Dr. Hackney STA Solutions pg 188

Dr. Hackney STA Solutions pg 188 - 11-16 Solutions Manual...

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Unformatted text preview: 11-16 Solutions Manual for Statistical Inference 11.39 Let Ai be the set Ai = , : (^ + x0i ) - ( + x0i ) ^ ^ ^ S 1 (x0i - x)2 + tn-2,/2m . n Sxx Then P (m Ai ) is the probability of simultaneous coverage, and using the Bonferroni Ini=1 equality (1.2.10) we have m m P (m Ai ) i=1 i=1 P (Ai ) - (m - 1) = i=1 1- - (m - 1) = 1 - . m 11.41 Assume that we have observed data (y1 , x1 ), (y2 , x2 ), . . . , (yn-1 , xn-1 ) and we have xn but not yn . Let (yi |xi ) denote the density of Yi , a n(a + bxi , 2 ). a. The expected complete-data log likelihood is n n-1 E i=1 log (Yi |xi ) = i=1 log (yi |xi ) + E log (Y |xn ), where the expectation is respect to the distribution (y|xn ) with the current values of the parameter estimates. Thus we need to evaluate 1 1 2 E log (Y |xn ) = E - log(21 ) - 2 (Y - 1 )2 , 2 21 2 where Y n(0 , 0 ). We have 2 E(Y - 1 )2 = E([Y - 0 ] + [0 - 1 ])2 = 0 + [0 - 1 ]2 , since the cross term is zero. Putting this all together, the expected complete-data log likelihood is - n 1 2 log(21 ) - 2 2 21 n-1 [yi - (a1 + b1 xi )]2 - i=1 n 2 0 + [(a0 + b0 xn ) - (a1 + b1 xn )]2 2 21 2 0 2 21 n 1 2 = - log(21 ) - 2 2 21 [yi - (a1 + b1 xi )]2 - i=1 if we define yn = a0 + b0 xn . b. For fixed a0 and b0 , maximizing this likelihood gives the least squares estimates, while the 2 maximum with respect to 1 is 1 = ^2 n i=1 [yi 2 - (a1 + b1 xi )]2 + 0 . n So the EM algorithm is the following: At iteration t, we have estimates a(t) , ^(t) , and 2(t) . ^ b ^ (t) We then set yn = a(t) + ^(t) xn (which is essentially the E-step) and then the M-step is ^ b to calculate a(t+1) and ^(t+1) as the least squares estimators using (y1 , x1 ), (y2 , x2 ), . . . ^ b (t) (yn-1 , xn-1 ), (yn , xn ), and 1 ^ 2(t+1) = n i=1 [yi - (a(t+1) + b(t+1) xi )]2 + 0 n 2(t) . ...
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