Appendix
The appendix is provided as a reference for the reader with interest in how some
of the calculations under discussion were derived, or for use in the design of future
trials. Because the examples and applications are based on dichotomous outcomes, the
derivations in the appendices are presented in those terms. However, the derivations
depend only on the central limit theorem (that is, that the sample mean tends to a
normal distribution), and hence the results are equally applicable to continuous as well
as dichotomous outcomes.
1
Appendix
A. Formulae for deriving the study-wide alpha for a common control trial:
Let X=A-C, Y=B-C be bivariate normal N 0,0,1,1, ρ . Let Zx and Zy be critical values and
α x  2 Z x  and α y  2Z y  be the corresponding alpha levels, where  is the
cumulative normal,   1  , and  is the normal density. Then the study wide
α  α x  α y  .051* α x  α y   5.137 * α x * α y *   .245 * α x  α y *  2
with an error less than 2.5% for
.01  α  .1 .
Proof: The distribution of X,Y is given by
f(x, y ) 
1
2π 1 ρ2
e

1 1

x2 2ρxy  y 2
2 1ρ2


1
2π 1 ρ2
1 x ρy
 
2 1ρ2
e 




2

1
2π
e
1
 y2
2
 = Prob (Type I error) = Prob (|X|>Zx and/or |Y|>Zy)
= Prob (|X|>Zx)+Prob(|Y|>Zy)-Prob(|X|>Zx and |Y|)>Zy)
 
= 2Z x   2 Z y  2 Prob(X>Zx and Y>Zy)-2Prob (X>Zx and Y<-Zy)
 
 2Z x   2 Z y  2 Prob(X>Zx and Y>Zy), since >0.
Now Prob(X>Zx and Y>Zy)

 y 
xdxdy = 

 Z  ρy 

x
dy
2
 1 ρ 


 y  


1  y  ρ -1 1  y / 2   -1 1   x / 2 

dy , and the result is then obtained by
2

2 0 
1

ρ


Zy


=
Zx

2
ρy , 1ρ
Zy
2
(1)
integrating numerically over a range of alphas and rho and fitting the resulting data so
q.e.d.
the fitted values were within 2.5% of the data.
B. Formulae for deriving the study-wide beta for a common control trial:
Under the alternative
X, Y ~ N , ,1,1, ρ , then the study wide


β  β x * β y  .224 * (β x  β y )  .059 * β x * β y  .179 * (β 2 x  β 2 y ) * ρ


 .138 * (β x  β y )  .867 * β x * β y  .622 * (β 2 x  β 2 y ) * ρ 2
with an error less than 2.5% for .1 
(2)
  .4 .
Proof:
β =Prob(X<Zx and Y<Zy|Δ). Now f  x, y  
1
=
2π 1  ρ 2
e
1  x  Δ  ρ  y  Δ  
 

2
1 ρ 2


β =   y  Δ  
=
=
Zy
Zx



Zy


Zy 


x
2π 1  ρ 2

e

1 1
 x   2  2ρ  x    y     y   2
2 1 ρ 2

2
1

Δ  ρ  y  Δ , 1ρ 2
Z
  y  
1
1  2  y  Δ 2
so
e
2π
x dxdy
   ρy    
dy

1  ρ2

(let y  y   )
-1
 -1

 Z  Δ  ρy 
x
dy   y    x   ρ y  dy , and the result is then
0 


1  ρ 2 
1 ρ2



 y 
obtained by integrating numerically over a range of betas and rho and fitting the
resulting data so the fitted values were within 2.5% of the data. q.e.d.
3
Application: Let the observations be ai, bj, ck and set X  A  C , Y  B  C . Then
Var X =Var Y 
Vara Varc
. Set X  X/ VarX and Y  Y/ Var Y . Then X and Y satisfy

N
kN
the criteria of the lemma. Note that if |E(a-c)|=  ,    /
(1) for
x =  y
for a given study wide  , and (3) for



Vara Varc

. Thus solving
N
kN
β x = β y for a given study wide β ,
the sample size, N(2+k), is given by solving β y    ( x / 2)   /
1
Vara Varc 
.

N
kN 
E.g., if θ T is the expected rate under treatment and θ C the expected control rate,
  θ T  θ C and    /
θ T 1 θ T  θ C 1 θ C 

.
N
kN
C. Formulae deriving the expected correlation (log of the relative risks) between the
comparison of each intervention to the common control, in a given subgroup. These
calculations are derived under the subgroup null, that is, presuming outcomes in each
subgroup are similar to the overall outcomes.
Consider the case of a binary outcome and suppose the overall rates are
 C , 1 and  2 .
Suppose baseline criteria identify subgroups (by the 1:1:k allocation) of size n, n and
kn. Often subgroups are chosen because of a presumed greater or lesser underlying
risk. Thus, under the subgroup null, the rates in the subgroups can be represented by
θ C   C , θ1   C , and θ 2   C . Then, under the subgroup null,
4
~
~
θ 1 θ C   ~

θ 1 θ1 

 , θ1 ~N  θ1, 1
θC ~N  θ C , C
 and similarly for θ2 , and the relative risks are
n
kn




S1=
θ1

θ

~
~
~
~
 1 and S2= 2  2 . Let X=log( S1 )=log( θ1 )-log( θ2 ) and Y=log( S 2 ). To a
θ C C
θ C C


first order of approximation, if W~ μ,σ 2 , log(W)~(log ( μ ),
σ2
μ2
) so, under the null X~
1  θ1 1  θ C

θ1
kθ C
N(log ( 1 )-log(  C ),
) and similarly for Y.
n
Then under the subgroup null, corr(X,Y)=

1
kOR1  1kOR 2  1
1  θC
kθ C
 1  θ1 1  θ C


kθ C
 θ1
 1  θ 2 1  θ C


kθ C
 θ 2



where OR is the odds ratio for the effect of treatment versus
the common control in the subgroups. Thus the corr(X,Y) decreases approximately
Let R  corr(X, Y) when
linearly from 1 to 0 as k increases from 0.
 1
(that is
when the risk in the subgroup is the same as the overall risk), and write
  1 1/  .
Then, as a function of  , under the null, corr(X, Y)( )  R  T , where
R and T can be estimated from the entire sample.
Assuming then that the set of
subgroups evaluated has risks centrally distributed about
 1
, the expected
correlation of the pairs of log relative risks across the subgroups would be R.
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D.
Calculating the probability for the magnitude of correlation observed between
treatment subgroups in common control trials (exemplified for SCD-HeFT):
The transformation W=
1
1
2
(loge(1+r)-loge(1-r)) has variance σ w 
[11]. Using this
2
n 3
transformation, the probability of observing the magnitude of correlation (r= 0.92) as
was seen in SCD-HeFT, when the expected value should have been 0.55, was
 (1.94)  0.052 .
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