properties of the expected value of the Poisson function Proposition. If a random variable $X$ with a Poisson distribution, show that:
i) $EX^{2}=\lambda E(X+1)$
ii) If $ \lambda =1$ then $E(|X-1|)$.
I have tried to do part $i)$, but I am having trouble concluding.
Since $\frac{d}{dt} M_{X}(t)=E[e^{tx}]=e^{-\lambda}e^{\lambda e^{t}}$, now let;
$E(X^{2})=\frac{d^{2}}{dt^{2}}M_{X}(t)=\lambda\frac{d}{dt}(e^{t}M_{X}(t))=\lambda \left[e^{t}M_{x}(t)+e^{t}\frac{d}{dt}M_{X}(t) \right]=\lambda[e^{t}\left (  M_{X}(t)+\frac{d}{dt}M_{X}(t)    \right)]=\lambda \left[e^{t} \left(\sum_{k=0}^{\infty}e^{tk}e^{-\lambda}\frac{\lambda ^{k}}{k!}  +\lambda        \right) \right]$
I would like to get a suggestion or alternative solution
 A: $\newcommand{\e}{\operatorname E}$
\begin{align}
\e(X^2) = {} & \sum_{x=0}^\infty x^2 \Pr(X=x) \\[8pt]
= {} & \sum_{x=0}^\infty x^2 \cdot \frac{\lambda^x e^{-\lambda}}{x!} \\[8pt]
= {} & \sum_{x=1}^\infty x^2 \cdot \frac{\lambda^x e^{-\lambda}}{x!} \\
& \text{This is valid because the} \\
& \text{term thus discarded is 0.} \\[8pt]
= {} & \sum_{x=1}^\infty x\cdot \frac{\lambda^x e^{-\lambda}}{(x-1)!} = \lambda \sum_{x=1}^\infty x\cdot\frac{\lambda^{x-1}e^{-\lambda}}{(x-1)!} \\
& \ldots\text{and this step would have been} \\
& \qquad\text{incorrect if applied when } x=0. \\[8pt]
= {} & \lambda \sum_{y=0}^\infty (y+1) \frac{\lambda^y e^{-\lambda}}{y!} \\
& \text{where } y = x-1 \\[8pt]
= {} & \lambda \sum_{y=0}^\infty (y+1) \Pr(X=y) \\[8pt]
= {} & \lambda \e(X+1).
\end{align}
A: The PGF $G_X(t)=Et^X=e^{\lambda(t-1)}$ is easier to analyze: $EX=G_X^\prime(1)=\lambda$ and $EX(X_1)=\lambda^2$, so $EX^2=\lambda(\lambda+1)$ as i) said. For ii) note$$E|X-1|=E(X-1+2\delta_{X0})=\lambda-1+2P(X=0)=\lambda-1+2e^{-\lambda},$$which for $\lambda=1$ is $2/e$ as @callculus said.
