Page 177 - Applied Probability

8. The Polygenic Model
162
random sample z 1,...,z l in a manner that leaves the loglikelihood
invariant up to a known multiplicative constant. Show that this re-
quirement can be expressed formally as
k
k

− ln | det Ω|− k tr Ω −1 1 (y j − µ)(y j − µ) t
2 2 k
j=1
 
l
l l −1 1 t
 
= −c ln | det Ω|− tr Ω (z j − µ)(z j − µ)
2 2 l
j=1 
for some constant c. Matching terms involving ln | det Ω| forces the
k
choice c = .Given c, we then take
l
l k
1 t 1 t
(z j − µ)(z j − µ) = (y j − µ)(y j − µ) .
l k
j=1 j=1
Prove that this last equality holds for all µ if and only if ¯ z =¯ y and
l k
1 t 1 t
(z j − ¯ z)(z j − ¯ z) = (y j − ¯ y)(y j − ¯ y) = S. (8.18)
l k
j=1 j=1
Until this point, we have not specified the reduced sample size l.If
each y j has m components, we claim that we can take l = m +1.
This claim is based on constructing m+1 vectors v 1 ,...,v m+1 in R m
satisfying
m+1

= 0 (8.19)
v j
j=1
m+1
m t
v j v j = I m×m ,
m +1
j=1
where I m×m is the m × m identity matrix. Given these vectors and
t
given the Cholesky decomposition S = MM of the sample variance
√
of the sequence y 1 ,...,y k , we define z j = mMv j +¯ y. Show that this
construction yields the sample mean equality ¯ z =¯ y and the sample
variance equality (8.18).
Thus, it remains to construct the sequence v 1 ,...,v m+1 . Although
geometrically the vectors form the vertices of a regular tetrahedron,
we proceed in a purely algebraic fashion. For m = 1, the trivial choice
v 1 = (1) and v 2 =(−1) clearly meets the stated requirements. If
v 1 ,... ,v m work in R m−1 , then verify by induction that the vectors
√
−2
v j 1 − m
w j = −1 1 ≤ j ≤ m
−m