The Fundamental Solution to the Wright-fisher Equation
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Chen, Linan, and Daniel W. Stroock. “The Fundamental Solution
to the Wright--Fisher Equation.” SIAM Journal on Mathematical
Analysis 42.2 (2010): 539-567. ©2010 Society for Industrial and
Applied Mathematics.
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http://dx.doi.org/10.1137/090764207
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c 2010 Society for Industrial and Applied Mathematics
SIAM J. MATH. ANAL.
Vol. 42, No. 2, pp. 539–567
THE FUNDAMENTAL SOLUTION TO THE WRIGHT–FISHER
EQUATION∗
LINAN CHEN† AND DANIEL W. STROOCK†
Abstract. The Wright–Fisher equation, which was introduced as a model to study demography
in the presence of diffusion, has had a renaissance as a model for the migration of alleles in the genome.
Our goal in this paper is to give a careful analysis of the fundamental solution to the Wright–Fisher
equation, with particular emphasis on its behavior for a short time near the boundary.
Key words. Wright–Fisher equation, estimates on the fundamental solution
AMS subject classifications. 35K20, 35K65, 35Q92
DOI. 10.1137/090764207
Introduction. The aim of this article is to study the fundamental solution
p(x, y, t) to the Cauchy initial value problem for the Wright–Fisher equation, that
is, the fundamental solution for the equation
(0.1)
∂t u(x, t) = x(1 − x)∂x2 u(x, t) in (0, 1) × (0, ∞) with boundary values
lim u(x, t) = ϕ(x) for x ∈ (0, 1) and u(0, t) = 0 = u(1, t) for t ∈ (0, ∞).
t0
Our interest in this equation is the outgrowth of questions asked by Nick Patterson,
who uses it to model the distribution and migration of genes in his work at the
Broad Institute. Patterson initially sought help from Charles Fefferman, and it was
Fefferman who relayed Patterson’s questions to the second author. Using much more
purely analytic technology, the same questions have been addressed by Charles Epstein
and Rafe Mazzeo, who are preparing a paper (cf. [2]) on the topic. Earlier work on
the same equation can be found in [8] and [5].
The challenge here comes from the degeneracy of the elliptic operator x(1 − x)∂x2
at the boundary {0, 1}. In order to analyze p(x, y, t) for x and y near the boundary,
we will compare it to the fundamental solution to the Cauchy initial value problem
(0.2)
∂t u(x, t) = x∂x2 u(x, t) in (0, ∞) × (0, ∞) with boundary values
u(0, t) = 0 and lim u(x, t) = ϕ(x) for x ∈ (0, ∞),
t0
and in order to provide a context for this problem, we will devote the rest of this
introduction to an examination of some easier but related equations.
First consider the Cauchy initial value problem:
(0.3)
∂t u(x, t) = x2 ∂x2 u(x, t) in (0, ∞) × (0, ∞) with boundary values
u(0, t) = 0 and lim u(x, t) = ϕ(x) for x ∈ (0, ∞).
t0
The fundamental solution for this problem is the density for the distribution of the
solution to the Itô stochastic differential equation:
√
dX(t, x) = 2X(t, x)dB(t) with X(0, x) = x,
∗ Received by the editors July 7, 2009; accepted for publication (in revised form) December 21,
2009; published electronically March 19, 2010.
http://www.siam.org/journals/sima/42-2/76420.html
† Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139
(lnchen@math.mit.edu, dws@math.mit.edu).
539
Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.
540
LINAN CHEN AND DANIEL W. STROOCK
where {B(t) : t ≥ 0} is a standard,
R-valued Brownian motion. As is well known, the
√
solution is X(t, x) = x exp( 2B(t) − t), and so
1
1
P X(t, x) ≤ y = P B(t) ≤ 2− 2 log xy + t = √
2πt
1
2− 2 (log
y
x +t)
ξ2
e− 2t dξ.
−∞
Hence, the fundamental solution to (0.3) is
p(x, y, t) = y −2 p̄(x, y, t) where p̄(x, y, t) =
1 y 2
xy
exp −
log
+ t2
4πt
4t
x
.
Isolating the factor y −2 is natural since the operator x2 ∂x2 is formally self-adjoint with
2
respect to dy
y 2 , and therefore one should expect that y p(x, y, t) is symmetric, which
indeed it is. Furthermore, it should be noted that the spacial boundary condition is
invisible here since X(t, 0) ≡ 0 and X(t, x) > 0 ∀ t ≥ 0 if x > 0. Finally, elementary
calculus shows that although p̄(x, y, t) is smooth on (0, ∞)3 , spacial derivatives of
p̄(x, y, t) become unbounded as x = y
0.
Next, consider the Cauchy initial value problem:
(0.4)
∂t u(x, t) = x∂x2 u(x, t) + 12 ∂x u(x, t) in (0, ∞) × (0, ∞) with boundary values
u(0, t) = 0 and lim u(x, t) = ϕ(x) for x ∈ (0, ∞).
t0
When one ignores the spacial boundary condition, the fundamental solution to this
problem is the density for the distribution of the solution to
dX(t, x) =
2|X(t, x)| dB(t) + 12 dt
with X(0, x) = x.
