Page 136 - Elements of Distribution Theory

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

122 Moments and Cumulants

and
4 (n − 1)(n − 2)(n − 3) 4(n − 1)
4
E X ¯ = E(X 1 ) + E(X 1 )E X 3
n 3 3 1
n n
6(n − 1)(n − 2) 2 2 3(n − 1) 2 2 1 4

+ E(X 1 ) E X 1 + E X 1 + E X 1
n 3 n 3 n 3
4 6 2 2 4
= E(X 1 ) + E(X 1 ) E X 1 − E(X 1 )
n
1 4 2 2 3 2 2

+ 11E(X 1 ) − 18E(X 1 ) E X 1 + 4E(X 1 )E X 1 + 3E X 1
n 2
1 4 2 2 3 2 2 4

+ E X 1 + 12E(X 1 ) E X 1 − 4E(X 1 )E X 1 − 3E X 1 − 6E(X 1 ) .
n 3
Example 4.25 (Standard exponential distribution). Let X 1 , X 2 ,..., X n denote indepen-
dent, identically distributed random variables, each with a standard exponential distribution.
The cumulant-generating function of the standard exponential distribution is − log(1 − t),
|t| < 1; see Example 4.11. Hence, the cumulants of the distribution are given by κ r =
¯ n
(r − 1)!, r = 1, 2,.... It follows from (4.4) that the cumulants of X n = X j /n are
j=1
given by
1
¯
κ r (X n ) = (r − 1)!, r = 1, 2,....
n r−1
The moments of the standard exponential distribution are given by
∞

r r
E X = x exp(−x) dx = r!, r = 1, 2,....
1
0
¯
¯
Hence, the first four moments of X n are given by E(X n ) = 1,
2 1 3 3 2
E X ¯ n = 1 + , E X ¯ n = 1 + + ,
n n n 2
and
4 6 11 6
E X ¯ n = 1 + + + .
n n 2 n 3
Expressions for moments and cumulants of a sample mean can also be applied to sample
moments of the form
n
1 m
X , m = 1, 2,...
j
n
j=1
by simply redefining the cumulants and moments in the theorem as the cumulants and
m
moments, respectively, of X . This is generally simpler to carry out with the moments,
1
m
m
since the moments of X are given by E(X ), E(X 2m ),....
1 1 1
Example 4.26 (Standard exponential distribution). As in Example 4.25, let X 1 ,
X 2 ,..., X n denote independent, identically distributed, standard exponential random
variables and consider the cumulants of
n
1 2
T n = X .
j
n
j=1