So we have the following question.
$Y$ is a discrete r.v with the positive integers as values. The p.m.f. of $Y$ is of the form:
$P(Y=k) = c \cdot \frac{1}{k!}, k = 1,2...$
a) what is the value of c?
$U_1, U_2, \dots$ are I.I.D r.v.s with $U_i \epsilon U(0,1)$ $U_1, U_2,...$are also independent of Y.
b) Set
$M = max(U_1, U_2, ..., U_Y)$
Show that for any $t$ in the interval $[0,1]$
$P(M\le t) = g_y(t)$
where $g_y(t)$ is the p.g.f. of $Y$.
So what I've done is calculated the c using the sum of p.m.f = 1 i.e.
$\sum_{k=1}^\infty p_x(k) = 1$
so
$\sum_{k=1}^\infty c \cdot \frac{1}{k!} = 1$
$c \sum_{k=1}^\infty \frac{1}{k!}$ = $c ([\sum_{k=0}^\infty \frac{1}{k!}] -1)$
this gives
$c = \frac{1}{e-1}$
Now for the b) part I first calculate the p.g.f:
$g_y(t) = \sum_{k=1}^\infty t^k \cdot \frac{1}{e-1} \frac{1}{k!}$
$= \frac{1}{e-1} ([\sum_{k=0}^\infty \frac{t^k}{k!}] - 1)$
$= \frac{e^t - 1}{e -1}$
Now this is where i get uncertain. I then compare to $P(M \le t)$ which is due to independence:
$P(M \le t) = P(Max(U_1, U_2, ..., U_Y) \le t) = \prod_{i=1}^Y P(U_i \le t)$
As $U_i \epsilon U(0,1)$ we get $P(U_i \le t) = t and Therefore:
$\prod_{i=1}^Y P(U_i \le t) = t^Y$
Which is not the same as $g_y(t)$.
Can anyone explain where my reasoning is wrong? Thanks!