Page 103 - Applied Probability

5. Genetic Identity Coefficients
and by symmetry
1
2
1
1
2
1
2
2
=Φ({G ,G }{G }{G })+Φ({G ,G }{G }{G })
Ψ 8
i
j
i
i
j
i
j
j
1
1
2
2
2
1
2
1
i
j
j
i
j
i
i
j
1
1
2
2
4Φ({G ,G }{G }{G })
= +Φ({G ,G }{G }{G })+Φ({G ,G }{G }{G }) 87
j
i
j
i
since S 8 = S ∗ ∪ S ∗ ∪ S ∗ ∪ S .
∗
10 11 13 14
It is straightforward to express the Ψ’s in terms of the ∆’s by conditioning
on which condensed identity state the four original genes of i and j occupy.
For instance,
1 1 1 1
Ψ 1 =∆ 1 + ∆ 3 + ∆ 5 + ∆ 7 + ∆ 8 . (5.1)
4 4 8 16
To verify equation (5.1), suppose the four genes of i and j occur in con-
densed identity state S 1 . Then the four randomly sampled genes fall in S 1
with probability 1. This accounts for the first term on the right of equation
1
(5.1). The second term ∆ 3 arises because if the four genes of i and j are
4
1
2
in S 3 , both G and G must be drawn from the lower left gene of S 3 to
j j
achieve state S 1 for the randomly sampled genes. Given condensed identity
2
1
state S 3 , G and G are so chosen with probability 1 . The term 1 ∆ 5 is
j j 4 4
1
accounted for similarly. The term ∆ 7 arises because if the four genes of i
8
and j are in S 7 , the four randomly sampled genes must all be drawn from
either the left-hand side of S 7 or the right-hand side of S 7 . Finally, the
term 1 ∆ 8 arises because if the four genes of i and j are in S 8 , the four
16
randomly sampled genes can only be drawn from the left-hand side of S 8 .
The remaining condensed identity states are incompatible with the random
condensed identity state S 1 . For example, there is no term involving ∆ 2 in
equation (5.1) since S 2 does not permit identity by descent between any
gene of i and any gene of j.
Similar reasoning leads to the complete system of equations
1 1 1 1
Ψ 1 =∆ 1 + ∆ 3 + ∆ 5 + ∆ 7 + ∆ 8
4 4 8 16
1 1 1 1 1 3 1
Ψ 2 =∆ 2 + ∆ 3 + ∆ 4 + ∆ 5 + ∆ 6 + ∆ 7 + ∆ 8 + ∆ 9
4 2 4 2 8 16 4
1 1 1
Ψ 3 = ∆ 3 + ∆ 7 + ∆ 8
2 4 8
1 1 1
Ψ 4 = ∆ 4 + ∆ 8 + ∆ 9
2 8 4
1 1 1
Ψ 5 = ∆ 5 + ∆ 7 + ∆ 8 (5.2)
2 4 8
1 1 1
Ψ 6 = ∆ 6 + ∆ 8 + ∆ 9
2 8 4
1
Ψ 7 = ∆ 7
4