Covariance of two Poisson Distributions Given two independent Variables X and Y with  
$ X \sim Poi_\lambda, Y \sim Poi_{2\lambda}$ 
and $Z = X + Y$
How to calculate the Covariance of X and Z?
I know $Z \sim Poi_{3\lambda}$
I have: $C(X, Z) = E(XZ) - E(X)E(Z)$ 
So I was trying to calculate E(XZ) first, but got stuck with:
$$E(XZ) = \sum_{n = 0, \frac{n}{k} \in N}^k \frac{\lambda^k \cdot (3\lambda)^{\frac{n}{k}}\cdot e^{-4}}{n! \cdot \frac{n}{k}!}$$
But I am not even sure, I haven't messed up already.
It should be way easier, if I'd be asured that X and Z are independent, so $E(XZ)$ would be $E(X) \cdot E(Z)$ and with that $C(X, Z) = 0$
 A: $$
\operatorname E(XZ)=\operatorname E(X^2+XY)=\operatorname EX^2+\operatorname E(XY)
$$
and, since $\operatorname{Var}X=\operatorname EX^2-(\operatorname EX)^2=\lambda$, we have that
$$
\operatorname E(XZ)=\lambda+3\lambda^2.
$$
Finally,
$$
\operatorname{Cov}(X,Z)=E(XZ)-\operatorname EX\operatorname EZ=\lambda+3\lambda^2-3\lambda^2=\lambda.
$$
Edit. As Michael Hoppe suggested, let $X$ and $Y$ be two random variables with $\operatorname EX^2<\infty$ and $\operatorname{Cov}(X,Y)=0$. Set $Z=X+Y$. Then
$$
\operatorname{Cov}(X,Z)=\operatorname{Cov}(X,X)+\operatorname{Cov}(X,Y)=\operatorname {Var}X.
$$
In our case, $\operatorname{Cov}(X,Y)=0$ since $X$ and $Y$ are independent and $\operatorname EX^2<\infty$ since the Poisson distribution has moments of all orders.
A: The followed makes more sense to me.
Cov(X,Z) = E(XZ)-E(X)E(Z)
         = E(X(X+Y)) - E(X)E(X+Y)
         = E(X^2)+E(XY)-E(X)^2-E(X)E(Y)
         = E(X^2)-E(X)^2 + E(XY)-E(X)E(Y)
         = Var(X) + Cov(X,Y)
Since X and Y are independent, Cov(X,Y)=0
So, Cov(X,Z) = Var(X)
