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Thread Subject:
pdf

Subject: pdf

From: george veropoulos

Date: 5 Dec, 2012 07:00:11

Message: 1 of 2

Dear friends

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 .


Thank you in advance

George Veropoulos

Subject: pdf

From: Roger Stafford

Date: 5 Dec, 2012 17:39:08

Message: 2 of 2

"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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