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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 $Prob(N=n)$.

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    $\begingroup$ 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? $\endgroup$ Jan 27, 2022 at 19:56
  • $\begingroup$ Hi @DilipSarwate, thank you for your comment. I tried that method but we do not know $Pr(N=n)$. $\endgroup$
    – T9h
    Jan 27, 2022 at 20:30
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    $\begingroup$ Since you don't know the marginal distribution of $N,$ what information are you hoping to use to find the answer? $\endgroup$
    – whuber
    Jan 27, 2022 at 22:24
  • $\begingroup$ @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 $Prob(Y=k)$. $\endgroup$
    – T9h
    Jan 27, 2022 at 22:34
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    $\begingroup$ Any such solution would be tantamount to knowing the full distribution of $N.$ $\endgroup$
    – whuber
    Jan 27, 2022 at 22:40

1 Answer 1

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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:

$$\begin{align} \mathbb{P}(N=n|Y=y) &= \frac{p(N=n,Y=y)}{\sum_n p(N=n,Y=y)} \\[6pt] &= \frac{\text{Pois}(y|n \lambda) \cdot \pi_N(n)}{\sum_{n=0}^\infty \text{Pois}(y|n \lambda) \cdot \pi_N(n)} \\[6pt] &= \frac{n^y \cdot e^{-n \lambda} \cdot \pi_N(n)}{\sum_{n=0}^\infty n^y \cdot e^{-n \lambda} \cdot \pi_N(n)}. \\[6pt] \end{align}$$

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  • $\begingroup$ Hi Ben, thanks for your answer, however, I have a few questions if you don't mind. $\endgroup$
    – T9h
    Jan 28, 2022 at 23:52

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