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/
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