Minimum probability that two Gaussian random variables are small 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 $1$?
 A: 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.
A: 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 $P(X\cap Y) \ge 2\alpha -1$.
