3 Independent Poisson Variables

Question

Suppose $X$, $Y$ and $Z$ are independent Poisson random variables, each with mean 1.

Find $E((X+Y)^2)$.

$E(X) = 1$

$E(Y)=1$

$E((X+Y)^2)$

$=E(X^2+2XY+Y^2)$

$=E(X^2) + E(Y^2) + E(2)E(X)E(Y)$, since $X$ and $Y$ are independent.

$= \sum_{x=0}^{\infty}x^2P(X=x)+ E(Y^2) + E(2)E(X)E(Y)$

$= \sum_{x=0}^{\infty}\frac{x^2e^{-1}1^x}{x!}+ E(Y^2) + E(2)E(X)E(Y)$

I can't simplify that summation.

Textbook Answer:

Okay, so I get they used the definition of $Var(X+Y)$ because we don't don't know $E(X^2)$ or $E(Y^2)$. But how is $Var(X+Y) = \mu =E(X+Y)??$

Answer 1

For your original approach, use $$\mathbb{E}(X^2) = \operatorname{Var}(X)+\mathbb{E}(X)^2$$

Hence, you end up with $\mathbb{E}(X^2)=1+1^2=2$. Same value for $\mathbb{E}(Y^2)$.

$$\mathbb{E}(X^2+Y^2+2XY)=2+2+2(1)(1)=6$$

We know the variance and the mean of a poisson distribution.Notice that for Poisson random variable, the variance and mean are both equal to $\lambda$.