Let $X\sim\mathrm N(\mu,\Sigma)$ with $\mu\in\mathbb R^d$ and $\Sigma\in\mathbb R^{d\times d}$ (non-diagonal). Is there an easy way to obtain the probability every coordinate of $X$ is positive? $$\mathbb P\left(X\ge 0\right)=\mathbb P\left(X_1\ge 0,\ldots,X_d\ge 0\right)$$ As far as I know, this problem typically involves heavy numerical computation for a generic region. However, I hoped there might be a simpler solution for the abovementioned region.
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No, you can't. If the solution you seek existed, we could calculate easily all cumulative multivariate normal probabilities (which currently cannot be calculated without numerical methods) – NN2 Oct 04 '24 at 12:05
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See https://math.stackexchange.com/q/379378/321264 and linked posts. – StubbornAtom Oct 04 '24 at 19:30
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Unfortunately, there couldn't be a simple formula in general, as you can always transfer $X$ to the collection of independent standard normal distributions by setting $$ Y=\Sigma^{-1/2}(X-\mu) $$ but this transformation will transfer your orthant $A=\{x\in\mathbb{R}^d: x_1,x_2,\dots,x_d\ge 0\}$ in fairly general ``cone" $$ A'=\left\{y\in\mathbb{R}^d:\ \sum_{j=1}^d c_{ij} y_j\ge d_i,\ i=1,2,\dots,d\right\} $$ for some $c,d$.
van der Wolf
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