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?

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

  • $\begingroup$ If $X$ is $n$-dimensional, where $n>1$, then that probability does not make sense. $\endgroup$ – Stefan Hansen May 2 '13 at 15:45
  • $\begingroup$ Presumably it means all components are $\ge 0$. $\endgroup$ – copper.hat May 2 '13 at 15:46
  • $\begingroup$ yes, i meant all the components. Hopefully my edit clarifies. $\endgroup$ – ezeidman May 2 '13 at 15:59

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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