Page 171 - Applied Probability

8. The Polygenic Model
156
recurrence
1
1
Cov(X l ,X j )
(8.13)
Cov(X i ,X j )=
Cov(X k ,X j )+
2
2
is precisely the recurrence obeyed by ordinary kinship coefficients.
Calculation of variances is a little more complicated. If i is a founder,
then the binomial distribution (8.12) implies Var(X i )=2n.If i has parents
k and l, then
Var(X i )=E[Var(X i | X k ,X l )] + Var[E(X i | X k ,X l )]
= E[Var(Y k→i | X k ,X l )] + E[Var(Y l→i | X k ,X l )]
1
+ Var[ (X k + X l )]
2
= E[Var(Y k→i | X k )] + E[Var(Y l→i | X l )] (8.14)
1 1 1
+ Var(X k )+ Cov(X k ,X l )+ Var(X l ).
4 2 4
To make further progress, we must compute E[Var(Y k→i | X k )]. With
this end in mind, suppose we label the 2n polygenes of k by the numbers
1,... , 2n, and let W m be +1 or −1 according as the mth polygene of
k is positive or negative. If we also let A m be the event that the mth
polygene of k is sampled in forming the gamete contribution Y k→i , then
2n 2n
X k = W m and Y k→i = W m . A moment’s reflection shows

m=1 m=1 1 A m
)= 1 and that
that Var(1 A r
4
2n−2

n−2 1
) = −
2n 4
Cov(1 A r , 1 A s
n
1
= − .
4(2n − 1)
t
Conditional on W =(W 1 ,... ,W 2n ) , we therefore calculate that
Var(Y k→i | X k )=Var(Y k→i | W)
+ 2
*
2n 2n

1 1 2 1
= + W − W m
m
4 4(2n − 1) 4(2n − 1)
m=1 m=1

1 1 1 2
= + 2n − X k
4 4(2n − 1) 4(2n − 1)
and consequently that
1 1 1

E[Var(Y k→i | X k )] = + 2n − Var(X k ).
4 4(2n − 1) 4(2n − 1)
Substituting this and a similar expression for E[Var(Y l→i | X l )] in equation
(8.14) produces the recurrence

1 1 1 1
Var(X i ) = + 2n + 1 − Var(X k )
2 2(2n − 1) 4 2n − 1