Unformatted text preview: Second Edition 7-21 7.53 Let a be a constant and suppose Cov0 (W, U ) > 0. Then Var0 (W + aU ) = Var0 W + a2 Var0 U + 2aCov0 (W, U ). Choose a -2Cov0 (W, U ) Var0 U, 0 . Then Var0 (W + aU ) < Var0 W , so W cannot be best unbiased. 7.55 All three parts can be solved by this general method. Suppose X f (x|) = c()m(x), a < x < . Then 1/c() = a m(x) dx, and the cdf of X is F (x) = c()/c(x), a < x < . Let Y = X(n) be the largest order statistic. Arguing as in Example 6.2.23 we see that Y is a complete sufficient statistic. Thus, any function T (Y ) that is an unbiased estimator of h() is the best unbiased estimator of h(). By Theorem 5.4.4 the pdf of Y is g(y|) = nm(y)c()n /c(y)n-1 , a < y < . Consider the equations f (x|) dx = 1
a T (y)g(y|) dy = h(), which are equivalent to m(x) dx =
a 1 c() and
a T (y)nm(y) h() dy = . c(y)n-1 c()n Differentiating both sides of these two equations with respect to and using the Fundamental Theorem of Calculus yields m() = - c () c()2 and T ()nm() c()n h () - h()nc()n-1 c () = . n-1 c() c()2n Change s to ys and solve these two equations for T (y) to get the best unbiased estimator of h() is h (y) . T (y) = h(y) + nm(y)c(y) For h() = r , h () = rr-1 . a. For this pdf, m(x) = 1 and c() = 1/. Hence T (y) = y r + ry r-1 n+r r = y . n(1/y) n b. If is the lower endpoint of the support, the smallest order statistic Y = X(1) is a complete sufficient statistic. Arguing as above yields the best unbiased estimator of h() is T (y) = h(y) - For this pdf, m(x) = e-x and c() = e . Hence T (y) = y r - ry r-1 ry r-1 = yr - . -y ey ne n h (y) . nm(y)c(y) c. For this pdf, m(x) = e-x and c() = 1/(e- - e-b ). Hence T (y) = y r - ry r-1 -y ry r-1 (1 - e-(b-y) ) (e - e-b ) = y r - . ne-y n ...
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- Spring '12
- Statistics, Trigraph, best unbiased estimator, complete sufficient statistic, ry r-1