Page 275 - Applied Probability

12. Models of Recombination
263

=0)
=0,N I 2
=0,N I 3
=0,N I 3
+ 2 Pr(N I 2
=0) − 2Pr(N I 1
1

(12.8)
=0} )
4
= − E (1 − 1 {N I 1
=0} )1 {N I 2
=0} (1 − 1 {N I 3
≤ 0.
On the other hand, if the map distance assigned to interval I j is d j , then one
can invoke Mather’s formula (12.1) and exchange avoidance probabilities
for recombination fractions in the above computation. This gives
y {1,2,3} − y {1,3}
1
= − 2[1 − 2M(d 2 )] + 2[1 − 2M(d 1 + d 2 )] (12.9)
8
!
+2[1 − 2M(d 2 + d 3 )] − 2[1 − 2M(d 1 + d 2 + d 3 )] .
Equality (12.9) and inequality (12.8) together imply that
M(d 1 + d 2 + d 3 ) − M(d 1 + d 2 ) − M(d 2 + d 3 )+ M(d 2 )
d 2 +d 3

= [M (d 1 + u) − M (u)]du
d 2
≤ 0.
Because this holds for all positive d j , the integrand M (d 1 +u)−M (u) ≤ 0,

and property (e) follows from the difference quotient definition of M (d)

[27].
Now suppose a chiasma process is determined by a stationary renewal
model with distribution function F(x) having density f(x)= F (x). Be-

cause of stationarity, the map length of the interval [a, a + b]is d = b .In
2µ
view of Mather’s formula (12.1) and the form of the equilibrium density
F (x), the corresponding recombination fraction is

∞
1 1 ∞ !
M(d) = 1 − [1 − F(x)]dx
2 µ b
1 1 ∞ ∞ !
= 1 − f(y)dydx .
2 µ 2µd x
Differentiating this expression twice with respect to d yields
M (d) = −2µf(2µd).

Thus, f(x) can be recovered via
1 x
f(x)= − M . (12.10)
2µ 2µ
Without loss of generality, we can always rescale distances so that µ = 1
2
in this formula.