution of all $X_\star$.
$$\begin{align}\mathsf P(Y\leqslant y) &= \mathsf P\Bigl(\max\bigl\{X_k:k\in\{1..N\}\bigr\}\leqslant y\Bigr) \\[1ex] &= \mathsf P(N=0)+ \sum_{n=1}^\infty \mathsf P(N=n)~\mathsf P\Bigl(\max\bigl\{X_k:k\in\{1..n\}\bigr\}\leqslant y\Bigm\vert N=n\Bigr) \\[1ex] &= \mathsf P(N=0)+\sum_{n=1}^\infty \mathsf P(N=n)\prod_{k=1}^n\mathsf P(X_k\leqslant y) \\[1ex] &= \sum_{n=0}^\infty \mathsf P(N=n)~\mathsf P(X_1\leqslant y)^n\\[1ex] & ~~~~\vdots \end{align}$$
Substitute the relevant pmf and CDF functions, then simplify. (HINT: use a well know Taylor Series).
A: It's quite obvious that $Y \mid N$ is a beta distribution; i.e., $$F_{Y \mid N}(y) = \Pr[Y \le y \mid N] = y^{N}, \quad 0 \le y \le 1.$$  Then it is trivial to see that if $N \sim \operatorname{Poisson}(\lambda)$, we have the unconditional CDF of $Y$ $$F_Y(y) = \Pr[Y \le y] = \sum_{n=0}^\infty \Pr[Y \le y \mid N = n]\Pr[N = n] = \sum_{n=0}^\infty y^n e^{-\lambda} \frac{\lambda^n}{n!} = e^{(y-1)\lambda}.$$  Note however that there is a subtlety here: although the support of $Y$ is on $[0,1]$, we have $F_Y(0) = e^{-\lambda} > 0$.  This means that the unconditional distribution of $Y$ is not fully continuous; there is a discrete probability mass at $Y = 0$, with probability $\Pr[Y = 0] = e^{-\lambda}$.  This is because $Y = 0$ with positive probability when $N = 0$ (there are no $X_i$s to maximize!), but if $N > 0$, then $Y$ has a continuous probability density, being the maximum of a nonzero number of continuous uniform distributions.
