Page 125 - Elements of Distribution Theory

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4.4 Cumulants 111

aX + b is given by exp(bt)M X (at), it is clear that repeated differentiation of this expression
with respect to t will lead to a fairly complicated expression; specifically, by Leibnitz’s rule
for differentiation (see Appendix 3), the nth moment of aX + b is given by
n
n n− j j j
b a E(X ), n = 0, 1,....
j
j=0
Similarly, the moment-generating of X + Y is given by M X (t)M Y (t) and differentiating this
function can lead to a complicated result.
In both cases, the situation is simplified if, instead of using the moment-generating
function itself, we use the log of the moment-generating function. For instance, if M aX+b
denotes the moment-generating function of aX + b, then

log M aX+b (t) = bt + log M X (at)
and the derivatives of log M aX+b (t)at t = 0havea relatively simple relationship to the
derivatives of log M X (at). Of course, these derivatives no longer represent the moments
of the distribution, although they will be functions of the moments; these functions of the
moments are called the cumulants of the distribution. In this section, the basic theory of
cumulants is presented.
Let X denote a real-valued random variable with moment-generating function M X (t),
|t| <δ. The cumulant-generating function of X is defined as

K X (t) = log M X (t), |t| <δ.
Since, by Theorem 4.8, M X has a power series expansion for t near 0, the cumulant-
generating function may be expanded

∞
κ j j

K X (t) = t , |t| <δ
j!
j=1
where κ 1 ,κ 2 ,... are constants that depend on the distribution of X. These constants are
called the cumulants of the distribution or, more simply, the cumulants of X. The constant
κ j will be called the jth cumulant of X;we may also write the jth cumulant of X as κ j (X).
Hence, the cumulants may be obtained by differentiation of K X ,in the same way that
the moments of a distribution may be obtained by differentiation of the moment-generating
function:
d j

κ j = K X (t) , j = 1, 2,....
dt j t=0
Example 4.17 (Standard normal distribution). Let Z denote a random variable with a
standard normal distribution. It is straightforward to show that the moment-generating
function of this distribution is given by
∞
1 1 2 2

M Z (t) = exp(tz)√ exp − z dz = exp(t /2), −∞ < t < ∞
(2π) 2
−∞
and, hence, the cumulant-generating function is given by
1 2
K Z (t) = t , −∞ < t < ∞.
2
It follows that κ 1 = 0, κ 2 = 1, and κ j = 0, j = 3, 4,....