tail probability of multivariate normal distribution

What is closed form equation for the tail probability of multivariate normal distribution ?

y ~ $\mathcal{N}$($0$, $\sigma^2I_n$ ) ; y $\in\mathbb{R}^n$

Pr(y > $\gamma$) = $\int_\gamma^\infty$ pr(y) d y ; $\gamma\in\mathbb{R}$

-Sun

-
You could ask yourself where your $\gamma$ lives? Is it in $\mathbb{R}$ (as the appearance of $\gamma$ as a bound of the integral seems to indicate) or in $\mathbb{R}^n$ (as the expression $\mathbb{y}>\gamma$ suggests)? This cannot be both... – Did Feb 4 '11 at 11:53
scope of $\gamma$ added... – user5978 Feb 4 '11 at 12:01
Then what does $\mathbb{y}>\gamma$ means? You cannot have it both ways. – Did Feb 4 '11 at 12:42
It might mean that each component of $\mathbf{y}$ is bigger than $\gamma$. That would make sense. – Raskolnikov Feb 4 '11 at 12:45
@Raskolnikov Sure, it might... Then one wonders about the one-dimensional integral on the RHS. My point is that the OP should make the question clearer. – Did Feb 4 '11 at 15:01

There are two ways to make sense of this problem: rectangular and radial.

The rectangular approach is to make $\gamma$ a vector and interpret $y > \gamma$ to mean that every component of $y$ is greater than the corresponding component of $\gamma$. This is a common problem, but it cannot be solved in closed form. See this article for numerical algorithms.

The radial approach is to ask for the probability that $|y| > \gamma$, i.e. the probability that $y$ is further than a distance $\gamma$ from the origin. That's an easier problem.

-

One way to define "tail probability" is by using the notion of rarity. Suppose an outcome has probability $p$ under some distribution $\mathcal{D}$. Its rarity is $$\Pr[\Pr[\mathcal{D}] \leq p].$$ So the rarity is the probability that the outcome of $\mathcal{D}$ is as rare as your particular outcome. In the case of a $N(0,1)$, for example, the rarity of an outcome $x$ is equal to the total weight of both tails extending from $\pm |x|$ outwards.

The nice thing is that the rarity of a random outcome of $\mathcal{D}$ (if it is continuous) is uniformly distributed (exercise). So it converts an arbitrary continuous random distribution to a uniform scale.

Given a multivariate normal distribution $N(\mu,\Sigma)$, we see that the probability of $x$ depends only on $(x-\mu)'\Sigma^{-1}(x-\mu)$. In the case where $\mu=0$ and $\Sigma=I$, we simply get $x'x$, which is a sum of squares. In the general case, there is an affine transformation which yields a $N(0,I)$ variable. We are thus led to the chi-square distribution.

The chi-square goodness-of-fit test, for example, starts with a multinomial distribution, approximates it using a multivariate distribution, and then outputs the rarity metric, which is chi-square distributed.

-