Page 129 - Elements of Distribution Theory

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4.4 Cumulants 115

Hence, we may write

m
m
log M X (t) = h(t;E(X),..., E(X )) + o(t )
for some function h. Since ϕ X (t) has the same expansion as M X (t), except with it replacing
t,we must have
m
m
log ϕ X (t) = h(it;E(X),..., E(X )) + o(t ).
It follows that
d m d m

= log ϕ X (t) .
dt m t=0 d(it) m t=0
log M X (t)
The result now follows from the fact that
d m d m
= i .
m
dt m log ϕ X (t) t=0 d(it) m log ϕ X (t) t=0
Therefore, the log of the characteristic function has an expansion in terms of the cumu-
lants that is similar to the expansion of the characterstic function itself in terms of moments.

Theorem 4.13. Let X denote a real-valued random variable and let ϕ X denote the charac-
m
teristic function of X. If E(|X| ) < ∞, then
m
j m
log(ϕ X (t)) = (it) κ j /j! + o(t ) as t → 0
j=1
where κ 1 ,κ 2 ,...,κ m denote the cumulants of X.
If, for some m = 1, 2,..., ϕ (2m) (0) exists then κ 1 ,κ 2 ,...,κ 2m all exist and are finite.

m
Proof. We have seen that if E(X )exists and is finite, then
m j
(it) j m
ϕ(t) = 1 + E(X ) + o(t )as t → 0.
j!
j=1
Since, for a complex number z,
d
j j d
log(1 + z) = (−1) z /j + o(|z| )as |z|→ 0,
j=1
for any d = 1, 2,..., it follows that log(ϕ(t)) may be expanded in a series of the form
m
j m
log(ϕ(t)) = (it) c j /j! + o(t )as t → 0,
j=1
for some constants c 1 , c 2 ,..., c m . Using the relationship between cumulants and moments,
it follows that these constants must be the cumulants; that is, c j = κ j , proving the first part
of the theorem.
The second part of the theorem follows from the fact that the existence of ϕ (2m) (0)
implies that all moments of order less than or equal to 2m exist and are finite. Since each
κ j , j = 1,..., 2m,isa function of these moments, the result follows.