Let $(X_1,X_2)\sim \mathcal{N}_{2}(\boldsymbol{\mu},\mathrm{\Sigma})$, where $\boldsymbol{\mu} = (\mu_1,\mu_2)^{T}$ and $\mathrm{\Sigma}$ is an appropriate variance matrix. Then how to find $\mathbb{P}(X_1>c\cap X_2>c)$?
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$\begingroup$ can't you just integrate the Gaussian over $x_1$ and $x_2$, both over the intervals $(\infty,c)$ and $(c,\infty)$? $\endgroup$– Carlo BeenakkerCommented Jul 25, 2021 at 12:45

$\begingroup$ @CarloBeenakker I think no. $\endgroup$– Satya PrakashCommented Jul 25, 2021 at 12:57

$\begingroup$ you mean "no" like "this integral has no closed form"  but that should not stop you from evaluating it numerically... $\endgroup$– Carlo BeenakkerCommented Jul 25, 2021 at 19:25

$\begingroup$ Is it correct to write $\mathbb{P}(X_1>c\cap X_2>c) = 1  \mathbb{P}(X_1\leq c)\mathbb{P}(X_1\leq c) + \mathbb{P}(X_1\leq c\cap X_2\leq c) $? $\endgroup$– Satya PrakashCommented Mar 23, 2022 at 14:28
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1 Answer
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Apparently, this probability cannot be expressed in closed form even when the means of the $X_i$'s are $0$ and their variances are $1$. At least, Mathematica cannot do anything with this probability:
 as opposed, say, to this:

$\begingroup$ you would need a closed form for the indefinite integral of $e^{x^2}\,\text{erf}(a+x)$, which is not available for $a\neq 0$. $\endgroup$ Commented Jul 25, 2021 at 19:25

$\begingroup$ @CarloBeenakker : Yes, we can of course integrate once, but not twice. $\endgroup$ Commented Jul 25, 2021 at 19:36