Re: Two intersecting gaussians?

From: Alan Miller (amiller_at_bigpond.net.au)
Date: 07/09/04 

Date: Fri, 09 Jul 2004 23:57:35 GMT

You want the probability that Z = (X - Y) > 0.
Z has a normal distribution with mean = Xmean - Ymean,
and variance = variance of X + variance of Y.
The probability that X > Y is then the probability that
a standardized normal variable with value = - Zmean / stdev(Z) is greater
than zero.

Cheers 

--
Alan Miller
Retired
Formerly with CSIRO,
Division of Mathematics & Statistics
"beliavsky@aol.com" <beliavsky@127.0.0.1:7501> wrote in message
news:40eef7d0$1_1@127.0.0.1...
>
> "Rumborak" <rumborsk2001@yahoo.com> wrote:
> >Hi all,
> >
> >I've been trying to find information on this online, but so far without
> >success. Here's my problem:
> >
> >I have two independent statistical variables (say X and Y) that produce
> >values with Gaussian distribution. They differ both in means and
variances.
> >Now, my question is: Is there a formula that tells me what the
probability
> >is of an X event having a higher value than a Y event?
> >I've been trying to do this analytically, but it fails at the integral of
> >the normal distribution (for which there is no easy, again integrable
> >solution).
> >
> >Anyone have any ideas?
> >
> >Rumborak
>
> You need the bivariate (cumulative) normal distribution, for which there
> is Fortran 77 code at
http://www.sci.wsu.edu/math/faculty/genz/papers/bvnn/node5.html
> and Fortran 90 code at http://users.bigpond.net.au/amiller/bivnorm.f90 .
>
>
>
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