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15. Diffusion Processes
greater accuracy can be achieved by solving Kolmogorov’s forward equa-
tion. The ideal of an exact solution is seldom attained in practice, even
for time-homogeneous problems. However, Kolmogorov’s forward equation
can be solved numerically by standard techniques for partial differential
equations. Here we would like to discuss a nonstandard method for finding
the distribution of X t that directly exploits the definition of a diffusion
process.
at n times
This method recursively computes the distribution of X t i
points labeled 0 <t 1 < ··· <t n = t. In the diffusion approximation to the
Wright-Fisher model, it is reasonable to let δt i = t i+1 −t i be one generation.
It is also convenient to supplement these points with the initial point t 0 =0.
∈ [a ij ,a i,j+1 ]
For each t i , we would like to compute the probability that X t i
. We will say more about these mesh points
for r i +1 points a i0 < ··· <a i,r i
∈ [a ij ,a i,j+1 ])
later. In the meanwhile, let p ij denote the probability Pr(X t i
∈ [a ij ,a i,j+1 ]). Our method
and c ij the center of probability E(X t i | X t i
carries forward approximations to both of these sequences starting from an
arbitrary distribution for X 0 .
In passing from time t i to time t i+1 , the diffusion process redistrib-
utes a certain amount of probability from interval [a ij ,a i,j+1 ] to interval
[a i+1,k ,a i+1,k+1 ]. Given the definition of a diffusion process and the no-
2
2
tation m(i, x)= x + µ(t i ,x)δt i and s (i, x)= σ (t i ,x)δt i , the amount
redistributed is approximately
p ij→i+1,k
2
a i,j+1 1 a i+1,k+1 [y−m(i,x)]
= e − 2s 2 (i,x) dyf(t i ,x) dx. (15.18)
2
2πs (i, x)
a ij a i+1,k
[a i+1,k+1 −m(i,x)]/s(i,x)
a i,j+1 1 z 2
= √ e − 2 dzf(t i ,x) dx.
2π [a i+1,k −m(i,x)]/s(i,x)
a ij
(Here and in the remainder of this section the equality sign indicates ap-
proximate equality.) In similar manner, the center of probability c ij→i+1,k
of the redistributed probability approximately satisfies
c ij→i+1,k p ij→i+1,k
a i,j+1 a i+1,k+1 [y−m(i,x)] 2

1 −
= ye 2s 2 (i,x) dyf(t i ,x) dx (15.19)
2
2πs (i, x)
a ij a i+1,k
a i+1,k+1 −m(i,x)
a i,j+1 1 s(i,x) z 2

= √ [m(i, x)+ s(i, x)z]e − 2 dzf(t i ,x) dx.
2π a i+1,k −m(i,x)
a ij
s(i,x)
Given these quantities, we calculate
r i −1

=
p i+1,k p ij→i+1,k
j=0