I have $\xi$ and $\eta$ with following properties: $E\xi = E\eta = 0$, $D\xi = D\eta = 1$. And the correlation coefficient: $\rho = \rho (\xi, \eta)$.

I want to prove the following inequality:

$$ E \max (\xi^2, \eta^2) \leq 1 + \sqrt{1 - \rho^2}.$$

I don't know how to start as r.v.'s are not independent and since I can't use standard approach:

$$ P( \max (\xi^2, \eta^2) \leq x) = P( \xi^2 \leq x, \ \eta^2 \leq x ) \neq P( \xi^2 \leq x)P( \eta^2 \leq x).$$

  • $\begingroup$ Have you showed the result when $\xi$ and $\eta$ are Gaussian? $\endgroup$ – Davide Giraudo Nov 19 '12 at 12:52
  • $\begingroup$ @DavideGiraudo: no I haven't. Actually this problem is from very beginning of my probability theory book even before continuous random variables but I think it holds for any distribution. $\endgroup$ – grozhd Nov 19 '12 at 13:05

As $\max\{a,b\}=\frac 12(a+b+|a-b|)$, we just have to show that $$E|X^2-Y^2|\leqslant 2\sqrt{1-\rho^2}.$$ We have by Cauchy-Schwarz inequality that \begin{align} E|X^2-Y^2|&=E|X-Y|\cdot |X+Y|\\ &\leqslant\sqrt{E(X-Y)^2}\sqrt{E(X+Y)^2}\\ &=\sqrt{2-2\rho}\sqrt{2+2\rho}\\ &=2\sqrt{1-\rho^2}. \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.