Data β (std. error)ˆ
Original data, before extra 94 baselines made missing 0.947 (0.020) Only using 194 with both baseline and 6-month observed 0.958 (0.024) Using all observations, as described in the text 0.947 (0.024)
Table A.3: Estimates of β using various subsets of data
A.4 Justification of using model in 3.4.3 to obtain conditional treatment estimates
We now develop the argument of the previous Section more formally to show the equivalance of REML (GLS) and analysis of covariance in a repeated measurements setting with baseline observation. Note first that it is sufficent to show the equivalence in the case of no missing data. With missing data assumed MAR, each patient simply contributes through the marginal distribution of their observations.
First a relationship between generalized least squares and covariate adjustment is shown in a generic multivariate normal setting. It is then shown how this applies to the repeated measure- ment setting with a baseline observation.
I. Generic result
Suppose that we have one observation Y from a T dimensional multivariate normal distribution with covariance matrix ΣΣΣ. Further, suppose that we can partition Y such that
E{Y} = E ½µ Y1 Y2 ¶¾ = µ 0 µµµ ¶
for Y1and Y2with dimensions p and q respectively, p + q = T , and µµµ unconstrained. Then the
generalized least squares estimator of µµµ for ΣΣΣ known is ˜ µ µ µ = Y2−ΣΣΣ21ΣΣΣ−111Y1 (A.18) for Σ Σ Σ = µ ΣΣΣ11 ΣΣΣ12 ΣΣΣ21 ΣΣΣ22 ¶ .
That is, ˜µµµ is the covariate adjusted estimator of µµµ with Y1as covariate.
Now suppose that ΣΣΣ must be estimated. Assume that there are an additional m independent observations, whose mean can be assumed 0 without loss of generality,
Zi= µ Z1i Z2i ¶ ∼ N(0,ΣΣΣ), i = 1, . . . m, where the partition conforms to that for Y. Define
A = µ A11 A12 A21 A22 ¶ = m
∑
i=1 ZiZTi .Applying standard REML theory to the complete dataset, Y and {Zi}, i = 1, . . . m, we find that
the REML estimator of ΣΣΣ11 is
ˆΣΣΣ11= 1 m + 1 ¡ A11+ Y1YT1 ¢ . Similary, the REML estimator of ΣΣΣ21is
ˆΣΣΣ21= A21ˆΣΣΣ11A−111,
which implies that the REML estimator of the regression coefficient in (A.18) can be written ˆΣΣΣ21ˆΣΣΣ−111 = A21ˆΣΣΣ11A−111 ˆΣΣΣ−111.
In the special case that Y1has dimension p = 1 this reduces to
ˆΣΣΣ21ˆΣΣΣ−111 = A21A−111
which is equivalent to the estimator that we would obtain from an analysis of covariance of Y2
and {Z2i} on Y1 and {Z1i}, i = 1, . . . , m. In conclusion, in the generic setting given here, the
REML (GLS) estimator of µµµ is identical to the covariate adjusted estimator, provided that Y1
has dimension 1.
II. Application to the repeated measurements setting
Suppose that the ith subject in a two group trial supplies a baseline measurement Yi0 and T
repeated measurements Y1i, . . . ,Y1T. There are n1and n2 subjects respectively in the two arms.
Assume that the subjects are ordered according to treatment group, so that we can define for t = 0, 1, . . . , T , for treatment group 1:
S1t = 1 n1 n1
∑
i=1 Yitand for treatment group 2:
S2t = n1 2 n
∑
i=1+n1 Yit.From these we define Dt= µ 1 n1+ 1 n2 ¶−1/2 (S1t− S2t), t = 0, 1, . . . , T,
the scaled mean treatment differences, for which
E[D0, D1, . . . , DT] = (0, δ1, . . . , δT),
where the δt are assumed to be unconstrained. Similarly we define
Mt= µ 1 n1+ n2 ¶−1/2 (n1S1+ n2S2), t = 0, . . . , T for which E[M0, M1, . . . , MT] = (ψ0, ψ1, . . . , ψT),
A.5 Summary 151 with the ψt assumed to be unconstrained.
Let X be the n × 2 design matrix for (Dt, Mt) that applies to any time point for t > 0 and choose
any n × (n − 2) matrix H that satisfies
HTX = 0 and HTH = I. Define Vt= HT Y1t Y2t ... Ynt , t = 0, 1, . . . , T.
The sets of derived variables {Dt, St, Vt} are mutually independent within time points and have
a common (unstructured) covariance matrix across times. The transformation from the original data to these is of full rank, and the treatment effects of interest are proportional to the δt’s.
As a final step we just need to map these quantities onto those in the part I. First, because the means of the Mtare saturated, and because of the mutual independence, these contribute nothing
to the estimation of the other parameters and so can be ignored. Second, we identify the Dt’s
above with the Yt’s in part I. Third, the ith element of Vt (i ∈ 1, . . . , n − 2) is identified with the
tth element of Ziin I (where m = n − 2). Fourth, and finally, the scaled treatment effects δt are
equated with the elements of µµµ in part I. We can now apply the result from part I to show that the baseline adjusted treatment difference for any time t > 0 will be identical to that obtained from a REML analysis applied to the whole data set treating all the observations including the baseline as responses, provided we ensure the same mean (across treatment groups) for baseline.
