Suppose $E[X_1] = 2$ and $Var(X_1) = 4$. Suppose $E[X_2] = 0$ and $Var(X_2) = 1$. Suppose also that $Cor(X_1,X_2) = \frac13$. I can calculate that the expected value and variance of $X_1+X_2$ is $2$ and that the variance of $X_1+X_2$ is $6 \frac13$ and that the expected value and variance of $X_1-X_2$ are $2$ and $3 \frac23$ respectively using $E[(a_1X_1 + a_2X_2)] = a_1E[X_1] + a_2E[X_2]$ and $Var (a_1X_1 + a_2X_2) = a_1^2Var(X_1) + a_2^2Var(X_2) + 2a_1a_2Cov(X_1,X_2)$

But how do we get the correlation of $X_1+X_2$ and $X_1-X_2$ ? the solution should be $0.623$, but I have no idea how to get there?


1 Answer 1



$$Cor(X_1+X_2, X_1-X_2) = \frac{Cov(X_1+X_2, X_1-X_2)}{\sqrt{Var(X_1+X_2)Var(X_1-X_2)}}\tag{1}$$

\begin{align}Cov(X_1+X_2, X_1-X_2)&=Cov(X_1,X_1)-Cov(X_1,X_2)\\&+Cov(X_2,X_1)-Cov(X_2,X_2)\\ &= Var(X_1)-Var(X_2)\end{align}

Also, compute $Var(X_1\pm X_2)$ and then substitute inside equation $(1)$ to evaluate the correlation.


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