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Integral of the bivariate normal distribution over an offset circle


Hello,

Does anyone know of a closed form to the integral of the bivariate
normal
distribution over an offset circle?
I'd rather avoid the integration.

Thanks,

Yair

yair@gmail.com wrote:
> Hello,

> Does anyone know of a closed form to the integral of the bivariate
> normal
> distribution over an offset circle?
> I'd rather avoid the integration.

And who could blame you?
On May 18, 3:49 am, "yair@gmail.com" <yair@gmail.com> wrote:

> Hello,

> Does anyone know of a closed form to the integral of the bivariate
> normal
> distribution over an offset circle?
> I'd rather avoid the integration.

Since the standard bivariate normal distribution is circular, it is
simpler to express the integral in polar coordinates, and you will
need only the integral of (exp(-x^2) dx) from 0 to 1, which can be
expressed in terms of the error function, for which free Fortran codes
exist.
On 2007-05-18 09:16:05 -0300, Beliavsky <beliav@aol.com> said:

> On May 18, 3:49 am, "yair@gmail.com" <yair@gmail.com> wrote:
>> Hello,

>> Does anyone know of a closed form to the integral of the bivariate
>> normal
>> distribution over an offset circle?
>> I'd rather avoid the integration.

> Since the standard bivariate normal distribution is circular, it is
> simpler to express the integral in polar coordinates, and you will
> need only the integral of (exp(-x^2) dx) from 0 to 1, which can be
> expressed in terms of the error function, for which free Fortran codes
> exist.

The gotcha in the original posting was the word "offset". The centre
of the bivariate normal and the centre of the circle do not coincide.

Net result is an awkward problem. It is awkward even for rectangular
regions. From time to time there are papers on problems like this in
the statistical literature. I do not recall seeing circles as the
region of interest.

If you know someone who is involved in aiming problems you might ask
them as this sure sounds like an error circle with an extra source of
error.

On May 18, 9:49 am, "yair@gmail.com" <yair@gmail.com> wrote:

> Hello,

> Does anyone know of a closed form to the integral of the bivariate
> normal
> distribution over an offset circle?
> I'd rather avoid the integration.

> Thanks,

> Yair

Nitpick: I guess you meant a disk, not a circle.
I think both would be hard problems to do analytically.
But since a bivariate gaussian can be factored, you can at least
transform this to 1d integral involving erf. 1d integrals aren't that
bad.

highegg

If either (i) you are working with a relatively small circles or (ii) you really do want an analytical
expression, then another possibilty is to work with a power-series expansion of the density about the centre of
the circle (in the first case to get an approximation). This is something that might be done as an extension of
the approach suggested by Kerridge&Cook, Biometrika 63,401-3,1976. The analytical expression would turn out to
be a double-infinite summation of Hermite polynomials. Expansion about the centre of the circle means that the
contribution of odd-power terms to the integral is zero, so that you might hope to get a reasonable
approximation with only a few terms. Of course other approaches may suit the OP better.

David Jones

How about changing the coordinates of the data to be based on the
centre of the offset circle, calculate, and interpret the distribution
integral back to original origin?

Terence wrote:
> How about changing the coordinates of the data to be based on the
> centre of the offset circle, calculate, and interpret the distribution
> integral back to original origin?

It isn't obvious that a change of variables will make this integral any
easier to solve.
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