Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

430 Sufficiency

Of course,E[Υ(Y 3 )] =θand var[Υ(Y 3 )]≤var(Y 1 /3), but Υ(Y 3 ) is not a statistic, as

it involvesθand cannot be used as an estimator ofθ. This illustrates the preceding

remark.

EXERCISES

7.3.1.In each of Exercises 7.2.1–7.2.4, show that the mle ofθis a function of the

sufficient statistic forθ.

7.3.2.LetY 1 <Y 2 <Y 3 <Y 4 <Y 5 be the order statistics of a random sample of size
5 from the uniform distribution having pdff(x;θ)=1/θ, 0 <x<θ, 0 <θ<∞,
zero elsewhere. Show that 2Y 3 is an unbiased estimator ofθ. Determine the joint
pdf ofY 3 and the sufficient statisticY 5 forθ. Find the conditional expectation
E(2Y 3 |y 5 )=φ(y 5 ). Compare the variances of 2Y 3 andφ(Y 5 ).
7.3.3.IfX 1 ,X 2 is a random sample of size 2 from a distribution having pdf
f(x;θ)=(1/θ)e−x/θ, 0 <x<∞, 0 <θ<∞, zero elsewhere, find the joint
pdf of the sufficient statisticY 1 =X 1 +X 2 forθandY 2 =X 2. Show thatY 2 is an
unbiased estimator ofθwith varianceθ^2 .FindE(Y 2 |y 1 )=φ(y 1 ) and the variance
ofφ(Y 1 ).

7.3.4.Letf(x, y)=(2/θ^2 )e−(x+y)/θ, 0 <x<y<∞, zero elsewhere, be the joint

pdf of the random variablesXandY.

(a)Show that the mean and the variance ofYare, respectively, 3θ/2and5θ^2 /4.

(b)Show thatE(Y|x)=x+θ. In accordance with the theory, the expected value

ofX+θis that ofY,namely,3θ/2, and the variance ofX+θis less than

that ofY. Show that the variance ofX+θis in factθ^2 /4.

7.3.5. In each of Exercises 7.2.1–7.2.3, compute the expected value of the given
sufficient statistic and, in each case, determine an unbiased estimator ofθthat is a
function of that sufficient statistic alone.

7.3.6. LetX 1 ,X 2 ,...,Xnbe a random sample from a Poisson distribution with
meanθ. Find the conditional expectationE(X 1 +2X 2 +3X 3 |

∑n

1 Xi).

7.4 CompletenessandUniqueness......................

LetX 1 ,X 2 ,...,Xnbe a random sample from the Poisson distribution that has pmf

f(x;θ)=

{ θxe−θ

x! x=0,^1 ,^2 ,...; θ>^0

0elsewhere.

From Exercise 7.2.2, we know thatY 1 =

∑n
i=1Xiis a sufficient statistic forθand
its pmf is

g 1 (y 1 ;θ)=

{

(nθ)y^1 e−nθ

y 1! y^1 =0,^1 ,^2 ,...

0elsewhere.