# 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?

• $\theta^{x_1}\theta^{x_2}=\theta^{x_1+x_2}$ Nov 14, 2014 at 19:07

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.

$\newcommand{\E}{\operatorname{E}}$ \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.

• Thank you Michael, the mgf of poisson is easy to find. My question is actually the second part.
– LPS
Nov 14, 2014 at 19:41
• "Perhaps I should say: it makes it a distribution whose moments are all the same as those of..." No, you should definitely say that "That makes it a Poisson distribution".
– Did
Nov 14, 2014 at 20:01