Sum of n iid random variables 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?
 A: 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. 
A: $\newcommand{\E}{\operatorname{E}}$
\begin{align}
M_{X_1+\cdots+X_n}(t) & = \E (e^{t(X_1+\cdots+X_n)}) = \E(e^{tX_1}\cdots e^{tX_n}) \\
& = \E(e^{tX_1})\cdots\E(e^{tX_n}) \qquad (\text{by independence}) \\
& = \left(\E (e^{tX_1}) \right)^n\qquad (\text{since the distributions are identical})
\end{align}
The mgf for the Poisson distribution is
$$
M_{X_1}(t) = \E(e^{tX_1}) = \sum_{x=0}^\infty e^{tx} \frac{\lambda^x e^{-\lambda}}{x!} = e^{-\lambda}\sum_{x=0}^\infty \frac{(e^t \lambda)^x}{x!} = e^{-\lambda} e^{e^t\lambda}  = e^{\lambda(e^t-1)}. \tag 1
$$
So the problem is to show that $(M_{X_1}(t))^n$ is the same as $(1)$ except with $n\lambda$ in place of $\lambda$.  That makes it a Poisson distribution with $n\lambda$ in place of $\lambda$.  So just apply a law of exponents
$$
\left( e^{\lambda(e^t-1)} \right)^n = e^{n\lambda(e^t-1)}.
$$
("That makes it a Poisson distribution with $n\lambda$ in place of $\lambda$."  Perhaps I should say: it makes it a distribution whose moments are all the same as those of a Poisson distribution with $n\lambda$ in place of $\lambda$.  If a distribution has the same moments as a Poisson distribution, is it the same distribution?  That's a subtler question, not usually expected of those who are assigned exercises like this one.)
You can also prove the result without MGFs by other methods.
