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Suppose that $X$ and $Y$ are random variables with the equal variance.

Show that $X-Y$ and $X+Y$ are uncorrelated.

I get I should use the equation $$E[XY] = E[X]E[Y]$$ For the first part I get $$E[(X-Y)(X+Y)] = E[X^2-Y^2] = E[X^2] - E[Y^2]$$ And I don't know how to follow. Someone has any ideas?

Thank you.

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    $\begingroup$ $(X-Y)(X+Y)=X^2-Y^2$, not $X^2+Y^2$ $\endgroup$
    – user940
    Mar 22, 2016 at 14:37
  • $\begingroup$ @ByronSchmuland That's right. Editing the mistyping. Thanks, Byron. $\endgroup$
    – Manuel
    Mar 22, 2016 at 14:41

5 Answers 5

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$\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$ Recall that covariance satisfies bilinearity, and so \begin{align*} \Cov(X-Y,X+Y) &=\Cov(X,X+Y)-\Cov(Y, X+Y)\\ &=\Cov(X,X)+\Cov(X,Y)-[\Cov(Y,X)+\Cov(Y,Y)]\\ &=\Var(X)-\Var(Y)\\ &=0 \end{align*} where we recall that $\Cov(X,X) = \Var(X)$, etc. and $\Var(X) = \Var(Y)$.
Hence the correlation is $0$.

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  • $\begingroup$ Wow, this is really neat I didn't know this. $\endgroup$
    – Manuel
    Mar 22, 2016 at 15:14
  • $\begingroup$ MSE is great! You learn something everyday. Glad I could help. $\endgroup$
    – Em.
    Mar 22, 2016 at 15:32
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$$\text{Cov}(X-Y,X+Y)=E[(X-Y)(X+Y)]=E[X^2]-E[Y^2]=\text{Var}(X)-\text{Var}(Y)=0$$

where WLOG i have assumed they have $0$ mean.

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  • $\begingroup$ Hi. Why can you assume they do have 0 mean? $\endgroup$
    – Manuel
    Mar 22, 2016 at 15:13
  • $\begingroup$ @Manuel Cov doesnt change when a constant is added to a r.v. $\endgroup$
    – JKnecht
    Mar 22, 2016 at 15:21
  • $\begingroup$ Sorry, I still don't get it. $\endgroup$
    – Manuel
    Mar 22, 2016 at 15:30
  • $\begingroup$ @Manuel $\text{Cov}(X+a, Y+b) = \text{Cov}(X,Y)$. So we can assume $X$ and $Y$ have 0 mean. $\endgroup$
    – JKnecht
    Mar 22, 2016 at 15:49
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To show $X+Y$ and $X-Y$ are uncorrelated, we need to show that $$E((X+Y)(X-Y))=E(X+Y)E(X-Y).\tag{1}$$ Using a calculation similar to yours, we obtain $$E((X+Y)(X-Y))=E(X^2)-E(Y^2).\tag{2}$$ But in general we have $\text{Var}(W)=E(W^2)-(E(W))^2$. So $E(X^2)=\text{Var}(X)+(E(X))^2$ and $E(X^2)=\text{Var}(X)+(E(X))^2$. Since the variances are the same, we conclude that $$E(X^2)-E(Y^2)=(E(X))^2-(E(Y))^2.\tag{3}$$

Now we compute $E(X+Y)E(X-Y)$. This one is easy. We get $$E(X+Y)E(X-Y)=(E(X)+E(Y))(E(X)-E(Y))=(E(X))^2-(E(Y))^2.\tag{4}.$$ Thus the left side of (1) and the right side of (1) are both equal to $(E(X))^2-(E(Y))^2$, and the result follows.

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Let $X$ and $Y$ be elements in an inner product space which have the same norm. Then $X+Y$ and $X-Y$ are orthogonal. Indeed, $$(X+Y,X-Y)=\|X|^2 -\|Y\|^2=0.$$

(same formula as $(a+b)(a-b) =a^2-b^2$).

Equivalently, the two diagonals of the parallelogram generated by $X$ and $Y$ are orthogonal (draw the picture and use elementary geometry).

Now, the set of square integrable random variables with zero mean is an inner product space, equipped with $(X,Y) = E[XY]$.

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Suppose that $$ \begin{pmatrix} X\\ Y \end{pmatrix} $$ is a random vector with the covariance matrix $$ \begin{pmatrix} \sigma^2&\sigma_{X,Y}\\ \sigma_{X,Y}&\sigma^2 \end{pmatrix}. $$ The covariance matrix of the random vector $$ \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \begin{pmatrix} X\\ Y \end{pmatrix} = \begin{pmatrix} X+Y\\ X-Y \end{pmatrix} $$ is given by $$ \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \begin{pmatrix} \sigma^2&\sigma_{X,Y}\\ \sigma_{X,Y}&\sigma^2 \end{pmatrix} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} = \begin{pmatrix} 2(\sigma^2+\sigma_{X,Y})&0\\ 0&2(\sigma^2-\sigma_{X,Y}) \end{pmatrix}. $$ This shows that $X+Y$ and $X-Y$ are uncorrelated and gives their variances.

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