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### 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.

```

```
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/
```
Associated Topics:
College Probability

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