*5.9The Logistics Distribution 193
DifferentiatingF(x)= 1 −1/(1+e(x−μ)/v) yields the density function
f(x)=e(x−μ)/v
v(1+e(x−μ)/v)^2, −∞<x<∞To obtain the mean of a logistics random variable,
E[X]=∫∞−∞xe(x−μ)/v
v(1+e(x−μ)/v)^2dxmake the substitutiony=(x−μ)/v. This yields
E[X]=v∫∞−∞yey
(1+ey)^2dy+μ∫∞−∞ey
(1+ey)^2dy=v∫∞−∞yey
(1+ey)^2dy+μ (5.9.1)where the preceding equality used thatey/((1+ey)^2 ) is the density function of a logistic
random variable with parametersμ=0,v=1 (such a random variable is called astandard
logistic) and thus integrates to 1. Now,
∫∞−∞yey
(1+ey)^2dy=∫ 0−∞yey
(1+ey)^2dy+∫∞0yey
(1+ey)^2dy=−∫∞0xe−x
(1+e−x)^2dx+∫∞0yey
(1+ey)^2dy=−∫∞0xex
(ex+1)^2dx+∫∞0yey
(1+ey)^2dy= 0 (5.9.2)where the second equality is obtained by making the substitutionx=−y, and the third
by multiplying the numerator and denominator bye^2 x. From Equations 5.9.1 and 5.9.2
we obtain
E[X]=μThusμis the mean of the logistic;vis called the dispersion parameter.