Calculate the Covariance of random variables that distribute normally

Question

$X_1$ and $X_2$ are two independent random variables that distribute normally with mean $μ$ and variance $σ^2$.

$Y_1 = X_1 + 2X_2$

$Y_2 = X_1 - 2X_2$

Calculate $Cov(Y_1,Y_2)$.

Well, I can tell that:

$\mathbb{E}[Y_1] = \mathbb{E}[X_1 + 2X_2] = \mathbb{E}[X_1] + 2\mathbb{E}[X_2] = μ+2μ = 3μ$ $\mathbb{E}[Y_2] = \mathbb{E}[X_1 - 2X_2] = \mathbb{E}[X_1] - 2\mathbb{E}[X_2] = μ-2μ = -μ$

$Cov(Y_1,Y_2) = \mathbb{E}[Y_1Y_2]-\mathbb{E}[Y_1]\mathbb{E}[Y_2] = \mathbb{E}[Y_1Y_2] +3μ^2$

What now? How do I calculate $\mathbb{E}[Y_1Y_2]$?

Answer 1

If you want to continue with your approach, just plug in the definitions of $Y_1:=X_1+2X_2$ and $Y_2:=X_1-2X_2$, multiply out, and collect terms.

Then use the fact that $E(X^2)=Var(X) + [E(X)]^2$ for any random variable $X$, and that if $X_1$ and $X_2$ are independent, then $E(X_1X_2)=E(X_1)E(X_2)$.

Answer 2

let ${{Y}_{1}}=\sum\limits_{i=1}^{n}{{{a}_{i}}}{{X}_{i}}\,$ and $\,{{Y}_{2}}=\sum\limits_{i=1}^{n}{{{b}_{i}}}{{X}_{i}}$. If $X_1,X_2,\cdots,X_n$ are independent then $$Cov({{Y}_{1}},{{Y}_{2}})=\sum\limits_{i=1}^{n}{{{a}_{i}}{{b}_{i}}\operatorname{var}({{X}_{i}})}$$ as a result

$$Cov(Y_1,Y_2)=Var(X_1)-4Var(X_2)=-3\sigma^2$$