Page 118 - Elements of Distribution Theory

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052184472Xc04 CUNY148/Severini May 24, 2005 2:39

104 Moments and Cumulants

Example 4.12 (Log-normal distribution). Let X denote a real-valued random variable
with an absolutely continuous distribution with density function
1 1 2
p(x) = √ exp{− [log(x)] }, x > 0;
x (2π) 2
this is an example of a log-normal distribution. Consider the integral
∞
1 1 2

exp(tx) √ exp − [log(x)] dx
0 x (2π) 2
1
∞ 2
= √ exp{tx − [log(x)] /2 − log(x)} dx.
(2π) 0
Since for any t > 0,
2
tx − [log(x)] /2 − log(x) ∼ tx as x →∞,
it follows that E[exp(tx)] =∞ for all t > 0 and, hence, the moment-generating function
of this distribution does not exist.

When moment-generating functions exist, they have many of the important properties
possessed by characteristic functions and Laplace transforms. The following theorem shows
that the moment-generating function can be expanded in a power series expansion and that
moments of the distribution can be obtained by differentiation.

Theorem 4.8. Let X denote a real-valued random variable with moment-generating func-
n
tion M X (t), |t| <δ, for some δ> 0. Then E[X ] exists and is finite for all n = 1, 2,...
and
∞
n n
M X (t) = t E(X )/n!, |t| <δ.
n=0
Furthermore,
(n)
n
E(X ) = M (0), n = 1, 2,....
X
Proof. Choose t
= 0in the interval (−δ, δ). Then E[exp{tX}] and E[exp{−tX}] are both
finite. Hence,

E[exp{|tX|}] = E[exp(tX)I {tX>0} ] + E[exp(−tX)I {tX≤0} ]
≤ E[exp{tX}] + E[exp{−tX}] < ∞.

Note that, since
∞ j
|tX|
exp{|tX|} = ,
j!
j=0
n!
n
|X| ≤ exp{|tX|}, n = 0, 1, 2,....
|t| n
n
n
It follows that E[|X| ] < ∞ for n = 1, 2,... and, hence, that E[X ]exists and is finite for
n = 1, 2,....