Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

7.3. Properties of a Sufficient Statistic 429

We illustrate this remark in the following example.

Example 7.3.2.LetX 1 ,X 2 ,X 3 be a random sample from an exponential distri-
bution with meanθ>0, so that the joint pdf is
(
1
θ

) 3

e−(x^1 +x^2 +x^3 )/θ, 0 <xi<∞,

i=1, 2 ,3, zero elsewhere. From the factorization theorem, we see thatY 1 =
X 1 +X 2 +X 3 is a sufficient statistic forθ.Ofcourse,

E(Y 1 )=E(X 1 +X 2 +X 3 )=3θ,

and thusY 1 /3=X is a function of the sufficient statistic that is an unbiased
estimator ofθ.
In addition, letY 2 =X 2 +X 3 andY 3 =X 3. The one-to-one transformation
defined by
x 1 =y 1 −y 2 ,x 2 =y 2 −y 3 ,x 3 =y 3
has Jacobian equal to 1 and the joint pdf ofY 1 ,Y 2 ,Y 3 is

g(y 1 ,y 2 ,y 3 ;θ)=

(

1

θ

) 3

e−y^1 /θ, 0 <y 3 <y 2 <y 1 <∞,

zero elsewhere. The marginal pdf ofY 1 andY 3 is found by integrating outy 2 to
obtain

g 13 (y 1 ,y 3 ;θ)=

(

1

θ

) 3

(y 1 −y 3 )e−y^1 /θ, 0 <y 3 <y 1 <∞,

zero elsewhere. The pdf ofY 3 alone is

g 3 (y 3 ;θ)=

1

θ

e−y^3 /θ, 0 <y 3 <∞,

zero elsewhere, since Y 3 =X 3 is an observation of a random sample from this
exponential distribution.
Accordingly, the conditional pdf ofY 1 ,givenY 3 =y 3 ,is

g 1 | 3 (y 1 |y 3 )=

g 13 (y 1 ,y 3 ;θ)

g 3 (y 3 ;θ)

=

(

1

θ

) 2

(y 1 −y 3 )e−(y^1 −y^3 )/θ, 0 <y 3 <y 1 <∞,

zero elsewhere. Thus

E

(

Y 1

3

∣

∣

∣

∣y^3

)

= E

(

Y 1 −Y 3

3

∣

∣

∣

∣y^3

)

+E

(

Y 3

3

∣

∣

∣

∣y^3

)

=

(

1

3

)∫∞

y 3

(

1

θ

) 2

(y 1 −y 3 )^2 e−(y^1 −y^3 )/θdy 1 +

y 3

3

=

(

1

3

)

Γ(3)θ^3

θ^2

+

y 3

3

=

2 θ

3

+

y 3

3

=Υ(y 3 ).