Page 174 - Applied Probability

8. The Polygenic Model
159
because of the functional relationship featured in Problem 1. (Hint:
1
i
−
2
(Y −A i µ)? Recall that a linear trans-
What is the distribution of Ω
i
formation of a multivariate normal variate is multivariate normal.)
3. Verify that the formulas (8.3) for the expected information matrix
2
continue to hold when the mean A(µ) and the covariance Ω(σ ) are
2
nonlinear functions of the underlying parameter vectors µ and σ ,
provided any appearance of A ∂ µ is replaced by ∂ A(µ) and any
∂µ i ∂µ i
2
appearance of Γ i is replaced by ∂ 2 Ω(σ ).
∂σ
i
4. Suppose all pedigrees from a sample have been amalgamated into a
single pedigree. For a trait vector Y with E(Y )= 0, consider the
covariance components model
r
2
Y i Y j = σ Γ kij + e ij , (8.16)
k
k=1
where the e ij are independent, identically distributed random errors.
t
Let U be the matrix YY , W k be the matrix Γ k , and e be the matrix
(e ij ), all written in lexicographical order as column vectors. Then the
model (8.16) can be written as
2
U = Wσ + e, (8.17)
where W =(W 1 ,... ,W r ). Show that the normal equations for esti-
t
2
mating σ reduce to one step of scoring starting from (0,... , 0, 1) .
This result is due to Robert Jennrich.
5. As an alternative to scoring in the polygenic model, one can imple-
ment the EM algorithm [7]. In the notation of the text, consider a
multivariate normal random vector Y with mean ν = Aµ and covari-
r 2 2
ance Ω = σ Γ k , where A is a fixed design matrix, the σ > 0,
k=1 k k
the Γ k are positive definite covariance matrices, and Γ r = I. Let the
complete data consist of independent, multivariate normal random
r
1
k
k
vectors X ,...,X such that Y = r k=1 X and such that X has
2 t
2
2
mean 1 {k=r} Aµ and covariance σ Γ k .If γ =(µ 1 ,...,µ p ,σ ,...,σ ) ,
r
1
k
and the observed data are amalgamated into a single pedigree with
m people, then prove the following assertions:
(a) The complete data loglikelihood is
r
1 2
ln f(X | γ)= − {ln det Γ k + m ln σ k
2
k=1
1 k k t −1 k k
+ 2 [X − E(X )] Γ k [X − E(X )]}.
σ
k