Page 337 - Applied Probability

15. Diffusion Processes
326
for X 0 = x 0 . The limiting value of η/(1−f) is the same as the deterministic
equilibrium. In the case of neutral evolution with f = 1 and η = 0, the
mean E(X t )= x 0 is constant. In these circumstances, equation (15.15)
reduces to
2
d E(X t ) − E(X )
t
Var(X t )=
dt 2N
2
x 0 − x − Var(X t )
0
= ,
2N
with solution

t
Var(X t )= x 0 (1 − x 0 ) 1 − e − 2N .
This expression tends to x 0 (1 − x 0 )as t tends to ∞, which is the variance
of the limiting random variable
,
= 1 with probability x 0
0 with probability 1 − x 0 .
X ∞
Fan and Lange [5] calculate Var(X t ) for the dominant case. In the recessive
case, this approach to E(X t ) and Var(X t ) breaks down because µ(t, x)is
quadratic rather than linear in x.
15.6 Equilibrium Distribution

In certain situations, a time-homogeneous diffusion process will tend to
equilibrium. To find the equilibrium distribution, we set the left-hand side
of Kolmogorov’s equation (15.5) equal to 0 and solve for the equilibrium
distribution f(x) = lim t→∞ f(t, x). Integrating the equation

d 1 d 2 2
0= − µ(x)f(x) + 2 σ (x)f(x) (15.16)
dx 2 dx
once gives

1 d 2
= −µ(x)f(x)+ σ (x)f(x)
k 1
2 dx
for some constant k 1 . The choice k 1 = 0 corresponds to the intuitively
reasonable condition of no probability flux at equilibrium. Dividing the no
2
flux equation by σ (x)f(x) yields
d 2 2µ(x)
ln[σ (x)f(x)] = .
2
dx σ (x)