Question

Answer the following in a mathematically sound argument (proof) using combinations of English / symbols as needed.

Find a sufficient statistic (in the rigorous sense) for 2 when X-Pois (a)

Answer 1

Let $X_1,X_2,\cdots,X_n$ be a random sample taken from $Pois(\lambda)$ .

The density function is given by

$f(x)=\left\{\begin{matrix} \frac{e^{-\lambda}\lambda^{x}}{x!}\, , \quad& x=0,1,2,\cdots \quad and \,\lambda>0\\ 0\, , \quad& Otherwise \end{matrix}\right.$

Then the joint density function of these n random variables is given by,

$\begin{align*} f(x_1,x_2,\cdots,x_n)&=\prod_{i=1}^{n}f(x_i)\\ &=\frac{e^{-n\lambda}\lambda^{\sum_{i=1}^{n}x_i}}{\prod_{i=1}^{n}x_i!} \\ &=e^{-n\lambda}\lambda^{\sum_{i=1}^{n}x_i}\frac{1}{\prod_{i=1}^{n}x_i!}\\ &=g(\lambda,T(\underline{x})) h(\underline{x})\end{align*}$

where is a function of $\lambda$ and statistic

$T(\underline{X})=\sum_{i=1}^{n}X_i$

and is a function which doesnot depends on $\lambda$ .

Using Neyman- Factorization theorem, if the density can be factorized in to the product of two functions such that one is a function of parameter and sample through a statistic $T(\underline{x})$ and the other doesnot depends on parameter, then the statistic $T(\underline{X})$ is sufficient for the parameter.

Here, in this case

$T(\underline{X})=\sum_{i=1}^{n}X_i$

is sufficient for $\lambda$ ​​​​​​​.