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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 $\mathbb{E}Z_1Z_2=-(1-2\epsilon)\leq 0.$

Then,

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

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?

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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 $\epsilon=1$.

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  • $\begingroup$ Thanks Didier. Can you give me a reference or quick proof of the formula? Thanks. $\endgroup$
    – Hedonist
    Feb 15, 2012 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, 2012 at 22:08
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    $\begingroup$ @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, 2012 at 22:25

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