Independence of Poisson distributed random variables

Question

Let $V, W, Z$ be independent Poisson distributed random variables. We now set $X:=V+W$ and $Y:=V+Z$.

• Compute $Cov(X,Y)$.
• Are $X$ and $Y$ independent?
• Find the probability mass function (PMF) of $X$ and $Y$.

I already have difficulties solving the first task. Covariance is defined as $Cov(X,Y)=E[XY]-E[X]E[Y]$. I already know that $E[X]=\lambda + \mu$ and $E[Y]=\lambda + \nu$. How do I get $E[XY]$? Do I have to solve: $\frac{(\lambda + \mu)^{k}}{k!}\cdot e^{-\lambda + \mu} \cdot \frac{(\lambda + \nu)^{l}}{l!}\cdot e^{-\lambda + \nu}$? If so, could somebody please give me a hint?

Best, Jolle

Answer 1

Hint: Use the Bilinearity of Covariance: For any random variables, $A,B,C,D$, we have ... $$\mathsf{Cov}(A+B,C+D)=\mathsf{Cov}(A,C)+\mathsf{Cov}(A,D)+\mathsf{Cov}(B,C)+\mathsf{Cov}(B,D)$$And, of course, you should know that the covariance of independent random variables is zero, so...

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As for (c)

$\mathsf P(X=x,Y=y)=\sum_{v=0}^{\min\{x,y\}}\mathsf P(V=v)\mathsf P(W=x-v)\mathsf P(Z=y-v)$

Answer 2

Hint: covariance is bilinear,so:

$$\mathsf{Cov}(X,Y)=\mathsf{Cov}(V+W,V+Z)=\mathsf{Cov}(V,V)+\mathsf{Cov}(V,Z)+\mathsf{Cov}(W,V)+\mathsf{Cov}(W,Z)=$$$$\mathsf{Var}(V)+0+0+0=\mathsf{Var}(V)$$