Page 170 - Applied Probability

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8. The Polygenic Model
for their number of positive polygenes i.
The hypergeometric polygenic model mimics the polygenic model well
in two regards. First, both models entail the same pattern of variances
and covariances among the relatives of a non-inbred pedigree. Second, as
n →∞ in the hypergeometric polygenic model, appropriately standardized
trait values within a pedigree tend to multivariate normality. We will verify
the first of these assertions, leaving the second for interested readers to
glean from the reference [20].
To compute the means, variances, and covariances of the trait values
within a pedigree, let X i denote the trait value of pedigree member i. When
i is a pedigree founder, E(X i ) = 0 by virtue of the binomial distribution
(8.12). If i has parents k and l in the pedigree, then we can decompose
X i = Y k→i + Y l→i
into a gamete contribution from k plus a gamete contribution from l. As-
suming that the parental trait means vanish, we infer that
E(X i ) = E[E(X i | X k ,X l )]
=E[E(Y k→i | X k )] + E[E(Y l→i | X l )]
1 1

=E X k +E X l
2 2
=0
and inductively conclude that all trait means in the pedigree vanish.
To compute trait covariances, let j be another member of the pedigree
who is not a descendant of i.If i is a founder, then X i and X j are inde-
pendent and consequently uncorrelated. If i has parents k and l, then
Cov(X i ,X j ) = E[Cov(X i ,X j | X k ,X l )]
+Cov[E(X i | X k ,X l ), E(X j | X k ,X l )]
1
=0 + Cov[ (X k + X l ), E(X j | X k ,X l )]
2
1
= Cov[X k , E(X j | X k ,X l )]
2
1
+ Cov[X l , E(X j | X k ,X l )]
2
1
= Cov[E(X k | X k ,X l ), E(X j | X k ,X l )]
2
1
+ Cov[E(X l | X k ,X l ), E(X j | X k ,X l )]
2
1 1
= Cov(X k ,X j )+ Cov(X l ,X j ).
2 2
Note that E[Cov(X i ,X j | X k ,X l )] = 0 in this calculation because X i and
X j are independent conditional on the parental values X k and X l . The