For $X \sim Pois(λ)$, find $\Bbb E(2^X)$, if it is finite.
Hint: Use these facts, $$\begin{align}\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}&=e^\lambda\\\sum_{k=0}^{\infty}\frac{(k+1)\lambda^k}{k!}&=e^\lambda+\lambda e^\lambda\end{align}$$
- The lecture ended before we had time to cover this section and I have no notes to work from. I am having trouble getting to the answer because I am getting confused with my work. Any help would be much appreciated.
- I have this so far but I am not sure if it is correct, $P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}$ for $x \in\{ 0, 1, 2, \ldots\}$
$$\begin{align}\mathbb E(2^X) &= \sum_{x=0}^{\infty} 2^x \frac{e^{-\lambda}\lambda^x}{x!}\\&= e^{-\lambda} \mathbb\sum_{x=0}^{\infty} \frac{(2\lambda)^x}{x!}\\&= e^{-\lambda} e^{2\lambda}\end{align}$$ So then $\Bbb E (2^X) = e^\lambda$
\begin{align} \operatorname E(2^X) = {} & \sum_{x=0}^\infty 2^x \Pr(X=x) = \sum_{x=0}^\infty 2^x \cdot\frac{\lambda^x e^{-\lambda}}{x!} \\[10pt] = {} & e^{-\lambda} \sum_{x=0}^\infty \frac{(2\lambda)^x}{x!} \\ & \text{The factor $e^{-\lambda}$ can be pulled out of} \\ & \text{the sum because it does not change} \\ & \text{as $x$ goes from $0$ to $\infty$.} \\[10pt] = {} & e^{-\lambda} \cdot e^{2\lambda} = e^{\lambda}. \end{align}