Fix $\epsilon, 0\leq \epsilon\leq 1/2.$ Let $Z_1,Z_2$ be zero mean, unit variance Gaussian random variables which are jointly Gaussian with $\mathbb{E}Z_1Z_2=-(1-2\epsilon)\leq 0.$


$$P(Z_1Z_2>0)\geq \epsilon.$$

This curious fact popped out of some calculations I was doing using the Central Limit Theorem. I wonder if it can be proved in some easy way by directly working with the double integral.

I also believe that the inequality is strict when $0<\epsilon<1/2.$ Can this be shown?


Since $P(Z_1Z_2>0)=\frac12-\frac1{\pi}\arcsin(1-2\epsilon)$, indeed $P(Z_1Z_2>0)\geqslant\epsilon$, and the inequality is strict except when $\epsilon=0$, $\epsilon=\frac12$ and $\epsilon=1$.

| cite | improve this answer | |
  • $\begingroup$ Thanks Didier. Can you give me a reference or quick proof of the formula? Thanks. $\endgroup$ – Hedonist Feb 15 '12 at 22:07
  • $\begingroup$ @DidierPiau: I think the case of $\epsilon = 1/2$ should be added to the values at which equality is attained. $\endgroup$ – cardinal Feb 15 '12 at 22:08
  • 1
    $\begingroup$ @Hedonist: Here's one way. $Z_1 Z_2 = \rho X_1^2 + \sqrt{1-\rho^2} X_1 X_2$ in distribution where $X_1$ and $X_2$ are iid standard normal and $\rho = -(1-2\epsilon)$. Then $\mathbb P(Z_1 Z_2 > 0 ) = \mathbb P(X_2/|X_1| > -\rho/\sqrt{1-\rho^2})$ and $X_2/|X_1|$ is standard Cauchy. A tiny bit of algebra then gets you the conclusion. $\endgroup$ – cardinal Feb 15 '12 at 22:25

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.