# Probability that multi-dimensional random variable is positive?

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?

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

• If $X$ is $n$-dimensional, where $n>1$, then that probability does not make sense. Commented May 2, 2013 at 15:45
• Presumably it means all components are $\ge 0$. Commented May 2, 2013 at 15:46
• yes, i meant all the components. Hopefully my edit clarifies. Commented May 2, 2013 at 15:59
• 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. Commented May 18, 2020 at 19:36

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