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Topic: pdf
Replies: 1   Last Post: Dec 5, 2012 12:39 PM

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 Roger Stafford Posts: 5,929 Registered: 12/7/04
Re: pdf
Posted: Dec 5, 2012 12:39 PM

"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

Date Subject Author
12/5/12 george veropoulos
12/5/12 Roger Stafford