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Let $X,Y$ be two centered Gaussian random variables each with variance at most $1$. Note that we do not assume independence. I would like to minimize $$\mathbb{P}(|X|\leq 1, |Y|\leq 1).$$ Is it true that the latter quantity is minimized when $X,Y$ are independent and both have variance $1$?

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  • $\begingroup$ Doesn't this follow simply from the Gaussian Correlation Inequality? en.wikipedia.org/wiki/Gaussian_correlation_inequality $\endgroup$ – Suvrit Nov 15 '17 at 0:40
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    $\begingroup$ You don't say if $X$ and $Y$ have a joint Gaussian distribution, or just are are individually Gaussian (which is a lot more general). $\endgroup$ – Brendan McKay Nov 15 '17 at 0:48
  • $\begingroup$ @Suvrit, as I see it, the Gaussian correlation inequality assumes a multinormal distribution; while here, the only assumption is that the marginal distributions are normal. $\endgroup$ – Matt F. Nov 15 '17 at 1:27
  • $\begingroup$ mathoverflow.net/questions/172310/… $\endgroup$ – fedja Nov 15 '17 at 2:25
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The minimum value is simply $2\alpha-1 = 0.365379$ where $\alpha = \Phi(1)-\Phi(-1) = P(|X|<1)$ where $X \sim N(0,1)$. This can be achieved by translating the percentile of $X$ (considering the percentile $\mod 1$) to produce the percentile of $Y$. For example, let $T=\Phi(X)+\alpha \mod 1$ and then $Y=\Phi^{-1}(T)$. Other translations work, too.

This is optimal because $P(|X|<1)= \alpha = P(|Y|<1)$ and $1 \ge P(X \cup Y) = P(X) + P(Y)-P(X \cap Y) = 2 \alpha - P(X \cap Y)$ so $P(X\cap Y) \ge 2\alpha -1$.

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The fraction in the central square is not minimized with independent standard normals.

Start with that distribution. Take a pair of points $(x,y)$ and $(u,v)$ with $|x|<1, |y|<1, |u|>1, |v|>1,\ p(x,y)>p(u,v)$. Reduce the probabilities near those points by $p(u,v)$, and increase the probabilities near $(x,v)$ and $(u,y)$ by the same amount.

Repeating this will maintain the marginal distributions, while decreasing the part of the distribution in the central square from 46.6% to 36.5%. It may also be possible to reduce the percentage further.

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