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Date: Wed, 5 Dec 2012 17:39:08 +0000 (UTC)
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"george veropoulos" <veropgr@yahoo.gr> wrote in message <k9mrdr\$35v\$1@newscl01ah.mathworks.com>...
> I  have  a  variable  ?  tha  is  function  of  randon  variable  ? (?=F(?), F  is  a  complex
> function !)
> (the  variable  ?   is  uniformly  distributed   in the interval [-? ?])
> ?y  question  is   who  i  cant  find  the  distribution  of   ?  variable .
- - - - - - - - - -
The text of your message did not come through very well, George.  I am guessing that you have a (complicated) function x = F(u) where u is a random variable uniformly distributed on the interval [a,b], and you wish to know the probability density of x over its corresponding span.  Is that correct?

If we assume F is monotone increasing, then its inverse would be a function G where u = G(x).  The probability density would then be:

p(x) = dG(x)/dx * 1/(b-a) .

This means that you have to be able to find the derivative of the inverse of F.

As an example, suppose x = F(u) = u^2 and [a,b] = [2,5].  Then the inverse function would be

u = G(x) = sqrt(x)

and the density would be

p(x) = dG(x)/dx * 1/(b-a) = 1/(2*sqrt(x)) * 1/3 = 1/(6*sqrt(x))

To verify this, take the integral of p(x) with respect to x from x = F(2) to x = F(5):

int(1/(6*sqrt(x)),'x',2^2,5^2) = 1/3*sqrt(5^2)-1/3*sqrt(2^2) = 1 ,

and indeed 1 is the value it should have.

Roger Stafford
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