# What is the probability that a multivariate Normal RV lies within a sphere of radius R?

I am currently using different procedures to estimate the probability that a $D$-dimensional Gaussian random variable with mean $\mu$ and covariance $\Sigma$ lies within a sphere of radius $R$ that is centered about the origin. That is, I am estimating $P(|| X ||_2 < R)$ where $X \sim N(\mu, \Sigma)$ and $X \in \mathbb{R}^D$.

I am wondering whether there is a way to obtain the exact value of this probability analytically (i.e. without using numerical integration or Monte Carlo)? I currently have two basic approaches to follow:

Approach 1

Find a way to analytically evaluate the integral:

$\int_{x \in S} (2\pi)^{-\frac{D}{2}}|\Sigma|^{-\frac{1}{2}} \exp(-\frac{1}{2} (x-\mu)^T \Sigma^{-1} (x-\mu) dx$

over the spherical region:

$S = \{||x|| < R \} = \{x^Tx < R^2\}$

Approach 2

Exploit the fact that if

$x \sim N(\mu,\Sigma)$

then

$(x-\mu)^T \Sigma^{-1} (x-\mu) \sim \chi^2(D)$

This implies that

$P( (x-\mu)^T \Sigma^{-1} (x-\mu) < R^2 ) = P(\chi^2(D) < R^2)$

which is very simple to evaluate...

I am hoping that there is a way to use this fact in order to evaluate:

$P( x^T x < R^2)$

-
 Does this help? – NRH May 16 '12 at 19:37 @NRH Could you elaborate? I understand the representation, though I'm not sure how I could exploit it within this context. – Berk Ustun May 16 '12 at 21:04 Approach 2 makes a mistake: it is correct only when $\mu=\mathbf{0}$. Even for $D=2$, zero correlation, and unit variances the integral is nasty: it evaluates to integrals of erf applied to trig functions, plus a Bessel function. If you need an approximation, saddlepoint methods are attractive, especially as $D$ grows. – whuber♦ May 16 '12 at 21:24 @BerkUstun, basically, the answer I linked to says that you wont find a very explicit expression for general $\mu$ and $\Sigma$ for the density (or distribution function) of $||X||^2$, which is what you seek. There is a book reference in the linked answer, which should be consulted for details. – NRH May 17 '12 at 6:34

$\mathrm P(\chi^2(D)<R^2) \neq \mathrm P(\|X\|^2_2 < R^2)$ The distributions are not the same. The second approach is easy because you standardized the random variable.