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| &gt; d $.
I may have made some sloppy mistakes, but I think this is the gist of it:

$\{(x,y) | x^2+y^2\leq d\} = \{(r,\theta) | r\leq d\}$,

So

$\int \int \frac{1}{2\pi} \exp\{-\frac{1}{2}(x^2+y^2)\} dy dx$

$=\intop\limits_0^{2\pi}\intop\limits_0^d \frac{1}{2\pi}\exp\{-\frac{1}{2}r^2\}r dr d\theta=1-e^{-\frac{1}{2}d^2}$
I may have made some sloppy mistakes, but I think this is the gist of it:

$\{(x,y) | x^2+y^2\leq d\} = \{(r,\theta) | r\leq d\}$,

So

$\int \int \frac{1}{2\pi} \exp\{-\frac{1}{2}(x^2+y^2)\} dy dx$

$=\intop\limits_0^{2\pi}\intop\limits_0^d \frac{1}{2\pi}\exp\{-\frac{1}{2}r^2\}r dr d\theta=1-e^{-\frac{1}{2}d^2}$