I have a question that I'm really struggling with (below): It's hard to understand exactly what is the question actually states. does this mean the number of trials itself is a distribution with a mean of lambda? Any direction on how to solve would really be appreciated.
Let $X$ be a random variable defined as the sum of $N$ independent Bernoulli trials in which the probability of each Bernoulli taking the value $1$ is given by $p$. The number of Bernoulli trials $N$ is itself a random variable that behaves according to a Poisson distribution function with the parameter $\lambda$.
(a) Derive the conditional distribution function of $X$ given $N = n$ and state your reasoning behind your derivation.
(b) Derive the joint distribution function of $X$ and $N$ and state your reasoning behind your derivation.
We are told that $X = X_1+\dotsb+X_N$, where $N\sim\text{Pois}(\lambda)$.
a) Essentially, you are asked to compute or give $P(X = k|N = n)$. If I tell you that $N = n$, then the sum of $n$ independent Bernoulli trials follows a binomial distribution with $(n,p)$. Hence, $X|N \sim \text{Bin}(n,p)$.
b) I believe that you are essentially being asked to compute $P(X = k, N = n) = P(X = k|N = n)P(N = n).$