Section 7.7 (p.417)

 

1.         E[ (1/n)SXik ] = (1/n) E[ SXik ] = (1/n) n E[ Xk ] = E[ Xk ]

 

2.         (1/n) SXi2 is an unbiased estimator of E(X2),  is an unbiased estimator of Var(X). Therefore, (1/n) SXi2 +  is an unbiased estimator of [E(X)]2.

 

4.         Follow the hint, we have

Therefore,

Therefore, we have d(x) = 2X.

 

5.         From equation (8) on p.415, the MSE of S02 is (2n-1)/n2 s4 and the MSE of S12 is 2/(n-1) s4. Since (2n-1)/n2 < 2/(n-1) for every positive integer n, it follows that the MSE of S02 is smaller than the MSE of S12 for all possible m and s2.

 

7.         E[d(X)] = p   Þ   S d(x) pqx = p   Þ   S d(x) qx = 1

It follows that d(0) + d(1)q + d(2)q2 + … = 1 for all possible q. Therefore the unbiased estimator d(X) is d(0) = 1 and d(X) = 0 for X > 0.

 

9.         Let X be the number of failures that will occur before exactly k successes have been obtained. From Section 5.5 we know that X has a negative binomial distribution with parameters k and p. Also note that N = X + k. Now we want to show E[(k-1)/(N-1)] = p.

 

The last summation is 1 because it is the sum of the probabilities of a negative binomial distribution with parameters k-1 and p.

 

11.       (a) Let X be the value of a certain characteristic among the total population and Y be the stratum number. Then

m = E(X) = E[ E(X|Y)] = S E(X|Y=i ) Pr(Y=i ) = S mi pi

 

            (b)

Note that n=Sni, we can let . And set . Solve theses equations we get

Note: The ni obtained above are not necessarily integers. The answer should be the closest integer.

 

12.       (a) E[d0(T)] = E[E(d|T)] = E(d) = q

            (b) From Theorem 1. of Sec. 6.9, R(q, d0) £ R(q, d).

R(q, d).= E[(d-q)2] = Varq(d) because E(d)=q. For the same reason, R(q, d0).= Varq(d0). Therefore Varq(d0) £ Varq(d).

 

13.       Let F(y) be the cdf of Yn. Then F(y) = Pr(Yn£y) = Pr(X1£y)… Pr(Xn£y) = (y/q)n.

Therefore Yn has pdf f(y) =F¢(y) = nyn-1/qn.