# Integrate bivariate normal distribution over circular region

Context: Need to compute the probability that a 2D Gaussian random walk falls within distance $d$ of some point $p$ on the next step. (Assume the covariance $\Sigma$ is the identity matrix $I$.)

Considering that there's no analytic expression for the CDF of a multivariate normal (according to Wikipedia), I expect that the same is true here... If that's the case, then how can I go about approximating such an integral?

The best thing I can think of is to approximate the circular area with a portion of an annulus, which would allow me to use the CDF of the univariate standard normal distribution ($F$). So if the RW is at $S_t$ at time $t$, the probability that it falls within the annulus which touches the circular region at the next time step is

$$2 [F(|S_t-p|+d) - F(|S_t-p|-d)]$$

if $|S_t-p| > d$.

• The proof that the Gaussian distribution is a proper density uses a switch to polar coordinates, so my first guess is (given the distribution is itself circular) we could be able to get a closed form. Haven't done any calculation yet though – MichaelChirico May 8 '15 at 1:38
• **given the region of integration itself is circular – MichaelChirico May 8 '15 at 1:45
• keyword: offset circle probability – Stéphane Laurent Aug 29 '17 at 13:50

$\{(x,y) | x^2+y^2\leq d\} = \{(r,\theta) | r\leq d\}$,
$\int \int \frac{1}{2\pi} \exp\{-\frac{1}{2}(x^2+y^2)\} dy dx$
$=\intop\limits_0^{2\pi}\intop\limits_0^d \frac{1}{2\pi}\exp\{-\frac{1}{2}r^2\}r dr d\theta=1-e^{-\frac{1}{2}d^2}$
• The circular region isn't centered at the origin ($S_t$) - That's what makes this problem hard – sirallen May 8 '15 at 1:54