1. A study i the Netherlads followed me ad wome for up to 21 years. At three year itervals, participats aswered questios about respiratory symptoms ad smokig status. Pulmoary fuctio was determied by forced expiratory volume i oe secod (FEV1) at each observatio time. The data i this questio cosist of participats who did ot chage smokig status over the duratio of the study. There are 32 former smokers ad 101 curret smokers for a total of 133 participats. Although it was iteded for FEV1 to be recorded at all time poits (baselie, every three years through year 15, ad at year 19), ot all idividuals have FEV at every time poit. Some possibly helpful descriptive statistics ad plots are show below. Former Smoker Curret Smoker Time (N=32) (N=101) 0 3.52 (23) 3.23 (85) 3 3.58 (27) 3.12 (95) 6 3.26 (28) 3.09 (89) 9 3.17 (30) 2.87 (85) 12 3.14 (29) 2.80 (81) 15 2.87 (24) 2.68 (73) 19 2.91 (28) 2.50 (74) Table 1: Mea FEV1 (ad sample size) by smokig status ad time. Curret Smokers Former Smokers FEV (L/sec) 1 2 3 4 5 FEV (L/sec) 1 2 3 4 5 0 5 10 15 Time 0 5 10 15 Time Figure 1: Mea trajectory (solid lie) ad idividual trajectories (dashed lie) by smokig status.
The results from fittig the model E(Y ij X) =β 0 + β 1 t ij + β 2 S i usig ordiary least squares are show below. Here i idicates the subject (1 to 133), j idexes the time measuremets o a specific subject, ad S i is the biary idicator of smokig status (1=curret smoker).. regress fev year smoke Source SS df MS Number of obs = 771 -------------+------------------------------ F( 2, 768) = 87.11 Model 54.6251219 2 27.3125609 Prob > F = 0.0000 Residual 240.805055 768.313548249 R-squared = 0.1849 -------------+------------------------------ Adj R-squared = 0.1828 Total 295.430177 770.383675555 Root MSE =.55995 ------------------------------------------------------------------------------ fev Coef. Std. Err. t P> t [95% Cof. Iterval] -------------+---------------------------------------------------------------- year -.038589.0032995-11.70 0.000 -.0450661 -.0321119 smoke -.3106707.0469242-6.62 0.000 -.4027857 -.2185558 _cos 3.562053.0510095 69.83 0.000 3.461918 3.662188 ------------------------------------------------------------------------------ (a) For each of the followig, describe the effect o the value if the regressio were ru usig GEE with a idepedet workig covariace matrix. (Chose from: stay the same, icrease, decrease, differ but could be i either directio, or caot be determied). Provide a brief justificatio for your choice. i. ˆβ0 Solutio: All of these poit estimates will stay the same because the equatios used to solve for the estimates are the same usig OLS ( ˆβ =(X T X) 1 (X T Y )) ad GEE with workig idepedece ( ˆβ =(X T W 1 X) 1 (X T W 1 Y ), W = I) ii. ˆβ1 Solutio: All poit estimates stay the same. iii. ˆβ2 Solutio: All poit estimates stay the same. Page 2
iv. Stadard error of ˆβ 0 Solutio: The stadard errors icrease for estimates of betas that must be made betwee subjects (effect of smokig ad mea at baselie) due to loss of idepedet subjects whe ot assumig that all measuremets are idepedet. v. Stadard error of ˆβ 1 Solutio: The stadard errors decrease for estimates of betas that ca be made withi subject (i.e. tred over time) due to positive correlatio of measuremets withi idividuals. vi. Stadard error of ˆβ 2 Solutio: The stadard errors icrease for estimates of betas that must be made betwee subjects (effect of smokig ad mea at baselie) due to loss of idepedet subjects whe ot assumig that all measuremets are idepedet. (b) A statistics studet brigs output from Stata ad expresses cocer that the results are slightly differet whe usig differet workig covariace matrices i GEE. Explai to this statistics studet why the results may be differet usig differet workig covariace structures. Be sure