Dr. Hackney STA Solutions pg 101

Dr. Hackney STA Solutions pg 101 - 7-4 Solutions Manual for...

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Unformatted text preview: 7-4 Solutions Manual for Statistical Inference and thus ^ E = n n-1 and ^ Var = n2 (n - 1) (n - 2) 2 2 0 as n . b. Because X beta(, 1), E X = /( + 1) and the method of moments estimator is the solution to 1 i Xi ~ Xi = = . n i +1 n- i Xi 7.12 Xi iid Bernoulli(), 0 1/2. a. method of moments: EX = = 1 n Xi = X i ~ = X. MLE: In Example 7.2.7, we showed that L(|x) is increasing for x and is decreasing for x. Remember that 0 1/2 in this exercise. Therefore, when X 1/2, X is is the overall maximum of L(|x). When X > 1/2, L(|x) is an the MLE of , because X increasing function of on [0, 1/2] and obtains its maximum at the upper bound of which ^ is 1/2. So the MLE is = min X, 1/2 . ~ ~ ~ ~ b. The MSE of is MSE() = Var + bias()2 = ((1 - )/n) + 02 = (1 - )/n. There is no ^ but an expression is simple formula for MSE(), n ^ MSE() = ^ E( - )2 = y=0 [n/2] ^ ( - )2 n y (1 - )n-y y n = y=0 y - n 2 n y (1 - )n-y + y y=[n/2]+1 1 - 2 2 n y (1 - )n-y , y where Y = i Xi binomial(n, ) and [n/2] = n/2, if n is even, and [n/2] = (n - 1)/2, if n is odd. c. Using the notation used in (b), we have n ~ MSE() = E(X - )2 = y=0 y - n 2 n y (1 - )n-y . y Therefore, n ~ ^ MSE() - MSE() = y=[n/2]+1 n y - n 2 - 1 - 2 2 n y (1 - )n-y y n y (1 - )n-y . y = y=[n/2]+1 y 1 + - 2 n 2 y 1 - n 2 The facts that y/n > 1/2 in the sum and 1/2 imply that every term in the sum is positive. ^ ~ ^ ~ Therefore MSE() < MSE() for every in 0 < 1/2. (Note: MSE() = MSE() = 0 at = 0.) 7.13 L(|x) = i 1 e- 2 |xi -| = 21 e- 2 i |xi -| , so the MLE minimizes n 2 where x(1) , . . . , x(n) are the order statistics. For x(j) x(j+1) , n j n 1 1 i |xi - | = n i |x(i) - |, j |x(i) - | = i=1 i=1 ( - x(i) ) + i=j+1 (x(i) - ) = (2j - n) - i=1 x(i) + i=j+1 x(i) . ...
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