We have a discrete random variable $N$, and $X_1, X_2, … X_N$ are i.i.d Poisson random variables with parameter $\lambda$. Denote $Y = \sum_{i=1}^{N} X_i$. What I want to know is:
- If finding the conditional pmf $Prob(N=n|Y=k)$ feasible?
- And if so how to find it.
I am well aware that $Prob(Y=k|N=n)$ is a Poisson r.v with parameter $n\lambda$
Any help or hints would be appreciated!
Edit 1: Additional information: we do not know $Prob(N=n)$.
Best Answer
This is essentially a problem of Bayesian inference --- if you have a prior distribution for $N$ then you can find its posterior given an observation $Y=y$. To do this, first note that:
$$Y|N \sim \text{Pois}(N \lambda).$$
To obtain the posterior of interest, take a prior mass function $\pi_N$ and you then have:
$$\begin{align} \mathbb{P}(N=n|Y=y) &= \frac{p(N=n,Y=y)}{\sum_n p(N=n,Y=y)} \\[6pt] &= \frac{\text{Pois}(y|n \lambda) \cdot \pi_N(n)}{\sum_{n=0}^\infty \text{Pois}(y|n \lambda) \cdot \pi_N(n)} \\[6pt] &= \frac{n^y \cdot e^{-n \lambda} \cdot \pi_N(n)}{\sum_{n=0}^\infty n^y \cdot e^{-n \lambda} \cdot \pi_N(n)}. \\[6pt] \end{align}$$