Appendix S1. Proof of Independence between Z12,agonistic and
Z12,antagonistic
To prove that Z12,agonistic ?Z12,antagonistic , we show that
y = (Z12,agonistic , Z12,antagonistic )T ⇠ N2 (µy , ⌃y ) with the identity matrix
⌃y = ( 10 01 ). To this end, we use the theorem (4.3) in Dickhaus [1] on page 50:
Theorem 1 Let x ⇠ Np (µx , ⌃x ) and y = Ax + b, in which A is a
(q ⇥ p)-matrix with rank(A) = q  p. Then y ⇠ Nq (µy , ⌃y ) with µy = Aµx + b
and ⌃y = A⌃x AT .
We let x = (Z1 , Z2 )T be N2 (µx , ⌃x ) distributed with µx = ( 00 ) and
⌃x = ( 10 01 ). Then, y = (Z12,agonistic , Z12,antagonistic )T = Ax + b with
1 ) and b = ( 0 ).
A = p12 ( 11 -1
0
It can be seen that E(y) = µy = Aµx + b = ( 00 ) with variance
⌃y = A⌃x AT
✓
◆✓
◆✓
1
1
1
1 0
1
=p
1
0 1
1
2 1
✓
◆✓
◆
1 1
1
1
1
=
1
1
1
2 1
✓
◆
1 0
=
.
0 1
1
1
◆
1
p
2
Reference for Appendix S1
1. Dickhaus T. Simultaneous Statistical Inference: With Applications in the
Life Sciences. vol. 1. Heidelberg - New York: Springer; 2014.