A.5 Summary
1. When baseline and responses are missing, and assumed MAR, we need to fit a bivariate normal response model.
2. Patients from whom we only observe the response contribute the marginal normal like- lihood for the responses.
3. Patients from whom we only observe baseline contribute the marginal normal likelihood for the baselines.
4. Means and variances are estimated by default; other parameters can be estimated from these (see the example on p.147) and standard errors estimated using the delta method (e.g. (A.16). Alternatively, as described in Chapter 3, the model can often be parame- terised so the estimates of interest are obtained directly.
Appendix B
Prior eliciting questionnaire (Subsection
6.4.3)
Questionnaire on differences between responders and non-responders in reviewer trial James Carpenter & Stephen Evans, 14th Jan 2003 The trial
Compares the effect of reviewer training by face-to-face workshop or postal tuition pack with control (i.e. no intervention). Participating reviewers were sent three papers (one baseline and two post randomisation) each of which contained a number of errors. Principal outcomes are quality or review (a score summed over 7 items) and number of major errors identified. Here we focus on the quality of review.
Missing outcome data
In any trial, missing outcome data are likely. If non-responders differ systematically from responders then we may get biased estimates of the benefit of the intervention.
Many statisticians now advocate performing sensitivity analysis to allow for systematic differences between responders and non-responders.
We wish to use a Bayesian method that requires researchers to guess the magnitude of this systematic difference, even though it cannot be estimated from data. We will use this information both in the main analysis of the trial and in methodological work for a more statistical audience.
What we want to do
We wish to ask all the investigators to think about plausible values of the mean difference between responders and non-responders. We would like to stress that we are asking about the average differ- ences that you would expect to observe over many thousands of responders and non-responders, not the random differences that you might expect to observe between individual responses. Naturally you are not sure what the differences between responders and non-responders are (if any). Nonetheless you may feel that some differences are more plausible than others. We therefore ask you to enter your weight of belief for each of the possible differences shown in the table overleaf. On a scale of 0 (im- possible) to 100 (certainty), the more strongly you believe the plausibility of a particular difference, the greater should be your weight for that difference. Your weights should sum to 100.
If anything is not clear, or you have any comments then please contact James on (+44) (0)20 7927 2033 or by email on [email protected]
Please turn over
Questionnaire, page 1
Job title:...
Question 1: Have you seen the data? Circle your answer: YES / NO
Question 2: Suppose the mean review quality for reviewers who respond to the second and third (i.e. final) paper is 3, with standard deviation 0.5, so that about 95% of these responders have values between 2 and 4. [These numbers are completely invented.]
What is your expectation for the mean review quality for those who do not respond to the either the second or the third paper?
To help you, the hypothetical example shows a rather idiosyncratic statistician who is convinced that non-responders will differ on average from responders by 1/2 or 3/4 points, but is not sure whether they will be better or worse than responders.
Mean review quality of reviewers who do not respond to either the 2nd or the 3rd paper (minimum 0, maximum 5)
Non-responders
worse than same as better than
responders by responders responders by TOTAL 1 or more 0.75 0.5 0.25 0.25 0.5 0.75 1 or more Hypothetical
example 0 25 25 0 0 0 25 25 0 100
Your answers
PLEASE CHECK YOUR WEIGHTS SUM TO 100
THANK YOU FOR YOUR HELP
Appendix C
Code for examples
The code for each of the examples in Chapters 3–6 is given below. With the exception of two examples, all the code is SAS.
C.1 Code for Chapter 3
Example3.1
* Input placebo arm data only data first;
infile "..."; input id time fev; run;
proc mixed data=first; class id time;
model fev = time/ ddfm=kr htype=2 noint; repeated time/subject=id type=un ;
lsmeans time; run;
Example3.2
* Adjusted for baseline data second;
infile "...";
input fev fevbl treat; run;
proc reg data=second; model fev=treat; run;
proc reg data=second; model fev=treat fevbl; run;
Example3.3
* Adjusted for (=conditional on) baseline data thirda;
infile "...";
input fev fevbl treat; run;
prog reg data=thirda; model fev=fevbl treat; run;
* Data MCAR given baseline and treatment. * Note time indexes baseline and response data third;
infile "...";
input id time fev treat ; run;
*Treatment estimate marginal to baseline proc mixed data=third;
class time treat subject;
model fev = time*treat/s ddfm=kr htype=2; repeated time/subject=id type=un;
run;
Example3.4
data four;
infile "...";
input id treat time fev ; if time=1 then mytreat=0; else mytreat=treat;
run;
proc mixed data=four; class id time mytreat;
model fev = mytreat/ s htype=2 noint ddfm=kr; repeated time/subject=id type=un ;
lsmeans mytreat / diff; run;
C.1 Code for Chapter3 157 Example3.6
data five;
infile "...";
input id treat rinidc fev newtreat; run;
proc mixed data=five;
class id time newtreat;
model fev = newtreat/ s htype=2 noint ddfm=kr; repeated rinidc/subject=id type=un ;
lsmeans newtreat / diff; run;
Example3.7
This uses the same code as Example 3.6, but with a different data arrangement, shown in Ta- ble3.12.
Example3.8
data six;
infile "...";
input id sex age baseline treat newtreat resp ; run;
*Model with all exacerbations as separate responses proc mixed data=six maxit=50 ;
class id sex treat newtreat;
model resp = newtreat treat age sex baseline newtreat*treat newtreat*age
newtreat*sex newtreat*baseline /s Htype=2 Ddfm=Kr; repeated newtreat/Subject=Id Type=Un ;