I have a lottery in which a random number $$ tickets are sold, where $$ has PMF given by $$p_N(n) = \frac{\lambda^{n-1}e^{-\lambda}}{(n-1)!}, \quad n = 1, 2, \ldots$$ The tickets are numbered $1, 2, … , $, and a single winning ticket is drawn uniformly at random. Let $$ denote the number on the winning ticket.
I want to find the Expectation and Variance of X. My thinking so far for the expectation is to use the law of total expectation, i.e. say $E[X] = E[E[X \mid N]]$, and then use the fact $E[X] = \sum_{n=1}^{\infty} E[X \mid N=n]P(N=n)$. So surely $E[X \mid N=n] = \frac{n+1}{2}$ as it is uniform, then I should be able to evaluate the sum? Feels like I am missing something obvious, like something within the sum. Any pointers appreciated, thanks in advance.
Right, so you are given $$\Pr[N = n] = \frac{\lambda^{n-1} e^{-\lambda}}{(n-1)!},$$ and you already determined that $$\operatorname{E}[X \mid N = n] = \frac{n+1}{2}.$$ So then $$\operatorname{E}[X] = \sum_{n=1}^\infty \frac{n+1}{2} \frac{\lambda^{n-1} e^{-\lambda}}{(n-1)!}.$$ How would you compute this sum? Here's a hint: $$\frac{n+1}{2} = \frac{n-1}{2} + 1.$$