Suppose $X, Y$ are random variables with the equal variance. Show that $X-Y$ and $X+Y$ are uncorrelated. 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.
 A: $$\text{Cov}(X-Y,X+Y)=E[(X-Y)(X+Y)]=E[X^2]-E[Y^2]=\text{Var}(X)-\text{Var}(Y)=0$$
where WLOG i have assumed they have $0$ mean.
A: $\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$ 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 $0$.
A: 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.
A: 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 $(X,Y) = E[XY]$. 
A: 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.
