Page 135 - Elements of Distribution Theory

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

4.5 Moments and Cumulants of the Sample Mean 121

Another approach that is often useful is to use the fact that, for r = 2, 3,...,
r
n n n
¯ r 1 1
X = X j = ··· X j 1 X j 2 ··· X j r
n
n r n r
j=1 j 1 =1 j r =1
3
¯
and then take the expected value of resulting sum. For instance, consider E(X ). Since
n
n n n
¯ 3 1
X = X i X j X k ,
n
n 3
i=1 j=1 k=1
n n n
3 1
E X ¯ n = E(X i X j X k ).
n 3
i=1 j=1 k=1
In order to evaluate
n

E(X i X j X k ),
i, j,k=1
we must keep track of the number of terms in the sum in which i, j, k are unique, the number
in which exactly two indices are equal, and so on. It is straightforward to show that of the
3
n terms in
n n n

X i X j X k ,
i=1 j=1 k=1
n terms have all indices the same, 3n(n − 1) terms have exactly two indices the same, and
in n(n − 1)(n − 2) terms all indices are unique. Hence,
n

3 2 3
E(X i X j X k ) = nE X 1 + 3n(n − 1)E(X 1 )E X 1 + n(n − 1)(n − 2)E(X 1 ) .
i, j,k=1
It follows that
3 (n − 1)(n − 2) 3(n − 1) 1
3
3
E X ¯ = E(X 1 ) + E(X 1 )E X 2 + E X .
n 2 2 1 2 1
n n n
¯
The same approach can be used for any moment of X n , although the algebra becomes
tedious very quickly. The following theorem gives expressions for the first four moments;
the proof is left as an exercise.
Theorem 4.17. Let X 1 , X 2 ,..., X n denote independent, identically distributed, real-valued
4
random variables such that E(X ) < ∞ and let
1
n
1
¯
X n = X j .
n
j=1
¯
The moments of X n satisfy
2 n − 1 1 1
2
2
2
¯
E(X n ) = E(X 1 ), E X ¯ n = E(X 1 ) + E X 2 1 = E(X 1 ) + E X 1 2 − E(X 1 ) ,
n n n
3
3 (n − 1)(n − 2) 3(n − 1) E(X )
3
E X ¯ = E(X 1 ) + E(X 1 )E X 2 + 1
n 2 2 1 2
n n n
3 3 2 3 1 3 2 3
= E(X 1 ) + E(X 1 )E X −E(X 1 ) + 2E(X 1 ) −3E(X 1 )E X +E X 1 ,
1
1
n n 2