# Conditional binomial distribution.

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.

• – StubbornAtom Jan 3 at 14:20

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,

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

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