Question

Show that $X$and $Y$are identically distributed and not necessarily independent, then$Cov(X+Y,X-Y)=0.$

Short answer

It has been show that$Cov(X+Y,X-Y)=0$

Step-by-step solution

Given Information

Identically distributed and not necessarily independent variable$=X,Y$

Show$Cov(X+Y,X-Y)=0$

Explanation

We know that,

$\begin{array}{r}\mathrm{Cov}(\mathrm{X}+\mathrm{Y},\mathrm{X}-\mathrm{Y})=\mathrm{Cov}(\mathrm{X},\mathrm{X})+\mathrm{Cov}(\mathrm{X},-\mathrm{Y})+\mathrm{Cov}(\mathrm{Y},\mathrm{X})+\mathrm{Cov}(\mathrm{Y},-\mathrm{Y})\end{array}$

$=\mathrm{Var}\left(\mathrm{X}\right)-\mathrm{Cov}(\mathrm{X},\mathrm{Y})+\mathrm{Cov}(\mathrm{X},\mathrm{Y})-\mathrm{Var}\left(\mathrm{Y}\right)$

$=\mathrm{Var}\left(\mathrm{X}\right)-\mathrm{Var}\left(\mathrm{Y}\right)$

Now As $X$and $Y$are identically Distributed

$\Rightarrow Var\left(\mathrm{X}\right)=Var\left(\mathrm{Y}\right)={\sigma }^2$

And $Cov(X,Y)=Cov(Y,X)=0$

Final Answer

It has been shown that$Cov(X+Y,X-Y)=0$