# Let $X_1$ and $X_2$ be independent random variables Let $Y_1 = X_1 + X_2$ and $Y2 = 2X_1-2X_2$, find covariance of $Y_1,Y_2$

Let $X_1$ and $X_2$ be independent random variables Let $Y_1 = X_1 + X_2$ and $Y_2 = 2X_1-2X_2$, find covariance of $Y_1,Y_2$

I tried simplifying the covariance equation into $$E[Y_1Y_2] - E[X_1]\cdot E[X_2]$$ I found the expected values of $x_1$ to be $0$ but I don't know how to find $E[Y_1Y_2]$. I tried simplifying it to: $$E[(2X_1-X_2) (X_1+X_2)]$$ but that didn't really help. Any help would be greatly appreciated. Thanks.

$\newcommand{\v}{\operatorname{var}}\newcommand{\c}{\operatorname{cov}}$ \begin{align} \c(Y_1,Y_2) & = \c(X_1+X_2, 2X_1-2X_2) \\[10pt] & = \c(X_1,2X_1-2X_2) + \c(X_2, 2X_1-2X_2) \\[10pt] & = \Big(\c(X_1,2X_1) + \c(X_1,-2X_2)\Big) + \Big( \c(X_2,2X_1) + \c(X_2,-2X_2)\Big) \\[10pt] & = 2\c(X_1,X_1) - 2\c(X_1,X_2) + 2\c(X_2,X_1) - 2\c(X_2,X_2) \\[10pt] & = 2\v(X_1) - 0 + 0 - 2\v(X_2). \end{align}

• Doing it the way you showed, it looks like you'd get 0 as your answer but the answer is 1? Commented Jun 23, 2017 at 23:59
• Are the variances of $X_1$ and $X_2$ known to be equal? Commented Jun 24, 2017 at 1:34
• @Mraie : The answer depends on what the two variances are. You didn't tell us that. Commented Jun 24, 2017 at 3:14

I tried simplifying the covariance equation into $$E[Y_1Y_2] - E[X_1]\cdot E[X_2]$$ I found the expected values of $x_1$ to be $0$ but I don't know how to find $E[Y_1Y_2]$. I tried simplifying it to: $$E[(2X_1-X_2) (X_1+X_2)]$$ but that didn't really help. Any help would be greatly appreciated.

(You have the 2 in the wrong place.)

Well, doing it this way we get:

$$\def\Cov{\mathsf {Cov}}\def\E{\mathsf {E}}\def\Var{\mathsf {Var}} {\Cov(X_1+X_2,2X_1-2X_2) ~}{= 2\Cov(X_1+X_2,X_1-X_2) \\ =2 \Big(\E\big((X_1+X_2)(X_1-X_2)\big)-\E(X_1+X_2)\,\E(X_1-X_2)\Big) \\ = 2\big(\E(X_1^2)-\E(X_2^2)-\E(X_1)^2+\E(X_2)^2\big) \\ = 2\big(\Var(X_1)-\Var(X_2)\big) }$$

Which is the same result Michael Hardy obtained by no coincidence.

Remark: The independence of the random variables was not invoked, and is in fact not necessary for the result to hold.