Sufficiency of two Poisson disributions If $X_1,X_2$ constitute a random sample of size n=2 from a Poisson Population show that the mean of the sample is a sufficient estimator of the parameter $\lambda$  .  
Since the sum of Poissons is also Poisson with parameter $\lambda+\lambda=2\lambda$
Then $E[Y]=2\lambda$.  where Y constitute of two Poisson distributions each with parameter $\lambda$.  
Do I have to show that $2\lambda$ is sufficient for $\lambda$.
I don't understand what the estimator is  here?
 A: You're not showing sufficiency at all.  The only thing your calculation shows is that $\bar X = (X_1 + X_2)/2$ is an unbiased estimator of $\lambda$.
To show that $\bar X$ is a sufficient statistic, you have to do something else.  The most straightforward approach is to use the Factorization Theorem, which seeks the expression of the joint distribution for $X_1, X_2$ as $$f_{X_1, X_2}(x_1, x_2) = g(T(x_1, x_2) \mid \lambda) h(x_1, x_2),$$ for some function $T$ that depends only on the sample; and in your case, $T(x_1, x_2) = \bar x = (x_1 + x_2)/2$.
A: Probably you mean that the sample average is sufficient statistics for $\lambda$. 
So, if $X_1, X_2$ are independent draws from Poisson distribution with parameter $\lambda$ then the answer is "yes". 
Check here: http://en.wikipedia.org/wiki/Sufficient_statistic

Using Fisher–Neyman factorization theorem:
$P\{X=x\}=P\{X_1=x_1,X_2=x_2\}=P\{X_1=x_1\}P\{X_2=x_2\}=\lambda^{x_1}e^{-\lambda}/x_1!*\lambda^{x_2}e^{-\lambda}/x_2!=\lambda^{(x_1+x_2)/2}e^{2-2\lambda}/(x_1!x_2!)=g_\lambda(T(x))h(x)$
where $h(x)=(x_1!x_2!)^{-1}$ and $g_\lambda(T(x))=\lambda^{(x_1+x_2)/2}e^{2-2\lambda}$
