




Fortran Programming Language









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 20070518 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
highegg wrote: > 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 powerseries 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 63,4013,1976. The analytical expression would turn out to be a doubleinfinite summation of Hermite polynomials. Expansion about the centre of the circle means that the contribution of oddpower 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.  And now a word from our sponsor  Want to have instant messaging, and chat rooms, and discussion groups for your local users or business, you need dbabble!  See http://netwinsite.com/sponsor/sponsor_dbabble.htm 





