Page 140 - Elements of Distribution Theory

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126 Moments and Cumulants

This result can now be used to express the unconditional cumulants in terms of the con-
ditional ones. Although a general expression relating the conditional and unconditional
cumulants may be given, here we simply outline the approach that may be used.
Consider κ 1 . Note that
m j−1
∂ t
2 m
K (t) = K m (t, t /2,..., t /m!) . (4.6)
∂t j ( j − 1)!
j=1
denote the joint cumulant of order (i 1 ,..., i m )of(κ 1 (Y),...,κ m (Y)); we will
Let ¯κ i 1 ···i m
use the convention that trailing 0s in the subscript of ¯κ will be omitted so that, for example,
¯ κ 10···0 will be written ¯κ 1 . Then evaluating (4.6) at t = 0 shows that

κ 1 = K (0) = ¯κ 1 ;
that is, the first cumulant of X is the first cumulant of κ 1 (Y). Of course, this is simply the
result that
E(X) = E[E(X|Y)].

Now consider the second cumulant of X.We may use the same approach as that used
above; the calculation is simplified if we keep in mind that any term in the expansion of
K (t)in terms of the derivatives of K m that includes a nonzero power of t will be 0 when

evaluated at t = 0. Hence, when differentiating the expression in (4.5), we only need to
consider

d ∂ 2 m ∂ 2 m
K m (t, t /2,..., t /m!) + K m (t, t /2,..., t /m!)t
dt ∂t 1 ∂t 2
t=0

∂ 2 2 m ∂ 2 m
= 2 K m (t, t /2,..., t /m!) + K m (t, t /2,..., t /m!) .
∂t ∂t 2
1 t=0 t=0
It follows that
κ 2 ≡ K (0) = ¯κ 2 + ¯κ 01 ;

that is,

Var(X) = Var[E(X|Y)] + Var[E(X|Y)].
The expressions for the higher-order cumulants follow in a similar manner. We may
obtain K (0) by taking the second derivative of

∂ 2 m ∂ 2 m
K m (t, t /2,..., t /m!) + K m (t, t /2,..., t /m!)t
∂t 1 ∂t 2
∂ 2 m 2
+ K m (t, t /2,..., t /m!)t /2
∂t 3
at t = 0. The result is

κ 3 = ¯κ 3 + 3¯κ 11 + ¯κ 001 .

Example 4.28 (Poisson random variable with random mean). Let X denote a Poisson
random variable with mean Y and suppose that Y is a random variable with a standard