# Expected value of 2 Poisson distributions

Let $X$ and $Y$ be independet Poisson random variables with parameters $\lambda$ and $\mu$.

I have to calculate $E((X+Y)^2)$ .

What I did: $E[(X+Y)^2]=E[X^2]+E[Y^2]+2EXEY$

I know that $2EXEY=2\lambda\mu$, but I don't know how to calculate the squared expected values.

Suppose that $X\sim\mathrm{Pois}(\lambda)$. Then $$\operatorname{Var}X=\operatorname EX^2-(\operatorname EX)^2=\lambda.$$ Hence, $$\operatorname EX^2=\lambda+\lambda^2.$$

Hint: $X+Y$ is a Poisson variable with parameter $\lambda+\mu$.

details:

indeed, $$Es^{X+Y} = Es^XEs^Y = \sum_k \exp(-\lambda)\frac{\lambda ^k}{k!} s^k \sum_l \exp(-\mu)\frac{\mu^l}{l!}s^l \\ = \exp (\lambda(s-1))\exp (\mu(s-1)) = \exp ((\lambda+\mu)(s-1))$$

Then $$E[(X+Y)^2] = Var(X+Y) + [E(X+Y)]^2 = (\lambda+\mu) + (\lambda+\mu)^2$$

$Var(X)=E[(X-\mu)^2]=E[X^2]-2E[X]\mu+\mu^2=E[X^2]-(E[X])^2$

$E[X^2]=Var(X)+(E[X])^2$

For Poisson distribution you have:

$E[X^2]=\lambda+\lambda^2$

Now do the same for Y and that's it...