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Testing treatment combinations versus the corresponding monotherapies in clinical trials
Ekkekhard Glimm, Novartis Pharma AG
8th Tartu Conference on Multivariate Statistics
Tartu, Estonia, 29 June 2007
2 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Setting the scene (I)
The problem
Two monotherapies available for the treatmentof a disease
Question: Does a combination / simultaneousadministration of the treatments („combination“) have a benefit?
Might be
• synergism ( positive interaction between monos)
• a way to overcome dose limitations of monos
3 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Setting the scene (II)
Ultimate task is to find if the best combination therapydose is better than the best dose of any of the monos.
Frequent problem in clinical trials (e.g. hypertension treatment)
A lot of literature on the topic:
Laska and Meisner (1989)
Sarkar, Snapinn and Wang (1995)
Hung (2000)
Chuang-Stein, Stryszak, Dmitrienko and Offen (2007)
4 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Setting the scene (III)
Limited goal in this talk:
Only two monotherapies
Optimal doses are known
Let A, B be the monotherapies, AB their combination.
Assume n individuals per treatment group with response
njABBAiNx iij ,,1,,,),,(~ 2
),max(:0 BAABH vs ),max(: BAABA
5 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Min Test (I)
Laska and Meisner (1989):
/21 AABn xxZ
/22 BABn xxZ
Reject H0 if min(Z1, Z2)>u1- (with u1- N(0,1)-quantile).
Note: Assumption of known is just for convenience, min-t-test is also possible. Same with equal n’s.
6 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Min Test (II)
Rejection probability of this test:
12125.0 22 , uur BAB
nAAB
nm
where .,. is the cdf of
1
1,
0
02
N
This test is uniformly most powerful
in the class of monotone tests (= tests whose teststatistic is a monotone function of Z1 and Z2).
7 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Min Test (III)
The Min test is „conservative“:
Let AB = B > A and
Then the null rejection probability is
The „least favorable constellation“ under H0 is δ→∞
with
But at nominal =0.05!
/2 AB
n
115.00 , uur m
.lim 0
mr
0122.000 mr
8 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Laska and Meisner Min Test (IV)
Min test rejection probability if AB=B>A
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
reje
cti
on
pro
b
Is there a way to alleviate this conservatism?
9 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
„Conditional“ tests
Tests uniformly more powerful (UMP) than the Min test can be derived, if we adjust the critical value based on the observed difference
/2 ABn xxV
In general such tests are of the form:
Reject if
To be UMP than the Min test, a sufficient condition is:
VcZZ ),min( 21
11 , uVcuVcV
and keeping is attainable. || somefor 1 VuVc
10 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Sarkar et al. test
Suggestion by Sarkar et al. (1995):
Reject if
k, d such that -level is kept.
The null rejection prob. r0 can be written as a function of bivariate normal cdfs.
The derivative can be written as a function of bivariate normal cdfs and pdfs.
Using these two components, we can let the computer search for d corresponding to given k (or vice versa).
dVu
dVkZZ
if,
if ,),min(
121
0r
11 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
What can be inferred about the derivative
. As δ from 0, for all δ < some δ+.
For δ→∞: → 0.
There is either no or one δ* where .
If there is, for < * and for > *,
so r0 has a maximum in *.
0)0(0 r
0r
0*0 r
0)(0 r
0)(0 r
0r
12 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Some remarks on computer implementation
k is fixed.
For given d, calculate at = 4.5
If this is <0, decrease d.
Stop if
Idea: If the conditions
hold, 0. = 4.5 is „close enough“ to .
This approach finds d within a few steps.
0r
60 100
r
11 , uVcuVcV
00r
13 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Modified Suggestion: linearized conditional test
Reject if
With this, it is also possible to write down the rejection prob r0 and its derivative
Need to find k,c and d. To limit options, k=0 and c= u1/d were assumed, so just search for d.
Same search algorithm as before.
For non-linear c(|V|), I did not try to work out r0
(maybe possible for special functions).
.0
r
dVu
dVVckZZ
if,
if ,),min(
121
14 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Rejection probability of conditional tests
Rejection probability is highest at max d which has k =
Once k > , power relatively quickly min test as d
0.01
0.02
0.03
0.04
0.05
0 0.5 1 1.5 2 2.5 3
reje
ctio
n p
rob
d=0.02
d=0.03
d=0.033, k=0
d=0.04
d=0.1
linear, d=0.047, k=0
min test
15 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
More Modifications
With both variants, we may want to allow
r0 approaches a value < as δ→∞.
