Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

(Jacob Rumans) #1
7.9. Sufficiency, Completeness, and Independence 461

(b)T 2 =

∑n− 1
i=1(Xi+1−Xi)

(^2) /S (^2).
(c)T 3 =(Xi−X)/S.
7.8.7. With random samples from each of the distributions given in Exercises
7.8.1(d), 7.8.2, and 7.8.3, define at least two ancillary statistics that are differ-
ent from the examples given in the text. These examples illustrate, respectively,
location-invariant, scale-invariant, and location- and scale-invariant statistics.


7.9 Sufficiency, Completeness, and Independence

We have noted that if we have a sufficient statisticY 1 for a parameterθ, θ∈Ω,
thenh(z|y 1 ), the conditional pdf of another statisticZ,givenY 1 =y 1 ,doesnot
depend uponθ.If,moreover,Y 1 andZare independent, the pdfg 2 (z)ofZis
such thatg 2 (z)=h(z|y 1 ), and henceg 2 (z) must not depend uponθeither. So the
independence of a statisticZand the sufficient statisticY 1 for a parameterθimply
that the distribution ofZdoes not depend uponθ∈Ω. That is,Zis an ancillary
statistic.
It is interesting to investigate a converse of that property. Suppose that the
distribution of an ancillary statisticZdoes not depend uponθ;thenareZand
the sufficient statisticY 1 forθindependent? To begin our search for the answer,
we know that the joint pdf ofY 1 andZisg 1 (y 1 ;θ)h(z|y 1 ), whereg 1 (y 1 ;θ)and
h(z|y 1 ) represent the marginal pdf ofY 1 and the conditional pdf ofZgivenY 1 =y 1 ,
respectively. Thus the marginal pdf ofZis
∫∞


−∞

g 1 (y 1 ;θ)h(z|y 1 )dy 1 =g 2 (z),

which, by hypothesis, does not depend uponθ. Because
∫∞


−∞

g 2 (z)g 1 (y 1 ;θ)dy 1 =g 2 (z),

if follows, by taking the difference of the last two integrals, that
∫∞


−∞

[g 2 (z)−h(z|y 1 )]g 1 (y 1 ;θ)dy 1 = 0 (7.9.1)

for allθ∈Ω. SinceY 1 is sufficient statistic forθ,h(z|y 1 ) does not depend uponθ.
By assumption,g 2 (z) and henceg 2 (z)−h(z|y 1 ) do not depend uponθ.Nowifthe
family{g 1 (y 1 ;θ):θ∈Ω}is complete, Equation (7.9.1) would require that


g 2 (z)−h(z|y 1 )=0 or g 2 (z)=h(z|y 1 ).

That is, the joint pdf ofY 1 andZmust be equal to


g 1 (y 1 ;θ)h(z|y 1 )=g 1 (y 1 ;θ)g 2 (z).

Accordingly,Y 1 andZare independent, and we have proved the following theorem,
which was considered in special cases by Neyman and Hogg and proved in general
by Basu.

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