Page 127 - Elements of Distribution Theory

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

4.4 Cumulants 113

Hence, it is clear that cumulants are functions of moments. The exact relationship may be
obtained by expanding

∞
j j
log 1 + t E(X )/j!
j=1
in a power series in t and then equating terms with an expansion for the cumulant generating
function. For example,

∞
1
j j 2 2 2
log 1 + t E(X )/j! = E(X)t + [E(X ) − E(X) ]t
2!
j=1
1 3 2 2
+ [E(X ) − E(X)E(X ) − 2(E(X )
3!
2 3
− E(X) )E(X)]t +· · ·
so that
2 2 3 2 2 2
κ 1 = E(X),κ 2 = E(X ) − E(X) ,κ 3 = E(X ) − E(X)E(X ) − 2[E(X ) − E(X) ]E(X).
Hence, the first cumulant is the mean of X, the second cumulant is the variance of X, and,
3
with a little algebra, it may be shown that the third cumulant is E[(X − E(X)) ], often called
the skewness of X.
The general form of the relationship between moments and cumulants is based on the
relationship between the coefficients of a power series and the coefficients in an expansion
of the log of that power series. Consider a function
∞
α j j

α(t) = t
j!
j=0
defined for t in a neighborhood of 0. Suppose that α(t) > 0 for t near 0 and write
∞
β j j

β(t) = log α(t) = t .
j!
j=0
Clearly, the coefficients β 1 ,β 2 ,... are functions of α 1 ,α 2 ,.... The following result can be
used to determine an expression for α r in terms of β 1 ,...,β r ; conversely, an expression
for β r can be given in terms of α 1 ,...,α r .
Lemma4.1. Definethefunctionsα(t)andβ(t)andthecoefficientsα 1 ,α 2 ,...andβ 1 ,β 2 ,...
as above. Then α 0 = exp(β 0 ) and
r
r
α r+1 = β j+1 α r− j , r = 0, 1,....
j
j=0
Proof. Note that
d j d j

α j = α(t) and β j = β(t) , j = 0, 1,....
dt j t=0 dt j t=0
The result relating α 0 and β 0 follows immediately.
Since α(t) = exp{β(t)},

α (t) = exp{β(t)}β (t) = α(t)β (t).