Given $$ X \sim \operatorname {Erlang}(n,1)$$ and $$ Y|X=x \sim\operatorname {Po}(x) .$$ How do I find the characteristic function of $Y$?
I did start with this: $$ \phi_{Y|X=x}(t) = E[e^{itY} | X=x] = e^{x(e^{it}-1)} $$by the Po(x)-distribution. $$ \phi_{Y}(t) = E[e^{itY}] = E[E[e^{itY}|X]] = E[e^{X(e^{it}-1)}] $$
But to solve that expected value gives me a really complicated integral which I can't solve. Am I doing something wrong?
UPDATE: I followed your hint with the integral, but I'm unable to solve it still: $$ E[e^{X(e^{it}-1)}] = \int_{0}^{\infty} e^{x(e^{it}-1)}f_x(x) dx = \frac{1}{(n-1)!} \int_{0}^{\infty} e^{x(e^{it}-2)} x^{n-1} = \frac{1}{(n-1)!} \{\frac{e^{x(e^{it}-2)}}{e^{n}}[(ex)^{n-1} - (n-1)(ax)^{n-2} + (n-1)(n-2)(ax)^{n-3} -... + (-1)^{n-1}(n-1)!]\}_0^\infty $$ But I have no idea how to simplify that to the final result you spoke of... Or am I doing any step wrong?
You are correct, here is a hint to solve the integral
$$ \int e^{ax} x^{n} = \frac{e^{ax}}{a^{n+1}} \left[ (ax)^n - n(ax)^{n-1}+ n(n-1)(ax)^{n-2} - \dots + (-1)^nn! \right] $$ and your final result should be $$ \frac{(-1)^n}{(e^{it}-2)^n} $$