Conditional probability mass function of number of Poisson random variable given their sum values

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:

1. If finding the conditional pmf $$Prob(N=n|Y=k)$$ feasible?
2. 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 $$Prob(N=n)$$.

• Can you use Bayes formula to write $\Pr(N=n\mid Y=k)$ in terms of $\Pr(Y=k\mid N=n)$, $\Pr(Y=k)$ and $\Pr(N=n)$? You already know (and have included in your question) the values of 2 of the probabilities mentioned in the previous sentence. What do you know about the third? Jan 27, 2022 at 19:56
• Hi @DilipSarwate, thank you for your comment. I tried that method but we do not know $Pr(N=n)$.
– T9h
Jan 27, 2022 at 20:30
• Since you don't know the marginal distribution of $N,$ what information are you hoping to use to find the answer?
– whuber
Jan 27, 2022 at 22:24
• @whuber I hope there will be ideas to derive the probability of the event $Prob(N=n \cap Y=k)$ by decomposition to other equivalent events since we do know $Prob(Y=k)$.
– T9h
Jan 27, 2022 at 22:34
• Any such solution would be tantamount to knowing the full distribution of $N.$
– whuber
Jan 27, 2022 at 22:40

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}