Path: news.mathworks.com!not-for-mail
From: <HIDDEN>
Newsgroups: comp.soft-sys.matlab
Subject: Re: pdf
Date: Wed, 5 Dec 2012 17:39:08 +0000 (UTC)
Organization: The MathWorks, Inc.
Lines: 29
Message-ID: <k9o0rs$e0m$1@newscl01ah.mathworks.com>
References: <k9mrdr$35v$1@newscl01ah.mathworks.com>
Reply-To: <HIDDEN>
NNTP-Posting-Host: www-03-blr.mathworks.com
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
X-Trace: newscl01ah.mathworks.com 1354729148 14358 172.30.248.48 (5 Dec 2012 17:39:08 GMT)
X-Complaints-To: news@mathworks.com
NNTP-Posting-Date: Wed, 5 Dec 2012 17:39:08 +0000 (UTC)
X-Newsreader: MATLAB Central Newsreader 1187260
Xref: news.mathworks.com comp.soft-sys.matlab:784232

"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