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


My answer: 

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

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