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If X and Y are independent random variables with equal variances, find Cov(X+Y, X-Y).

I am confused on how to do this? I feel like I am over thinking this question.

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  • $\begingroup$ Read about the bilinearity of the covariance function. Or, just bull your way through it. Put $W=X+Y, Z=X-Y$ and write $$\operatorname{cov}((W,Z)=E[WZ]-E[W]E[Z]$$ and note that the linearity of expectation allows you to compute $E[W]=E[X+Y]$ and $E[Z] = E[X-Y]$ easily while expanding out $WZ=(X+Y)(X-Y)=X^2-Y^2$ allows you to use linearity of expectation to write $E[WZ]=E[X^2]-E[Y^2]$, etc. $\endgroup$ – Dilip Sarwate Oct 19 '14 at 23:07
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I suppose that it should be $X,Y$ are independent random variables. Then:

$$Cov(X+Y,X-Y)=E((X+Y)(X-Y))-E(X+Y)E(X-Y)=\\=E(X^2-Y^2)-(E(X)+E(Y))(E(X)-E(Y))=\\=E(X^2)-E(Y^2)-E(X)^2+E(Y^2)=\\=E(X^2)-E(x)^2-(E(Y^2)-E(Y)^2)=Var(X)-Var(Y)=0$$

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We have $$\begin{align}\operatorname{Cov}(X+Y,X-Y)&=\operatorname{Cov}(X,X-Y)+\operatorname{Cov}(Y,X-Y)\\&=\operatorname{Cov}(X,X)-\operatorname{Cov}(X,Y)+\operatorname{Cov}(Y,X)-\operatorname{Cov}(Y,Y)\\&=\operatorname{Var}(X)-\operatorname{Var}(Y).\end{align}$$

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