to iclude (a) why the poit estimates may be differet, (b) why the stadard errors may be differet, ad (c) if these differeces represet a substatial problem with the GEE method. You may assume that the statistics studet ca uderstad mathematical otatio, but you should iclude writte explaatio as well. Solutio: Estimates i GEE come from solvig the equatio ˆβ = (X T W 1 X) 1 (X T W 1 Y ), where W is the workig covariace matrix. Therefore, usig a differet form of the workig covariace matrix will lead to slightly differet estimates. The stadard errors of ˆβ are give by (X T W 1 X) 1 X T W 1 (rr T )W 1 X(X T W 1 X) 1, where r is the vector of residuals. Thus the stadard errors will also differ based o the choice of workig covariace matrix. These differeces do ot represet a problem with the GEE method although choosig a workig covariace structure that is close to the truth will be more efficiet, all choices will result i estimates ad stadard errors with the correct iferetial properties (e.g., ubiased estimates ad omial type I error rate). Page 3
(c) A collaborator suggests usig a radom effects model istead of GEE. This collaborator is satisfied with the model of a liear time effect ad a fixed effect of smokig that was used above. The collaborator suggests icludig a radom itercept ad a radom time effect (radom slope). There are three possible models: oe with radom itercepts, oe with idepedet radom itercepts ad slopes, ad oe with possibly correlated radom itercepts ad slopes. Which of these models is most aalogous to the GEE model usig workig exchageable? Explai your choice. Solutio: The model with radom itercepts oly is aalogous to GEE with a exchageable correlatio structure; the robust stadard errors do ot formally assume that this model is maitaied, but the exchageable covariace structure assumes a costat correlatio betwee measuremets o the same idividual, which is exactly what is assumed with the radom itercept model. (d) The STATA output from fittig the model with possibly correlated radom itercepts ad slopes is show below. As alluded to previously, this model is: Y ij = β 0 + β 1 t ij + β 2 S i +b 0i +b 1i t ij + ij As before, i idicates the subject (1 to 133), j idexes the time measuremets o a specific subject, ad S i is the biary idicator of smokig status (1=curret smoker).. xtmixed fev year smoke id: year, cov(ustructured) [some output omitted] ------------------------------------------------------------------------------ fev Coef. Std. Err. z P> z [95% Cof. Iterval] -------------+---------------------------------------------------------------- year -.0371543.0015181-24.47 0.000 -.0401298 -.0341789 smoke -.3250446.1083222-3.00 0.003 -.5373523 -.1127369 _cos 3.546401.096068 36.92 0.000 3.358111 3.734691 ------------------------------------------------------------------------------ ------------------------------------------------------------------------------ Radom-effects Parameters Estimate Std. Err. [95% Cof. Iterval] -----------------------------+------------------------------------------------ id: Ustructured sd(year).009866.0019306.0067232.0144781 sd(_cos).5502334.0368449.4825569.6274014 corr(year,_cos) -.31308.1480349 -.5687205 -.0022833 -----------------------------+------------------------------------------------ sd(residual).2043253.0063677.1922183.2171948 ------------------------------------------------------------------------------ Page 4
