Section 6.9 (p.376)

 

1.         The sufficient statistic for q is max(X1, …, Xn) (Example 4 of Sec. 6.7). Since d1 is not a function of the sufficient statistic, alone, it is inadmissible.

 

2.         E(Xi) = q/2, Var(Xi) = q2/12.

Therefore, R(q,d1) = E[ (d1-q)2 ] = Var (d1) = q2/3n.

 

3.         It follows from Sec. 3.9 (p.159 – 161) that the pdf of Yn is

(a)

 

            (b) For n=2, R(q,d1) = R(q,d2) = q2/6

 

            (c) R(q,d1)  R(q,d2)

 

4.        

which has a minimum if c = (n+2)/(n+1). Let c* be (n+2)/(n+1), then c*Yn dominates every other estimator cYn.

 

5.         From Exercise 6 of Sec. 6.7, the sufficient statistic for a is P Xi. Since is not a function of P Xi alone, it is inadmissible.

 

6.         (a) R(b, d) = R(b, 3) = (b - 3)2.

 

            (b) If the unknown parameter b = 3, then R(3, d) = (3 - 3)2 = 0. There are no other estimator d1 can dominate d when b = 3 (i.e. R(3, d1) £ R(3, d)) unless d1 = d. Therefore d is admissible.

 

7.         From Example 1 of Sec. 6.7, the sufficient statistic for q is S Xi. Since the proportion  is not a function of S Xi alone, it is inadmissible.