Poisson-Binomial Distribution

Theorem

Let

• $N\sim \text{Po}(\mu )$ and conditional on $N=n$
• $M\mid N=n\sim \text{Bin}(n,p)$.

Then $M$ (unconditional on $N$) is Poisson distributed with parameter $\mu p$.

Interpretation

If you have a Poisson number of trials and each succeeds with probability $p$, the number of successes is still Poisson — just with a scaled-down rate.

Proof

$$\begin{array}{rl}P(M=k)& =\sum _{n=k}^{\infty }P(M=k\mid N=n)P(N=n)\\ & =\sum _{n=k}^{\infty }\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}\cdot \frac{e^{-\mu }{\mu }^n}{n!}\\ & =\frac{e^{-\mu }{p}^k}{k!}\sum _{n=k}^{\infty }\frac{(\mu (1-p){)}^{n-k}}{(n-k)!}{\mu }^k\\ & =\frac{e^{-\mu }(\mu p{)}^k}{k!}\sum _{j=0}^{\infty }\frac{(\mu (1-p){)}^j}{j!}\\ & =\frac{e^{-\mu }(\mu p{)}^k}{k!}{e}^{\mu (1-p)}\\ & =\frac{e^{-\mu p}(\mu p{)}^k}{k!}\end{array}$$

Thus $M\sim \text{Po}(\mu p)$.

Related

• Poisson Distribution
• Sum of Poisson Random Variables