Logistic Distribution

Date: 10/12/2002 at 21:29:57
From: Bob
Subject: Probabilities

Hi.

I'm trying to solve the following but keep getting stalled on how to 
illustrate it. 

Let X have a logistic distribution with the following probability 
distribution function (p.d.f):

       	    e^-x
  f(x) = ---------- , 	-y < x < y
         (1+e^-x)^2

Show that

            1
  Y =  -----------         
         1 + e^-X

has a U(0,1) distribution. 

Thanks in advance. 

Date: 10/13/2002 at 06:23:26
From: Doctor Anthony
Subject: Re: Probabilities

When x = -inf, y = 0.

When x = +inf, y = 1.  

Let G(y) be the cumulative distribution of y:

   G(y) = P(Y < y) = P[1/(1+e^-X) < y)]

                   = P[1/y < 1+e^-X]

                   = P[1/y - 1 < e^-X]

                   = P[(1-y)/y < e^-X]

                   = P[e^X < y/(1-y)]
                   
                   = P[X < ln[y/(1-y)]

                        ln[y/(1-y)]
                   = INT            [e^-x/(1+e^-x)^2 dx]
                       -inf

Let

     1+e^-x = u

   -e^-x.dx = du   

so we have  

    INT[-du/u^2] = 1/u

and so

                         ln[y/(1-y)]
      G(y) =  [1/(1+e^-x)]
                        -inf 

           = [1/(1 + e^(-ln[y/(1-y)]) - 0]

           = [1/(1 + e^ln[(1-y)/y]) - 0]

           = [1/(1+ (1-y)/y)]

           = [y/(y + 1 - y)]                  

           =  y   from 0 to 1

and this is the cumulative distribution function for the rectangular 
distribution on the interval 0 < y < 1.

In this particular problem we could also have proceeded as follows:

  We showed that x=-inf, y=0    

                 x=+inf, y=1


   y = 1/(1 + e^-x)  = [1 + e^-x]^-1

                  dy = -1(1 + e^-x)^-2 (-e^-x)].dx

                             e^x
                     =   ------------  .dx
                         (1 + e^-x)^2

                     =  f(x).dx  = pdf of x distribution

and therefore dy is the pdf of the y distribution and it represents 
the uniform distribution on [0,1]

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/