How can I show that the pgf of the sum of n independent discrete random variables is equal to the product of their probability generating functions?

Question

How could I show that if $S = X_1 + X_2 +...+ X_n$ then $E(z^{S}) = E(z^{X_1})E(z^{X_2})...E(z^{X_n})$

Where E is the expectation.

Thanks very much in advance.

Answer 1

Let $X$ and $Y$ be independent random variables. Then \begin{align} E(z^{X + Y}) &= \sum\limits_{k =0}^\infty \mathbb{P}(X+Y = k)z^k\\ &= \sum\limits_{k =0}^\infty \sum\limits_{r = 0}^k \mathbb{P}(X = r \wedge Y = k-r)z^k \\ &= \sum\limits_{k =0}^\infty \sum\limits_{r = 0}^k \mathbb{P}(X = r)\mathbb{P}(Y = k-r)z^rz^{k -r} \\ &= \left(\sum\limits_{k = 0}^\infty \mathbb{P}(X= k)z^k\right) \left(\sum\limits_{k = 0}^\infty \mathbb{P}(Y= k)z^k\right) \\ &= E(z^X) E(z^Y) \end{align}

Now use induction on $n$.