|
|
cumulative bivariate normal distribution as a function of radius
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?
|
|
0
|
|
|
|
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?
- - - - - - - - - -
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
|
|
0
|
|
|
|
Reply
|
Roger
|
8/18/2010 5:32:09 PM
|
|
|
1 Replies
579 Views
(page loaded in 0.026 seconds)
|
|
|
|
|
|
|
|
|