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Let $X$ be a random variable that follows a Poisson distribution with the mean value $m$.

Let $Y$ be the random variable which conditional probability by $X = n$ follows a binomial distribution with parameters $n,p$.

Prove that:

$$ p(Y = k) = \frac{(pm)^k e^{-mp}}{k!} $$.

$X$ follows a Poisson distribution, which means: $p(X = n) = \frac{m^n}{n!}e^{-m}$.

and $p(Y = k | X = n) = C^k_n p^k (1 - p)^{n-k}$ because it is a binomial distribution.

We have: $$p(Y = k | X = n) = \frac{ p(Y = k \text{ and } X = n) }{p(X = n) }$$

I don't know how to proceed to get the result.

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Law of total probability $$P(Y=k)=\sum_{j=0}^{\infty}{P(Y=k|X=j)P(X=j)}$$

$$P(Y=k|X=j)= {j \choose k}p^{k}(1-p)^{j-k} $$

$$P(X=j)=e^{-m}\frac{m^j}{j!}$$

$$P(Y=k)=\sum_{j=0}^{\infty}{{j \choose k}p^{k}(1-p)^{j-k}e^{-m}\frac{m^j}{j!}}=\sum_{j=k}^{\infty}{{j \choose k}p^{k}(1-p)^{j-k}e^{-m}\frac{m^j}{j!}}$$

$$\sum_{j=k}^{\infty}{\frac{j!}{k!(j-k)!}p^{k}(1-p)^{j-k}e^{-m}\frac{m^j}{j!}}=\frac{p^ke^{-m}}{k!}\sum_{j=k}^{\infty}{\frac{1}{(j-k)!}(1-p)^{j-k}m^j}$$

Change of variable $i=j-k$, and use the definition of exponential in terms of series

$$\sum_{j=k}^{\infty}{\frac{1}{(j-k)!}(1-p)^{j-k}m^j}=\sum_{i=0}^{\infty}{\frac{1}{i!}(1-p)^{i}m^{i+k}}=m^ke^{(1-p)m}$$ Finally,

$$P(Y=k)=\frac{p^ke^{-m}}{k!}m^ke^{(1-p)m}=\frac{(pm)^k e^{-pm}}{k!}$$

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  • $\begingroup$ $P(Y=k)=\sum_{j=0}^{\infty}{{j \choose k}p^{k}(1-p)^{j-k}e^{-m}\frac{m^j}{j!}}=\sum_{j=k}^{\infty}{{j \choose k}p^{k}(1-p)^{j-k}e^{-m}\frac{m^j}{j!}}$. Would you please explain why the second sum starts from $k$ $\endgroup$ – Zouhair El Yaagoubi Jan 1 at 22:38
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    $\begingroup$ $${j \choose k}=0$$ if $k>j$ $\endgroup$ – Canardini Jan 1 at 22:39
  • $\begingroup$ Accepted answer, thank you so much for your help. $\endgroup$ – Zouhair El Yaagoubi Jan 1 at 22:40

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