cumulative bivariate normal distribution as a function of radius

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hi, im stuck on a problem and looking for a bit of help.  I have a bivariate normal distribution- mean= [0, 0] , and standard deviation equal in x and y directions.  Im trying to find a function or routine that will return the cumulative normal as a function of radius or distance form [0, 0].   i think mvncdf is only good to me for square regions.  any clue on how to do this?
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Reply sam 8/18/2010 2:35:23 PM

"sam Zughni" <samirzughni@gmail.com> wrote in message <i4gr3b$n37$1@fred.mathworks.com>...
> hi, im stuck on a problem and looking for a bit of help.  I have a bivariate normal distribution- mean= [0, 0] , and standard deviation equal in x and y directions.  Im trying to find a function or routine that will return the cumulative normal as a function of radius or distance form [0, 0].   i think mvncdf is only good to me for square regions.  any clue on how to do this?
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  You didn't say whether the two distributions are independent.  Assuming that they are and assuming that by "cumulative normal" you mean the probability that both distributions lie inside a circle with center (0,0) and radius r, then you can find it by doing the integration.  The answer will be

 P(x^2+y^2<=r^2) = 1 - exp(-r^2/(2*sigma^2))

  This comes from the double integral of p(x)*p(y) taken over the circle x^2+y^2<=r^2 where p(x) and p(y) are the two pdf's.  You change variables from x and y to polar coordinates and you have integrals that are easy to solve directly by calculus methods.

Roger Stafford
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Reply Roger 8/18/2010 5:32:09 PM


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