Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

588 Nonparametric and Robust Statistics

Basedonthisobservation,forssuch that−∞<s<∞,themgfofTis

E[exp{sT}]=E

⎡

⎣exp

⎧

⎨

⎩

∑n

j=1

sjsgn(Xij)

⎫

⎬

⎭

⎤

⎦

=

∏n

j=1

E[exp{sjsgn(Xij)}]

=

∏n

j=1

(

1

2

e−sj+

1

2

esj

)

=

1

2 n

∏n

j=1

(

e−sj+esj

)

. (10.3.6)

Because the mgf does not depend on the underlying symmetric pdff(x), the test
statisticTis distribution free underH 0. Although the pmf ofTcannot be obtained
in closed form, this mgf can be used to generate the pmf for a specifiedn;see
Exercise 10.3.1.
Because the sgn(Xij)s are mutually independent with mean zero, it follows that
EH 0 [T] = 0. Further, because the variance of sgn(Xij)is1,wehave

VarH 0 (T)=

∑n

j=1

VarH 0 (jsgn(Xij)) =

∑n

j=1

j^2 =n(n+ 1)(2n+1)/ 6.

We summarize these results in the following theorem:

Theorem 10.3.1. Assume that Model (10.2.1) is true for the random sample
X 1 ,...,Xn. Assume also that the pdff(x)is symmetric about 0. Then under
H 0 ,

Tis distribution free with a symmetric pmf (10.3.7)

EH 0 [T] = 0 (10.3.8)

VarH 0 (T)=

n(n+ 1)(2n+1)

6

(10.3.9)

q T

VarH 0 (T)

has an asymptoticallyN(0,1)distribution. (10.3.10)

Proof: The first part of (10.3.7) and the expressions (10.3.8) and (10.3.9) were
derived above. The asymptotic distribution ofTcertainly is plausible and its proof
can be found in more advanced books. To obtain the second part of (10.3.7), we
need to show that the distribution ofTis symmetric about 0. But by the mgf of
Y, (10.3.6), we have

E[exp{s(−T)}=E[exp{(−s)T}]=E[exp{sT}].

HenceT and−Thave the same distribution, soT is symmetrically distributed

about 0.