For each term below, (1) ote the estimate provided by this model ad (2) provide a iterpretatio suitable for a o-statisticia. Estimate ˆβ 0 3.55 Iterpretatio This is the estimated FEV1 at baselie for a typical former smoker. ˆβ 1-0.037 This is the estimated chage i FEV1 for each year i the study for a typical idividual, holdig smokig status costat. ˆβ 2-0.33 This is the estimated differece i mea FEV1 betwee curret ad former smokers. ˆσ b0 0.55 This represets a measure of the idividual variability i the baselie FEV1 values. ˆσ b1 0.0098 This represets a measure of the idividual variability i the slope. ˆρ b0,b 1-0.31 This is the estimated correlatio betwee the radom itercepts ad radom slopes, idicatig that idividuals with higher estimated itercepts have more egative slopes (faster rate of FEV1 declie). Page 5
(e) Adifferet collaborator suggests ceterig the time variable usig t ij = t ij 9. If the same model is ru usig this ew cetered time variable, describe what you kow about the results based o the results of the model with the ucetered time variable ad the descriptive statistics of the data. Provide justificatio for your respose. i. What will happe to the estimate ad stadard error of ˆβ 1? Solutio: The estimate of the slope ad the stadard error of this estimate will be uchaged due to the ceterig of the time values. ii. What will happe to the estimate ad stadard error of ˆβ 2? Solutio: The estimate of the slope ad the stadard error of this estimate will be uchaged due to the ceterig of the time values. iii. What will happe to the estimate ad stadard error of ˆβ 0? Solutio: The estimate of the itercept will be exactly ˆβ 0old 9( ˆβ 1 ), which is 3.212. The stadard error will differ depedig o variatio i the data; we would expect it to be smaller if the data are truly homoscedastic. iv. What will happe to the estimate of ˆσ b0? Solutio: The estimate of the variability of the radom itercept effect may be differet. If the data were truly homoscedastic, we would expect the variability of the radom itercept to be slightly less after early ceterig the time values. v. What will happe to the estimate of ˆσ b1? Solutio: The estimate of the variability of the radom slopes should remai uchaged. vi. What will happe to the estimate of ˆρ b0,b 1? Solutio: The correlatio betwee the radom effects will likely decrease i absolute value due to the itercept at t = 9. With the itercept at t =0, we saw that the radom itercept ad slope were egatively correlated, idicatig that higher that average values at t =0leadtosteeperslopes (egative values of b 1 ). At t =9,weexpectthiscorrelatiotobeless strog. (f) Fially, ot all participats completed the lug fuctio assessmet at each scheduled time poit. A collaborator turs to you ad asks, is this a problem? Briefly discuss the missig data mechaism assumed i the followig two approaches. Use Page 6
laguage appropriate for a scietific collaborator. i. A GEE approach? Solutio: The GEE approach assumes that the missig data mechaism is missig completely at radom; i this sceario this might be because the machie was broke o radom days. However, if the data were missig due to smokig status or rate of lug fuctio declie (i.e., ot missig completely at radom), the GEE approach could be problematic. I such asceario,estimatesmaybebiasedduetotheuobserveddata. ii. A radom effects model? Solutio: The radom effects model assumes that the missig data mechaism is missig at radom; i this sceario the missig mechaism might be related to smokig status (for istace) ad the radom effects model would be valid. However, if the data were missig based o uobserved factors (i.e., missig ot at radom), this would pose a problem similar to that described usig the GEE approach. Page 7
