If X ~ Poisson(λ), then find E((X+1)^2) I've tried to simplify using the infinite summation but to no avail. Any help would be greatly appreciated.
 A: The mean and variance of a Poisson distribution with parameter $\lambda$ are both $\lambda$.
$$\mathbb{E}[X]=\lambda \text{ and } \mathbb{Var}[X]=\lambda.$$
Since $\mathbb{Var}[X]=\mathbb{E}[X^2]-(\mathbb{E}[X])^2$, we can calculate the expected value of $X^2$:
$$\mathbb{E}[X^2]=\mathbb{Var}[X]+(\mathbb{E}[X])^2=
\lambda+\lambda^2.$$
Expectation is linear, so we have:
$$\begin{aligned}\mathbb{E}[(X+1)^2]&=\mathbb{E}[X^2+2X+1]\\
&=\mathbb{E}[X^2]+2\mathbb{E}[X]+\mathbb{E}[1]\\
&=\lambda+\lambda^2+2\lambda+1\\ &=\lambda^2+3\lambda+1.\end{aligned}$$
A: Since $E((X+1)^2)= E(X^2)+2E(X)+1$, and I assume you know $E(X)$, the only real problem is finding $E(X^2)$. By definition, we have:
$$\sum_{k=0}^\infty k^2e^{-\mu}\frac{\mu^k}{k!} = e^{-\mu}\mu\sum_{k=1}^\infty k\frac{\mu^{k-1}}{(k-1)!} = e^{-\mu}\mu\sum_{k=0}^\infty (k+1)\frac{\mu^k}{k!} = e^{-\mu}\mu\frac{d}{d\mu}\sum_{k=0}^\infty\frac{\mu^{k+1}}{k!} = e^{-\mu}\mu\frac{d}{d\mu}\mu e^{u} = e^{-\mu}\left(\mu e^\mu+\mu^2e^\mu\right) = \mu+\mu^2$$
A: Let $\lambda = \operatorname{E}[X]$ be the Poisson mean.
$$\begin{align}
\operatorname{E}[(X+1)^2]
&= \operatorname{E}[((X - \lambda) + (\lambda + 1))^2] \\
&= \operatorname{E}[(X-\lambda)^2 + 2(\lambda + 1)(X - \lambda) + (\lambda+1)^2] \\
&= \operatorname{E}[(X - \lambda)^2] + 2(\lambda + 1)\operatorname{E}[X - \lambda] + (\lambda + 1)^2 \\
&= \operatorname{Var}[X] + 0 + (\lambda+1)^2 \\
&= \lambda + \lambda^2 + 2\lambda + 1 \\
&= \lambda^2 + 3\lambda + 1.
\end{align}$$

*

*Subtract and add $\lambda$ within the square.

*Expand the square.

*Linearity of expectation, keeping in mind that $\lambda$ is a fixed constant.

*Definition of variance, and definition of $\operatorname{E}[X]$.

*Use the fact that the Poisson variance is equal to its mean, $\lambda$.

*Simplify.

A: Let $E[(X+1)^2]=E[X^2+2X+1]=E[X^2]+2E[X]+E[1]=E[X^2]+2E[X]$ since $X$ had a Poisson distribution $E[X]=\lambda$ and too $E[X^2]=\sum_{ x \in X}{x^2p(x)}$ you should can calculate $E[X^2]$
