Prove that $E(|(X+Y)(X-Y)|) \leq 2\sqrt{1-\rho²}$, where $\rho$ is correlation.

For two random variables $X,Y$ with mean $0$ and variance $1$, their correlation is $\rho$. We have to prove that

$$E(|(X+Y)(X-Y)|) \leq 2\sqrt{1-\rho^2}.$$

But, I can't understand how the $\rho$ will come under root.

Solution 1:

Hint: By Cauchy-Schwarz, we have $E\Big(\big|(X+Y)(X-Y)\big|\Big)\leq \sqrt{E\Big(|X+Y|^2\Big)\cdot E\Big(|X-Y|^2\Big)}$.

Solution 2:

Recall that $$\operatorname{Cov}(X,Y) = \mathbb E[XY] - \mathbb E[X]\mathbb E[Y] $$ and $$\rho = \frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var}(X)}\sqrt{\operatorname{Var}(y)}}.$$ Now as $\mathbb E[X]=\mathbb E[Y] = 0$, we have $$\operatorname{Cov}(X,Y) = \mathbb E[XY],$$ and as $\operatorname{Var}(X)=\operatorname{Var}(Y)=1$, it follows that $$\rho=\operatorname{Cov}(X,Y)=\mathbb E[XY].$$ By Hölder's inequality, we have \begin{align} \mathbb E[|(X+Y)(X-Y)|] &\leqslant \left(\mathbb E[|X+Y|^2]\mathbb E[|X-Y|^2]\right)^{\frac12}\\ &=\left((\mathbb E[X^2]+2\mathbb E[XY] + \mathbb E[Y^2])(\mathbb E[X^2]-2\mathbb E[XY]+\mathbb E[Y^2])\right)^{\frac12}\\ &=\left((2+2\mathbb E[XY])(2-2\mathbb E[XY] )\right)^{\frac12}\\ &= \left(4(1+\rho)(1-\rho)\right)^{\frac12}\\ &= 2(1-\rho^2)^{\frac12}. \end{align}