Q.7.19 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

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

Step 1: Given Information

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

Show $Cov(X+Y,X-Y)=0$

Step 2: 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 $Cov(X,Y)=Cov(Y,X)=0$

Step 3: Final Answer

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