Page 175 - Applied Probability

8. The Polygenic Model
160
(b) Omitting irrelevant constants, the Q(γ | γ n ) function of the EM
algorithm is
Q(γ | γ n )
1 r−1 2 1 −1 t −1

= − {m ln σ + 2 [tr(Γ k Υ nk )+ ν Γ k ν nk ]}
nk
k
2 σ
k=1 k
m 2 1 t
− ln σ − [tr(Υ nr )+ (ν nr − Aµ) (ν nr − Aµ)],
r
2 2σ r 2
where ν nk is the conditional mean vector
2
=1 {k=r} Aµ + σ Γ k Ω −1 [y − Aµ]
ν k
k
and Υ nk is the conditional covariance matrix
2
2
= σ Γ k − σ Γ k Ω −1 2
Υ k σ Γ k
k k k
k
of X given Y = y evaluated at the current iterate γ n . (Hint:
k
Consider the concatenated random normal vector X and use
Y
fact (e) proved in the text.)
(c) The solution of the M step is
t
t
µ n+1 =(A A) −1 A ν nr
1 −1 t −1
2
σ = [tr(Γ Υ nk )+ ν Γ ν nk ], 1 ≤ k ≤ r − 1
n+1,k k nk k
m
1 t
2
σ = [tr(Υ nr )+ (ν nr − Aµ n+1 ) (ν nr − Aµ n+1 )].
n+1,r
m
In the above update, µ n+1 is the next iterate of the mean vector
µ and not a component of µ.
6. Continuing Problem 5, show that σ 2 ≥ 0 holds for all k and n if
nk
σ 2 ≥ 0 holds initially for all k.If all σ 2 > 0, show that all σ 2 > 0.
1k 1k nk
7. Continuing Problem 5, suppose that one or more of the covariance
matrices Γ k is singular. For instance, in modeling common household
effects, the corresponding missing data X k can be represented as
k
k
X = σ k M k W , where M k is a constant m×s matrix having exactly
one entry 1 and the remaining entries 0 in each row, and where W k
has s independent, standard normal components. Each component of
k
W corresponds to a different household; each row of M k chooses the
correct household for a given person. It follows from this description
that
2 k
σ Γ k = Var(X )
k
2
t
= σ M k M .
k
k