Prove XXXXXXXXXXHint: Step (a). E(mn(x) − m(x))2 ≥ E(mn(x) − E{mn(x)|X1,...Xn}) 2 + (E{mn(x)} − m(x))2 . Step (b). E{(mn(x) − E{mn(x)|X1,...Xn}) 2 } = 1 kn . Step (c). Observe that the function...

Prove (6.2). Hint:

Step (a). E(mn(x) − m(x))2 ≥ E(mn(x) − E{mn(x)|X1,...Xn}) 2 + (E{mn(x)} − m(x))2 .

Step (b). E{(mn(x) − E{mn(x)|X1,...Xn}) 2 } = 1 kn .

Step (c). Observe that the function E{mn(x)|X1,...,Xn} = 1 kn kn i=1 X(i,n)(x)

is a monotone increasing function of x, therefore

              E{mn(x)|X1,...,Xn} ≥ 1 kn kn i=1 X(i,n)(0).

Let X∗ 1 ,...,X∗ n be the ordered sample of X1,...,Xn, then X(i,n)(0) = X∗ i , and so

                         E{mn(x)} ≥ E  1 kn kn i=1 X∗ i  = αkn .

Thus

Step (d). αkn = 1 2 kn n + 1 .