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