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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 $.

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  • $\begingroup$ 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 $\endgroup$ – MichaelChirico May 8 '15 at 1:38
  • $\begingroup$ **given the region of integration itself is circular $\endgroup$ – MichaelChirico May 8 '15 at 1:45
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    $\begingroup$ keyword: offset circle probability $\endgroup$ – Stéphane Laurent Aug 29 '17 at 13:50
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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

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

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

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    $\begingroup$ The circular region isn't centered at the origin ($S_t$) - That's what makes this problem hard $\endgroup$ – sirallen May 8 '15 at 1:54
  • $\begingroup$ I see. then perhaps use a tangent plane approximation? $\endgroup$ – MichaelChirico May 8 '15 at 2:12

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