Resulting test is no longer UMP than min test, but
we gain more power (in the vicinity of H0) for small δ.
Here, k, d, c are even easier to find:
The max r0 is at δ where
1ucVcV
.00 r
16 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
More Modifications: „Maximin test“
Idea: Find c(|V|) such that r0(0)=r0(∞).
This test maximizes the minimum rejection probability among all conditional tests.
Results:
Sarkar test with k=0: d=0.08025, c=1.767
r0(0)=r0(∞)=0.0386.
Linearized test with k=0: d=0.2125, c=8.539, c=1.81 r0(0)=r0(∞)=0.0348.
17 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Rejection probability of „maximin“ test (k=0)
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
delta
reje
ctio
n pr
ob
min test
d=0.033
minimax
linear minimax
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
delta
reje
ctio
n pr
ob
min test
d=0.033
maximin
linear maximin
18 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
A remark on the power of conditional tests
Suppose
These tests do not dominate each other.
As and δ increase, the tests with large k, d overtake tests with small k, d.
Unfortunately, „real gains“ coincide with low power: Power gain over Min test (all at δ = 0):
- =0.8: k=0: 10.4%; Min test: 8.7% (max absolute gain, 1.73%)
- =2: k=0: 49.2%; Min test: 48.3%
- =2.8: k=1.3: 79.9%; Min test: 79.6%
- =3: k=1.5: 85.4%; Min test: 85.2%
./2
BABn
19 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
A few remarks on generalizations
Unequal n‘s: No problem. • The in the bivariate Normal distribution changes, so k, c and d
change, but approach remains the same.
Estimated s instead of known : In principle, same approach.
• Rejection prob a sum of bivariate t- rather than normal cdfs.
• Basic idea for constructing a UMP conditional test works the same.
• k, c and d can be found by a grid search.
> 2 monos: Again, in principle same approach, but gets messy: • more than one δ to be considered.
• Generalization of rule „if |V|< d“ to V1, ..., Vg not obvious.
20 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Contentious issues about conditional tests
If we allow k<0, it is possible that we identify the combi as superior, although its observed average is lower than the better of the monos.
→ This can be avoided by requiring k 0.
Non-monotonicity: It can happen thatrejects, but does not, although
ABAB xxx ,,
**,, ABAB xxx
**, AABB xxxx
(However, we should keep in mind: The power never only depends on ).),max( BAAB
21 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Conclusions
„Conditional“ non-monotone tests are UMP than the Laska-Meisner min test.
There is not that much to be gained:
The power depends on .
Even for modest n, the region where the min test‘s r0m<< is very small. E.g. n=8, (B A)/ =1 has r0m = 0.0471.
d is also very small. Only if the monotherapies arereally similar, this makes a difference.
/2 AB
n
22 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Conclusions
Power profile is primarily driven by choice of k, irrespective of the variant of the conditional test.
Gains over the Min test are in the „wrong places“:• They are where power is low (10%). Here, small values of k
are best.
• At powers that matter to the pharma industry, „biggest“ gains are achieved for large k, but are generally very small (<<1%).
k and d are easy to obtain with a relatively simple search algorithm on a computer.
In practice, we‘ll rarely experience a difference from the Min test (with k=0, P( |V|<d |=0)=2.6%).
23 | Testing treatment combinations | Ekkehard Glimm | 29 June 2007
Literature
Laska, E.M. and Meisner, M. (1989): Testing whether an identified treatment is best. Biometrics 45, 1139-1151.
Hung, H.M.J. (2000): Evaluation of a combination drug with multiple doses in unbalanced factorial design clinical trials. Statistics in Medicine 19, 2079-2087.
Sarkar, S.K., Snapinn, S., and Wang, W. (1995): On improving the min test for the analysis of combination drug trials. Journal of Statistical Computation and Simulation 51, 197-213.
Chuang-Stein, C., Stryszak, P., Dmitrienko, A., Offen, W. (2007): Challenge of multiple co-primary endpoints: a new approach. Statistics in Medicine 26, 1181-1192.