$X\sim\mathrm{Poisson}(\lambda_1 = 5)$ and $Y\sim\mathrm{Poisson}(\lambda_2 = 15)$. Let $Z = X + Y$ . Compute $\mathrm{Corr}(X, Z)$.

Question

Let $X$ and $Y$ be independent random variables such that $X\sim\mathrm{Poisson}(\lambda_1 = 5)$ and $Y\sim\mathrm{Poisson}(\lambda_2 = 15)$. Let $Z = X + Y$ . Compute $\mathrm{Corr}(X, Z)$.

Answer:

$\mathrm{Var}(X) = \lambda_1 = 5$

$\mathrm{Var}(Y) = \lambda_2 = 15$

$\mathrm{Cov}(Z) = \mathrm{Cov}(X, X+Y) = \mathrm{Cov}(X, X) = V(X) = 5$

$\mathrm{Var}(Z) = \mathrm{Var}(X + Y) = V(X) + V(Y) = 5 + 15 = 20$

$\mathrm{Corr}(X, Z) = \dfrac{\mathrm{Cov}(X, Z)}{\sqrt{\mathrm{Var}(X)} \sqrt{\mathrm{Var}(Z)}} = \dfrac{5}{\sqrt{5 \cdot 20}} = 0.5$

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I get most of the solution but I'm confused on how they got $\mathrm{Cov}(Z)$

$\mathrm{Cov}(Z) = \mathrm{Cov}(X, X + Y) = \mathrm{Cov}(X + Y, X) = \mathrm{Cov}(X, X) + \mathrm{Cov}(X, Y) \neq \mathrm{Cov}(X, X)$

Could someone explain how they got $\mathrm{Cov}(Z)$?

Answer 1

It does not concern $\mathsf{Cov}(Z)$ (which is even not properly defined) but $\mathsf{Cov}(X,Z)$.

It is found by using that covariance is bilinear:$$\mathsf{Cov}(X,Z)=\mathsf{Cov}(X,X+Y)=\mathsf{Cov}(X,X)+\mathsf{Cov}(X,Y)=\mathsf{Var}X+0=\mathsf{Var}X=5$$

The second equality is on base of linearity. The third on base of definition and on base of independence of $X$ and $Y$.

Answer 2

$Cov (X,X+Y)=EX(X+Y)-(EX)E((X+Y))$ You can write this as $EX^{2}+EXY-((EX)^{2}+(EX)(EY))=EX^{2}-(EX)^{2}$ since $EXY=EXEY$ by independence.