*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]=
∫∞
−∞
x
e(x−μ)/v
v(1+e(x−μ)/v)^2
dx
make the substitutiony=(x−μ)/v. This yields
E[X]=v
∫∞
−∞
yey
(1+ey)^2
dy+μ
∫∞
−∞
ey
(1+ey)^2
dy
=v
∫∞
−∞
yey
(1+ey)^2
dy+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)^2
dy=
∫ 0
−∞
yey
(1+ey)^2
dy+
∫∞
0
yey
(1+ey)^2
dy
=−
∫∞
0
xe−x
(1+e−x)^2
dx+
∫∞
0
yey
(1+ey)^2
dy
=−
∫∞
0
xex
(ex+1)^2
dx+
∫∞
0
yey
(1+ey)^2
dy
= 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]=μ
Thusμis the mean of the logistic;vis called the dispersion parameter.