Poisson Distribution Example_Sufficient

Example)

Let $X_{1}, X_{2}$ be a random sample of size 2 from Poisson distribution $f(x_{1}=\lambda^{x_{1}} \cdot \frac {e^{-\lambda}}{x_{1}!})$ . Show $T=X_{1}+ X_{2}$ 

$\triangleright$  Solution

The joint distribution of the random sample is $\prod_{i=1}^{2}=f(x_{i}|\lambda)=\frac {\lambda^{x_{1}+x_{2}}}{x_{1}!x_{2}!} \cdot e^{-2\lambda}$ 

The joint probability of $X_{1}=x_{1}, X_{2}=x_{2}$ and $T=X_{1}+X_{2}=t$ is

$\rightarrow f(x_{1}, x_{2}, t|\lambda)=\frac {\lambda^t}{x_{1}!x_{2}!} \cdot e^{-2\lambda}$ 

We know $T=X_{1}+X_{2}$ has a Poisson distribution with parameter $2\lambda$

$\rightarrow g(t|\lambda)=\frac {(2\lambda)^t}{t!}\cdot e^{-2\lambda}$  

Consequently, the conditional distribution of the sample, given T=t is

$\rightarrow f(x_{1}, x_{2}|T=t; \lambda)=\frac {f(x_{1}, x_{2}, t|\lambda)}{g(t|\lambda)}=\frac {\lambda^t}{x_{1}!x_{2}!}\cdot e^{-2\lambda}/ \frac {(2\lambda)^t}{t!}\cdot e^{-2\lambda}$ 

$\therefore$  This does not depend on the parameter $\lambda$ as the $\lambda$ is cancelled out.