Answer Key for STAT 200B HW No. 8

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1 Answer Key for STAT 200B HW No. 8 May 8, 2007 Problem 3.42 p. 708 The values of Ȳ for x 00, 0, 20, 30 are 5/40, 0, 20/50, and, respectively. From Corollary 3.5 it follows that MLE exists i G is identiable on the design points for which Ȳ > 0. Hence: a the space of linear functions is identiable, because p dim G 2 < 3 d, hence MLE exists; b in this case G is identiable and saturated, p d, MLE exists; c in this case, p > d, so G is not identiable, MLE does not exist. Problem 3.48 p. 722 Solution to this problem is similar to Ex. 3.0 on pp As before, Ȳ.2, Ȳ2.4, Ȳ3.6 for x, x 2 0, x 3, respectively. a Using eq. 0 on p. 699 we get the MLE equations for the parameters of the canonical regression function β 0 and β : n exp ˆβ0 + ˆβ x n Ȳ n x exp ˆβ0 + ˆβ x n x Ȳ, where n, n 2 40, n 3 : exp ˆβ0 ˆβ exp + 40 exp ˆβ0 ˆβ0 ˆβ + exp + exp ˆβ0 + ˆβ ˆβ0 + ˆβ 64 24, b X 0, N ˆβ 00 ˆβ 0 0 ˆθ 0 ˆθ 20 ˆθ 30 0 ˆσ 0 2 ˆθ0 C e 0 ˆµ 2 20 ˆµ2 30 Ŵ 0 diag Nˆσ 0 2 ˆV 0 40 e 0 ˆσ 2 20 ˆσ2 30, ˆµ 0 C ˆθ0 40, [ X t / 0 Ŵ 0 X 0 /20 ],

2 Z 0 ˆV 0 X t N.6 Ȳ ˆµ 0.2 ˆβ ˆβ Z 0..2 c MLE of θ is ˆθ X ˆβ X , and ˆλ exp ˆθ d exp exp.08 + exp exp exp Problem 3.49 p. 722 a For Poisson MLE ˆλ ˆµ ˆσ 2 exp ˆθ, hence the Newton-Raphson iterate is: ˆλ exp ˆθ Z ȳ ˆλ ˆλ ˆθ + ˆθ + Z. b Starting at ˆθ 0 0 and applying the iterative updates derived in a we get convergence of the log-lielihood after 5 iterations to values of ˆθ and ˆλ 5 exp ˆθ 5.. c log-lielihood function: l θ n [ θȳ C θ] + log r Y ; n θ 0 exp θ + log 0. Taylor expansion up to 2 nd term of C θ exp θ around ˆθ 0 0: C θ C ˆθ0 + C ˆθ0 θ ˆθ 0 + ˆθ0 θ 2 C ˆθ 0 + θ + 2 θ2. Quadratic approximation of l θ around ˆθ 0 0: l θ θ 0 + θ + 2 θ2 + log 0. Fig. shows the plot of the log-lielihood and its 2 nd order approximation. 2

3 log lielihood theta Figure : Plot of log-lielihood blac and its 2 nd order approx. red around 0. Solid vertical line - location of MLE, dashed vertical - 0. Problem 3.54 a p. 734 a θ 0, x 2, θ 0, x 2, 2β 4, hence the nominal 95%CI is given by 2 ˆβ 4 ± z.975 SE 2 ˆβ 4.356,.368 from table The corresponding odds ratio is exp 2 ˆβ with 95%CI.43, The odds of getting CHD in the younger group of both sexes on average more than double when the cholesterol level increases from to b with SE 2 ˆβ ˆβ 5 θ, x 2, θ, x 2, 2β 4 + 2β 5, 2, 2 V C 2 2 2, 2 t Thus, the nominal 95%CI for 2β 4 + 2β 5 is ± z.975 SE 2 ˆβ ˆβ 5.343,.59, with the corresponding odds-ratio nominal 95%CI of.7,.68. The odds of getting CHD does not signicantly increase for the older group when increasing the cholesterol level from to

4 c Test H 0 : 2β 4 + 2β 5 0 vs. H : 2β 4 + 2β 5 0. From b the Wald statistic is calculated to be W 2 ˆβ 4 +2 ˆβ , with resulting p-value of 2 Φ W.7, hence we accept the null hypothesis. SE2 ˆβ 4 +2 ˆβ 5 Problem 3.6 p. 735 a ˆτ 2 h 2 h ˆβ h 2 ˆτ. Hence, AV ˆτ 2 gradient of h. By the chain rule h t AV ˆβ h h 2 τ τ h 2 τ h, h h, where denotes the where h denotes the gradient of t h. Furthermore, AV ˆτ h AV ˆβ h AV ˆτ 2 h 2 τ 2 t h τ AV ˆβ h τ h 2 τ 2 AV ˆτ. b From a SE ˆτ 2 AV ˆτ 2 h 2 τ AV ˆτ h 2 τ SE ˆτ. Problem 3.68 p. 748 a Wald test for H 0 : β 3 0 vs. H : β 3 0: ˆβ 3 0 W p 2 [ Φ 2.8].005, SE ˆβ strongly reject the null. b p-value for the lielihood-ratio test is computed based on the dierence between model and sub-model the one for which β 3 0 deviances: D D G 0 D G 8.3, which in this case is approx. distributed as χ 2 distr. with dim G dim G 0 deg. of freedom leading to p-value of χ which is close to the one derived in a. Problem 3.76 p. 749 a In the full model there are 3 parameters and 3 observations, thus the linear space G has to obey dim G 3 to be identiable. One particular example for a basis of G is {I X, I X 2, I X 3} leading to design matrix: On the other hand the one-dimensional sub-model G 0 has to satisfy the constraints π.5, and π 2 π 3 which can be enforced by choosing a basis of G 0 s.t. g G 0 : g 0, g 2 g 3, leading to a design matrix as the one below for example: b The MLE for model G leads to the following estimates for the probabilities of success in each group: ˆπ.6, ˆπ 2.5, ˆπ 3.3, while for G 0 we get the estimates ˆπ 0.5, ˆπ0 2 ˆπ0 3 2 ˆπ 2 + ˆπ 3 4

5 .4, with corresponding deviance dierence calculated according to eqs. 44 and 45: D 2 [ Y log ˆπ ˆπ 0 + n Y log ˆπ ] ˆπ The corresponding p-value for testing goodness-of-t between the 2 models is χ , so we reject the simpler model in favor for the saturated one. 5

Answer Key for STAT 200B HW No. 7

Answer Key for STAT 200B HW No. 7 May 5, 2007 Problem 2.2 p. 649 Assuming binomial 2-sample model ˆπ =.75, ˆπ 2 =.6. a ˆτ = ˆπ 2 ˆπ =.5. From Ex. 2.5a on page 644: ˆπ ˆπ + ˆπ 2 ˆπ 2.75.25.6.4 = + =.087;