I am curious about orthant probabilities for the multivariate normal distribution for any finite dimension $n$. While Wikipedia currently doesn't seem mention these quantities the Wolfram Mathworld entry for the bivariate normal gives
$$\frac{1}{4} + \frac{1}{2 \pi} \left( \sin^{-1} \rho \right)$$
as the quadrant probability and similarly the trivariate normal octant probability is given by
$$\frac{1}{8} + \frac{1}{4 \pi} \left( \sin^{-1} \rho_{1,2} + \sin^{-1} \rho_{1,3} + \sin^{-1} \rho_{2,3} \right).$$
The more general case appears to be difficult. It was asked about here in Multivariate Normal Orthant Probability, but has equivalently been asked about in math.se in Probability that multi-dimensional random variable is positive? and Multivariate gaussian integral over positive reals.
Sometimes it is easier to evaluate a candidate guess to a problem than to rigorously derive something from first principles and definitions. Looking at the bivariate and trivariate cases led me to guess an n-dimensional generalization of these equations for the orthant probability.
$$\frac{1}{2^n} + \frac{1}{2 \pi (n-1)} \left( \sum_{\substack{i,j \in \{ 1, \cdots, n \} \\ i < j}} \sin^{-1} (\rho_{i,j}) \right)$$
I have started with some sanity checks.
- This equation agrees with the bivariate and trivariate cases.
- When there is no correlation we have $\frac{1}{2^n}$ probability equally for all orthants, which makes sense for a symmetric distribution.
The main irregularity that jumps out at me is the equation is undefined for $n=1$, but this is not necessarily a problem if we simply limit the generalization to $n \geq 2$. Of course in the $n=1$ case we should have $\frac{1}{2}$ probability either left or right of the mean, which agrees with the $\frac{1}{2^n}$ term.
Can we can rule this guess out?
Best Answer
Another check is what happens when you set one of the correlations to 1, effectively reducing the dimension from $n$ to $n-1$ (because two variables will be the same if the correlation is 1).
Examples of checks that work
Reducing from $n = 3$ to $n=2$
With $n=3$, if $\rho_{1,2}=1$ then $\rho_{2,3}=\rho_{1,3}=\rho$, so you get $$\frac{1}{8}+\frac{1}{2\pi(3-1)}\left(\sin^{-1}\rho+\sin^{-1}\rho+\frac{\pi}{2}\right)=\frac{1}{8}+\frac{1}{2\pi}\sin^{-1}\rho+ \frac{1}{8}$$ which is equal to the formula of the $n=2$ case.
Reducing from $n=2$ to $n=1$
And with $n=2$, setting the correlation to 1 reduces the formula to $1/2$, which is the $n=1$ case.
Counter example that does not work
Reducing from $n=4$ to $n=3$
With $n=4$, if you set $\rho_{1,2}=1$, you get three distinct $\rho$s: $\rho_{2,3}=\rho_{1,3}$, $\rho_{2,4}=\rho_{1,4}$ and $\rho_{3,4}$. Two of these are pairs; one isn't. So, the formula will reduce to something of the form $$A+B\left(2\sin^{-1}\rho_{2,3}+2\sin^{-1}\rho_{2,4}+\sin^{-1}\rho_{3,4}\right)$$ which can't match the $n=3$ formula.
So this general formula must be wrong.