Poisson Distribution

A discrete Random Variable is of Poisson distribution if its range is ${\mathbb{N}}_0$, and

• Parameters
  • $\lambda >0$
• PMF
  • $p(n)=e^{-\lambda }\frac{{\lambda }^n}{n!}$
  • The important part is ${\lambda }^n/n!$; $e^{-\lambda }$ is just a constant to make if sum to 1
  • Remember it as the expansion of the exponential $e^{\lambda }$
• Mean
  • $\lambda$
• Variance
  • $\lambda$
• MGF
  • $\exp (\lambda (e^t-1))$
• CF
  • $\exp (\lambda (e^{it}-1))$

Poisson Distribution as Approximation to Binomial Distribution

When $n$ is large and $p$ is small, Poisson distribution can be an approximation of a Binomial Distribution. This is because for $X\sim \mathrm{Binom}(n,p)$, we have

$$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}=\frac{n!}{k!(n-k)!}{\left(\frac{np}{n}\right)}^k{\left(1-\frac{np}{n}\right)}^n{\left(1-p\right)}^{-k}$$

When $n$ is large and $p$ is small, we have

$$\frac{n(n-1)\dots (n-k+1)}{n^k}\approx 1,{\left(1-\frac{np}{n}\right)}^n\approx e^{-np},(1-p{)}^{-k}\approx 1.$$

Therefore, we have

$$P(X=k)\approx e^{-np}\frac{(np{)}^k}{k!},$$

i.e., $X\dot{\sim }\mathrm{Poisson}(np)$. Or, $\mathrm{Binom}(n,\lambda /n)\stackrel{d}{⟶}\mathrm{Poisson}(\lambda )$.

• The Poisson approximation result can be shown to be valid under even more general conditions than those so far mentioned. For instance, suppose that n independent trials are to be performed, with the ith trial resulting in a success with probability $p_i,i=1,...,n$. Then it can be shown that if n is large and each $p_i$ is small, then the number of successful trials is approximately Poisson distributed with mean equal to ${\sum }_{i=1}^n{p}_i$. In fact, this result will sometimes remain true even when the trials are not independent, provided that their dependence is “weak”.

Splitting a Poisson Random Variable

Consider conducting $N\sim \mathrm{Poisson}(\lambda )$ trials. Let $X$ be the success count with a conditional distribution $X\mid N=n\sim \mathrm{Binom}(n,p)$. Let $Y=N-X$. We can show that $X$ and $Y$ are independent, and $X\sim \mathrm{Poisson}(\lambda p)$ and $Y\sim \mathrm{Poisson}(\lambda (1-p))$.

Since all r.v.s are discrete, we consider their PMFs. First, we have

$$p_{X,Y}(x,y)=p_{X,Y\mid N}(x,y\mid n=x+y)p_N(n=x+y)=1\{y=y\}{p}_{X\mid N}(x|n=x+y)p_N(n=x+y)=e^{-\lambda }\frac{{\lambda }^{x+y}}{(x+y)!}\cdot (\genfrac{}{}{0ex}{}{x+y}{x})p^x(1-p{)}^y.$$

Then, as $X$ is the marginal distribution of $(X,N)$, we have

$$p_X(x)=\sum _{n=0}^{\infty }{p}_{X,N}(x,n)=\sum _{n=0}^{\infty }{p}_{X\mid N}(x\mid n)p_N(n)=\sum _{n=x}^{\infty }\left(\genfrac{}{}{0ex}{}{n}{x}\right)p^x(1-p{)}^{n-x}{e}^{-\lambda }\frac{{\lambda }^n}{n!}=\frac{(p/(1-p){)}^x{e}^{-\lambda }}{x!}\sum _{n\ge x}\frac{(\lambda (1-p){)}^n}{(n-x)!}=\frac{(\lambda p{)}^x{e}^{-\lambda }}{x!}\sum _{n=0}^{\infty }\frac{(\lambda (1-p){)}^n}{n!}=\frac{(\lambda p{)}^x}{x!}{e}^{\lambda (1-p)-\lambda }=e^{-\lambda p}\frac{(\lambda p{)}^x}{x!}.$$

By symmetry, $p_Y(y)=e^{-\lambda (1-p)}(\lambda (1-p{)}^y)/y!$. Thus, $X\sim \mathrm{Poisson}(\lambda p)$ and $Y\sim \mathrm{Poisson}(\lambda (1-p))$, and,

$$p_X(x)p_Y(y)=e^{-\lambda }\frac{{\lambda }^{x+y}{p}^x(1-p{)}^y}{x!y!}=p_{X,Y}(x,y),$$