Let $X_1,X_2,\ldots,X_n$ be iid poisson random variables with mean $\lambda$ , then it can be verified using mgf that the sum $S=\sum\limits_{i=1}^n X_i$ is also poisson with mean $n\lambda$.
However, let $X_i$ be iid random variables having the pmf $$ f_X(x;\theta)=\frac{h(x)\theta^x}{\sum\limits_{y=0}^{\infty}h(y)\theta^y} ,x=0,1,2,\ldots$$ with $\theta >0$. How do we verify that $S=\sum\limits_{i=1}^n X_i$ is also a member of the same distributional family? Using mgf seems tedious or is there a trick to calculate mgf?
Best Answer
Let $g(\theta)= \dfrac{1}{\sum\limits_{y=0}^{\infty}h(y)\theta^y}$, which does not depend on $x$,
so $f_X(x;\theta)=g(\theta)h(x)\theta^x$, a function of $\theta$ multiplied by a function of $x$ multiplied by $\theta^x$ with its sum over $x$ being $1$. Then
$$\Pr(S=s)=\sum_{\sum_j x_j=s} \prod_i f_X(x_i; \theta) = \sum_{\sum_j x_j=s} \prod_i g(\theta)h(x_i)\theta^{x_i} = g(\theta)^n \left(\sum_{\sum_j x_j=s} \prod_i h(x_i) \right) \theta^s$$ which is of the same form of a function of $\theta$ multiplied by a function of $s$ multiplied by $\theta^s$, with its sum over $s$ being $1$.
So in that sense the distribution of the sum is the from the same general family of distributions over non-negative integers.