Page 116 - Applied Probability

6. Applications of Identity Coefficients
100

l kl p l = 0 for all k.
It follows that the optimal deviations satisfy
δ
Because E(X) = 0 and

0=
δ kl p l
l
k p k k p k = 1, we also find that

= δ kl p k p l
k l

= µ kl p k p l − α k p k p l − α l p k p l
k l k l k l

= −2 α k p k .
k

Using the fact α k p k = 0 just established, we now conclude that
k

0= δ kl p l
l

= µ kl p l − α k p l − α l p l
l l l

= µ kl p l − α k .
l

In other words, α k = µ kl p l .
l
The above calculations can be carried out in a more abstract setting.
Suppose Z 1 and Z 2 are independent random variables. Given a random
variable X with E(X) = 0, how can one choose functions h 1 and h 2 so that
2
E{[X − h 1 (Z 1 ) − h 2 (Z 2 )] } is minimized? This problem is easy to solve if
one observes that E[h 1 (Z 1 )] = E[h 2 (Z 2 )] = 0 should hold and that
Var[X − h 1 (Z 1 ) − h 2 (Z 2 )] = Var(X)+Var[h 1 (Z 1 )] + Var[h 2 (Z 2 )]
− 2Cov[X, h 1 (Z 1 )] − 2Cov[X, h 2 (Z 2 )]
= Var[X − h 1 (Z 1 )] + Var[X − h 2 (Z 2 )]
− Var(X).
Now it is well known that Var[X − h i (Z i )] is minimized by taking h i (Z i )
to be the conditional expectation E(X | Z i )of X given Z i [4]. In the
present case, X is the trait value, Z 1 is the maternal allele, and Z 2 is the
paternal allele. The solution E(X | Z i = a k )= µ kl p l coincides with
l
α k given above. It is natural to introduce the additive genetic variance
2
σ = 2 Var[E(X | Z i )] and the dominance genetic variance
a
σ 2 = Var[X − E(X | Z 1 ) − E(X | Z 2 )].
d
Next suppose i and j are relatives. It is of some interest to compute the
covariance Cov(X i ,X j ) between the trait values X i and X j of i and j. Let