Calculate $E[a^N]$ for Poisson random variable $N$

Question

Calculate $E[a^N]$ for a Poisson random variable $N$ with parameter $\lambda \cdot T$, where $a>0$ is a constant independent of $N$.

My answer:

\begin{align} E[a^N] &= \sum_{n=0}^{\infty} \frac{e^{-\lambda T}(\lambda T)^n}{n!} a ^n\\ &= \sum_{n=0}^{\infty} \frac{e^{-\lambda T}(a\lambda T)^n}{n!} \\ &= \sum_{n=0}^{\infty} e^{-\lambda T} e^{a\lambda T}\frac{e^{-a\lambda T}(a\lambda T)^n}{n!} \\ &= \sum_{n=0}^{\infty} {e^{-\lambda T(1-a)}} \frac{e^{-a\lambda T}(a \lambda T)^n}{n!}\\ &= {e^{-\lambda T(1-a)}} \sum_{n=0}^{\infty} \frac{e^{-a\lambda T}(a \lambda T)^n}{n!}\\ &= {e^{-\lambda T(1-a)}}\\ \end{align}

Is this right?

Answer 1

Yes, this is correct. In general, if $X$ is a random variable (typically a discrete random variable) then $\mathbb{E}[z^X]$ is referred to as the probability generating function (PGF for short) of $X$. Your answer for the Poisson random variable of mean $\lambda T$ is correct, as can be verified on Wikipedia.

Answer 2

\begin{align} \operatorname{E} \Big[a^N\Big] &= \sum_{n=0}^\infty \frac{e^{-\lambda T}(\lambda T)^n}{n!} a^n \\[10pt] &= \sum_{n=0}^\infty e^{-\lambda T(1-a)} \frac{e^{-a\lambda T}(a \lambda T)^n}{n!} \longleftarrow \text{I'm not sure what you did here.} \end{align}

I would do this: \begin{align} \operatorname{E} \Big[a^N\Big] &= \sum_{n=0}^\infty \frac{e^{-\lambda T}(\lambda T)^n}{n!} a^n \\[10pt] &= e^{-\lambda T} \sum_{n=0}^\infty \frac{(\lambda T)^n}{n!} a^n & \longleftarrow & \text{ This can be done because $e^{-\lambda T}$ does not} \\ & & & \text{ change as $n$ goes from $0$ to $\infty$.} \\[10pt] & = e^{-\lambda T} \sum_{n=0}^\infty \frac{(a\lambda T)^n}{n!} \\[10pt] & = e^{-\lambda T} e^{a\lambda T} = e^{-\lambda T(1-a)}. \end{align}