Page 241 - Elements of Distribution Theory

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052184472Xc07 CUNY148/Severini May 24, 2005 3:59

7.6 Ranks 227

so that

σ 2
c =− .
n − 1
2
To find σ , note that R 1 = j with probability 1/n. Hence,
1 n (n + 1)(2n + 1)
2 2
E R = j = .
1
n 6
j=1
2
2
Since E(R 1 ) = (n + 1)/2, it follows that σ = (n − 1)/12 and c =−(n + 1)/12.
Now consider the statistic T . The expected value of T is given by
n
n + 1
E(T ) = a j ;
2
j=1
the variance of T is given by
n n − 1 n + 1
n
2
2

2
Var(T ) = σ 2 a + 2c a i a j = a − a i a j .
j j
j=1 i< j 12 j=1 6 i< j
For instance, consider a j = j; when the data are collected in time order, the statistic
n
j=1 jR j may be used to test the hypothesis of a time trend in the data. If the data are in
fact independent and identically distributed, this statistic has mean
n(n + 1) 2
4
and variance
n
n
2
2
2
n + n 2 n + 1 3 n (n + 1)(n − 1)
j − j = .
12 12 144
j=1 j=1
Example 7.28 (Conditional expectation of a sum of uniform random variables). Let
X 1 ,..., X n denote independent, identically distributed random variables, each uniformly
distributed on (0, 1) and let a 1 ,..., a n denote a sequence of constants. Consider

n

E a j X j |R 1 ,..., R n
j=1
where (R 1 ,..., R n ) denotes the vector of ranks.
Let (r 1 ,...,r n ) denote a permutation of (1,..., n). According to Theorem 7.11, part (iv),

n n n

E a j X j |R 1 = r 1 ,..., R n = r n = E a j X (r j ) = a j E{X (r j ) }.
j=1 j=1 j=1
From Example 7.19, we know that X (m) has a beta distribution with parameters m and
n − m + 1; hence, it is straightforward to show that
m
E{X (m) }= .
n + 1