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Unformatted text preview: 74 Solutions Manual for Statistical Inference and thus ^ E = n n1 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  )ny y
n =
y=0 y  n 2 n y (1  )ny + y y=[n/2]+1 1  2 2 n y (1  )ny , 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  )ny . y Therefore,
n ~ ^ MSE()  MSE() =
y=[n/2]+1 n y  n 2  1  2 2 n y (1  )ny y n y (1  )ny . 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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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics

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