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If we have some multi-dimensional normal probability distribution, $X \sim \mathcal{N}(0,\Sigma)$, with zero mean and a known covariance matrix then what is the probability that every component of $X$ is greater than or equal to zero?

$$\mathbb{P}(\forall i.X_i \ge 0) = \: ?$$

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  • $\begingroup$ If $X$ is $n$-dimensional, where $n>1$, then that probability does not make sense. $\endgroup$ Commented May 2, 2013 at 15:45
  • $\begingroup$ Presumably it means all components are $\ge 0$. $\endgroup$ Commented May 2, 2013 at 15:46
  • $\begingroup$ yes, i meant all the components. Hopefully my edit clarifies. $\endgroup$ Commented May 2, 2013 at 15:59
  • $\begingroup$ Some discussion related to orthant probabilities of normal distribution can be found in these books: link.springer.com/book/10.1007/978-1-4613-9655-0, onlinelibrary.wiley.com/doi/book/10.1002/0471722065. $\endgroup$ Commented May 18, 2020 at 19:36

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Stuart and Ord (1994, Kendall's Advanced Theory of Statistics, volume 1, Section 15.10) report symbolic solutions for the standardised bivariate Normal orthant probability $P(X>0,Y>0)$ as:

$$\text{P2}=\frac{\sin ^{-1}(\rho )}{2 \pi }+\frac{1}{4}$$

... while the standardised trivariate Normal orthant probability $P(X>0,Y>0,Z>0)$ is:

$$\text{P3}=\frac{\sin ^{-1}\left(\rho _{\text{xy}}\right)+\sin ^{-1}\left(\rho _{\text{xz}}\right)+\sin ^{-1}\left(\rho _{\text{yz}}\right)}{4 \pi }+\frac{1}{8}$$

I have found these exact symbolic solutions to be useful when testing the accuracy of multivariate Normal cdf functions (which use numerical integration to solve ... unless they know about the special cases :) )

Apparently, no such simple result is available for higher than 3 dimensions ... but there are tables, and some special results reduced to smaller numbers of integrals ...

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