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

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


$(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) $

  • $\begingroup$ Does this help? $\endgroup$ – NRH May 16 '12 at 19:37
  • $\begingroup$ @NRH Could you elaborate? I understand the representation, though I'm not sure how I could exploit it within this context. $\endgroup$ – Berk U. May 16 '12 at 21:04
  • 1
    $\begingroup$ 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. $\endgroup$ – whuber May 16 '12 at 21:24
  • $\begingroup$ @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. $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ Was this meant as an answer? Can you elaborate a bit on why approach 2 is flawed and perhaps give any recommendations on what the original poster should do? $\endgroup$ – Macro May 17 '12 at 0:15
  • $\begingroup$ Yes. The OP has solved a different, simpler problem, with the second approach. Since the RV is nonstandard, the OP is really asking how to integrate a multivariate Gaussian random variable in closed form. $\endgroup$ – Emre May 17 '12 at 0:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.