Knowing that $\phi(t)$ is a characteristic function and $G(s)$ is a generating function, prove $G(\phi(t))$ is also a characteristic function.

Question

If $\phi(t)$ is a characteristic function of a random variable $X$ and if $G(s)$ is the generating function of a random variable $Y$, $Y \in \{0,1,2,\dots,\}$, prove that $G(\phi(t))$ is also a characteristic function, and find its corresponding random variable.

I've already done some work but get stuck. According to the definition of the generating function and the property of the characteristic function: $$ G(\phi(t))=E(\phi(t)^Y)=\sum_kP(Y=k)\phi(t)^k $$ $$ \phi(t)^k= E(e^{it\sum_{j=1}^{k}X_j})$$ And we also know $$ G(\phi(t))=G(0) + G^{'}(0)\phi(t)+\frac{G^{''}(0)}{2!}\phi(t)^2+\dots+\frac{G^{(k)}(0)}{k!}\phi(t)^k+\dots $$ Then $$ G(\phi(t))=\sum_{k}\frac{G^{(k)}(0)}{k!}E(e^{it\sum_{j=1}^{k}X_j}) $$

How can I show this $G(\phi(t))$ is a characteristic function?

Answer 1

Hint: Take a sequence of independent identically distributed random variables $(X_i)_{i \geq 1}$ such that

• $X_1=X$ in distribution,
• $Y$ is independent of $(X_i)_{i \geq 1}$.

Show that the characteristic function of the random variable

$$Z:=\sum_{i=1}^Y X_i$$

equals $G(\phi)$.