Divisio of Biostatistics Q2 Exam Day 1 Methods ad Applicatios Survival Questio Cosider a idepedet cesorship model i which failure time T is expoetial with hazard rate θ ad cesorig time C follows a arbitrary oparametric distributio free of θ. The observed data are {(x i,δ i ),i =1,...,} where x i =mi(t i,c i ) ad δ i = I(t i c i ). 1. Write out the full likelihood fuctio L f. Obtai the MLE, say ˆθ 1, of θ from the fuctio L f based o the observed data. The full likelihood fuctio o the basis of (x 1,δ 1 ),...,(x,δ ) is L f = [f(x i ; θ)s C (x i )] δ i [f C (x i )S(x i ; θ)] 1 δ i f(x i ; θ) δ i S(x i ; θ) 1 δ i = Score fuctio is l (θ) = i (δ i/θ x i ) MLE of θ is ˆθ 1 = i δ i i x i. (θe θx i ) δ i (e θx i ) 1 δ i 2. What is the asymptotic distributio of ˆθ 1? Provide a estimator of the asymptotic variace of ˆθ 1. We have l (θ) = i δ i/θ 2, by the asymptotic ormality of MLE, (ˆθ1 θ) D N (0, θ 2 i δ ) i 3. What is the role of the distributio of C i i the maximum likelihood derivatio of ˆθ 1, ad the asymptotic distributio of ˆθ 1? Explai with details. Uder idepedet cesorig, both the likelihood fuctio used to derive ˆθ 1 ad the asymptotic distributio of ˆθ 1 does ot ivolve C distributio. It is uisace ad does ot cotribute to the estimatio ad iferece of parameter θ i T distributio. I the followig, assume that cesorig time C also follows a expoetial distributio with hazard rate λ, ad λ = θ. 1
4. Obtai the MLE, say ˆθ 2, of θ based o the observed data. The full likelihood fuctio becomes L f = {θe θx i e λx i } δ i (λe λx i e θx i ) 1 δ i = θe 2θx i Score fuctio is l f = i (1/θ 2x i) MLE of θ is ˆθ 2 = 2. i x i 5. What is the asymptotic distributio of ˆθ 2? Be explicit with the distributioal parameters. l f = θ 2, so we have (ˆθ2 θ) D N (0, θ2 ) (X =mi(t,c) exp(2θ), so i x i/ p 1/2θ ad ˆθ 2 = 2 i x i p θ) 6. I this case, does the distributio of C i provide iformatio i estimatio of θ? Compare the asymptotic variaces of ˆθ 1 ad ˆθ 2, ad discuss the results (less tha 50 words). The distributio of C cotributes i the likelihood derivatio ad does provide iformatio i estimatio of θ. The ratio of the two asymptotic variace is V ar(ˆθ 2 ) V ar(ˆθ 1 ) = i δ i From Q4, we kow {δ i,i = 1,...,} are i.i.d. radom variables with Beroulli ( θ =1/2as λ = θ), thus i δ i 1/2 whe. θ+λ The asymptotic variace of ˆθ 2 is half of that of ˆθ 1 i large sample. It is reasoable because additioal iformatio o cesorig time C is used i the estimatio ad iferece of θ by ˆθ 2. 2
Solutio to Qual 2 Theory Questio #3 2011 September Exam We assume at least oe x i is ozero. (a) The log likelihood fuctio is give by log(l) =(log 2π) log(σ) 1 2σ 2 (y i βx i ) 2, ad log L β log L σ 2 log L β 2 = 1 σ 2 = 1 σ 2 x i (y i βx i ) = σ + 1 σ 3 (y i βx i ) 2 2 log L σ 2 = σ 2 3 σ 4 2 log L σ β x 2 i (y i βx i ) 2 = 2 σ 3 x i (y i βx i ) The solutio of the first two equatios is give by: ˆβ = x iy i, x2 i ad ˆσ 2 = 1 (y i ˆβx i ) 2. This solutio is uique ad at ( ˆβ, ˆσ) the2 2 matrix of secod order derivatives is egative defiite (at least oe of the diagoal etries is egative ad the determiat is positive) wheever at least oe x i is ozero. (b) Sice ˆβ is a liear fuctio of idepedet ormal radom variables, it is agai ormally distributed with mea β ad variace σ 2 /{ x2 i }. 1
(c) Variace of ˆβ is the smallest wheever the sum of the x 2 i is the largest. This is possible whe the x i are either +1 or 1. So, there are 2 10 differet possible choices for the x i that lead to the smallest variace for ˆβ. (d) Likelihood ratio test statistic is give by ad upo simplificatio, we get whe σ = 1 or Λ = sup β=1 L(β) L( ˆβ) Λ =exp{ 1 2 ( x 2 i )( ˆβ 1)} 2 2 log(λ) =( x 2 i )( ˆβ 1) 2. [Not asked: Clearly, whe H 0 is true, this statistic has a χ 2 distributio with 1 degree of freedom sice Var( ˆβ) =1/( x 2 i ), ad ˆβ is ormally distributed with mea β]. (e) We reject H 0 whe ˆβ is outside the iterval (1 (1.96/10), 1+(1.96/10)) or outside (0.804, 1.196). [Oe ca also describe the critical regio i terms of χ 2 (1) critical value.] (f) No because i Model 1, the alterative hypothesis correspods to the slope aloe while the itercept is assumed to be held at 0 whereas i the test for 0 mea for the differece Y i x i correspods to a test for 0 itercept where the slope parameter β is held at 1. 2