Page 138 - Elements of Distribution Theory

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

124 Moments and Cumulants

Theorem 4.18. Let X 1 , X 2 ,..., X n denote independent, identically distributed, real-
valued random variables such that all moments of X 1 exist and are finite. Let µ = E(X 1 )
and

n
1
¯
X n = X j .
n
j=1
Then, for k = 1, 2,...,
1 2k 1

2k−1
¯
¯
E[(X n − µ) ] = O and E[(X n − µ) ] = O as n →∞.
n k n k
Proof. The proof is by induction. For k = 1, the result follows immediately from Theorem
4.17. Assume that the result holds for k = 1, 2,..., m.For each j = 1, 2,..., let
¯
j
¯ µ j = E[(X n − µ) ].
¯
Note that, applying Lemma 4.1 to the moment- and cumulant-generating functions of X n −
¯
µ, the cumulants and central moments of X n are related by
r
r
¯
¯ µ r+1 = κ j+1 (X n )¯µ r− j , r = 0, 1,....
j
j=0
¯
j
Since, by (4.4), κ j+1 (X n ) = O(1/n ), and taking ¯µ 0 = 1,
r
1
¯ µ r+1 = ¯ µ r− j O .
n j
j=0
Consider r = 2m. Then, since the theorem is assumed to hold for k = 1, 2,..., m,
1
O( m−( j−1)/2 )if j = 1, 3,..., 2m − 1
¯ µ 2m− j = n 1 .
O( m− j/2 ) if j = 0, 2, 4,..., 2m
n
Hence,

1 1 1 1 1
¯ µ 2m+1 = O + O + O +· · · + O = O
n m n m+1 n m+1 n 2m n m
as n →∞.
Now consider r = 2m + 1. Then
1 1 1 1

¯ µ 2m+2 = O + O +· · · + O = O
n m n m+1 n 2m+1 n m
as n →∞.
It follows that the result holds for k = m + 1, proving the theorem.

4.6 Conditional Moments and Cumulants
Let X denote a real-valued random variable and let Y denote a random variable which
may be a vector. The conditional moments and cumulants of X given Y = y are simply
the moments and cumulants, respectively, of the conditional distribution of X given Y = y;