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

Question

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

Answer:

$\operatorname{Var}(X) = 5$

$\operatorname{Var}(Y) = 15$

$\operatorname{Cov}(Z) = \operatorname{Cov}(X, X+Y) = \operatorname{Cov}(X, X) = \operatorname{Var}(X) = 5$

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

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

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I'm just wondering if this statement is true for all covariance values or just if independent.

$\operatorname{Cov}(Z) = \operatorname{Cov}(X, X+Y) = \operatorname{Cov}(X, X) = \operatorname{Var}(X) = 5$

Couldn't find it on wiki

Answer 1

$$Cov(X,Z) = Cov(X,X+Y)$$ Let $E(X) = \mu_X$ and $E(Y) = \mu_Y$ then $$Cov(X,X+Y) = E(X-\mu_X)(X-\mu_X + Y - \mu_Y) $$ which is $$Cov(X,X+Y) = E((X-\mu_X)^2) + E(X-\mu_X)(Y-\mu_Y) $$ hence $$Cov(X,X+Y)= Var X + E(XY) - \mu_XEY -\mu_YEX +\mu_X\mu_Y $$ Due to independence of $X$ and $Y$, we have $$Cov(X,X+Y) = Var X + E(X)E(Y) - \mu_X\mu_Y -\mu_Y\mu_X +\mu_X\mu_Y$$ i.e. $$Cov(X,X+Y) = Var X + \mu_X\mu_Y- \mu_X\mu_Y -\mu_Y\mu_X +\mu_X\mu_Y = Var X$$

Answer 2

Lemma. $$\text{cov}(X, Y) = 0$$ if $X$ and $Y$ are independent.

Proof. $$\text{cov}(X, Y) = \mathop{\mathbb{E}}[XY] - \mathop{\mathbb{E}}[X]\mathop{\mathbb{E}}[Y] $$

which is 0 as the expected value of a product of random variables is the product of the expected values of those variables.

Original question

Now back to the original question:

$$\begin{align} \text{cov}(X, Z) &= \text{cov}(X,X-Y)\\ &= \text{cov}(X,X)-\text{cov}(X,Y)\\ &= \text{var}(X,X) - 0\\ &= 5 \end{align}$$