Although the coefficient of dB(t) is not Lipschitz continuous at 0, a theorem of Watanabe and Yamada (cf. Chapter 10 in [7]) says that this equation has an almost surely
unique solution and that this solution stays nonnegative if x ≥ 0. Further, using Itô’s
formula, one can check that the distribution of this solution is the same as the distri1
1
bution of (x 2 + 2− 2 B(t))2 . Thus, when one ignores the spacial boundary condition,
the fundamental solution is
x+y
1
1
xy
,
y − 2 (πt)− 2 e− t cosh 2
t2
where again we have isolated the factor which is the Radon-Nikodým derivative of the
dy
measure √
y with respect to which the diffusion operator is formally self-adjoint. Obviously, apart from this factor, the fundamental solution is smooth, in fact, analytic,
all the way to the boundary. On the other hand, if we impose the spacial boundary
condition, then the fundamental solution is the density of
1
√
1
1
y P B(t) ≤ 2 2 y 2 − x 2 and B(τ ) > − 2x for τ ∈ [0, t] ,
and an application of the reflection principle shows that this density is
x+y
1
1
xy
y − 2 p̄(x, y, t) where p̄(x, y, t) = (πt)− 2 e− t sinh 2
,
t2
which is smooth but has derivatives which become unbounded as x = y tends to 0.
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541
THE WRIGHT–FISHER EQUATION
So far as we know, there is no general theory which predicts the behavior displayed
by these two examples. Indeed, past experience with degenerate parabolic equations
inclined us to believe that, because x2 has a smooth, nonnegative extension to R,
whereas x does not, the heat equation for x2 ∂x2 should have smoother solutions than
the one for x∂x2 + 12 ∂x . Be that as it may, as we will see in the next section, the
fundamental solution to (0.2) has regularity properties which resemble those of the
fundamental solution to (0.4) when one ignores the spacial boundary condition, and
the reason for this is easy to understand. Namely, because the diffusion corresponding
to (0.2) gets absorbed when it hits 0, the density of its distribution is the same on
(0, ∞) as the density of the diffusion when the spacial boundary condition is ignored.
For potentially practical applications, the most useful results in this article are
likely to be the small time estimates for the Wright–Fisher fundamental solution in
Corollaries 4.4 and 4.5 (as well as the slight refinement in Theorem 6.4) and the expansion procedure, described in Theorem 6.2, for measures evolving under the Wright–
Fisher forward equation. Although the derivative estimates in section 5 may have
some mathematical interest, their practical value is probably not significant.
1. The model equation. Following a suggestion made by Charles Fefferman,
in this section we will study the one-point analogue of the Wright–Fisher equation.
Specifically, we write down that fundamental solution q(x, y, t) for the Cauchy initial
value problem:
(1.1)
∂t u = x∂x2 u in (0, ∞) × (0, ∞)
with u(0, t) = 0 and lim u(x, t) = ϕ(x) for (x, t) ∈ (0, ∞) × (0, ∞).
t0
Actually, prior to our own work on this topic, Fefferman wrote down an expression for
this fundamental solution as a Fourier transform, and Noam Elkies recognized that
Fefferman’s Fourier expression could be inverted to yield the formula at which we will
arrive by non-Fourier techniques. See the discussion following (6.2).
Because x∂x2 is formally self-adjoint with respect to y −1 dy, it is reasonable to
write q(x, y, t) = y −1 q̄(x, y, t) and to expect q̄(x, y, t) to be symmetric with respect to
x and y. In addition, an elementary scaling argument shows that q̄(x, y, t) = q̄( xt , yt , 1).
Taking a hint from the second example in the introduction, we seek q̄(x, y, 1) in the
form
q̄(x, y, 1) = e−x−y q(xy).
If we do, then in order that q be a solution to our evolution equation, we find that it
is necessary and sufficient that
(1.2)
ξq (ξ) − q(ξ) = 0,
ξ ∈ (0, ∞),
and we will show that the solution to (1.2) which makes
(1.3)
q(x, y, t) ≡ y −1 q̄(x, y, t),
where q̄(x, y, t) = e−
x+y
t
q
xy t2
into a fundamental solution is the one satisfying
(1.4)
lim q(ξ) = 0
ξ0
and
lim q (ξ) = 1.
ξ0
Before proceeding, it will be useful to have the following simple version of the
minimum principle. Namely, if ξw (ξ) − w(ξ) ≥ 0, w(0) = 0, and w > 0 in an
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542
LINAN CHEN AND DANIEL W. STROOCK
interval (0, δ) for some δ > 0, then w > 0 on (0, ∞). Indeed, if not, then there would
exist a ξ0 > 0 such that w > 0 on (0, ξ0 ) and w(ξ0 ) = 0. In particular, this would
mean that w achieves a strictly positive, maximum value at some ξ ∈ (0, ξ0 ), and this
would lead to the contradiction w (ξ) ≤ 0 < w(ξ) ≤ ξw (ξ).
Now suppose that q satisfies (1.2) and (1.4). Then
1
0 < q(ξ) ≤ ξe2ξ 2
for ξ ∈ (0, ∞).
1
To see this, apply the preceding to w = q and w = (1 + )ξe2ξ 2 − q(ξ) for > 0.
Given the estimate above, we know that the Laplace transform
f (λ) =
e−λξ q(ξ) dξ
(0,∞)
exists for all λ > 0. Moreover, by (1.2) and (1.4),
dξ
−λξ
=
f (μ) dμ =
e
q(ξ)
e−λξ q (ξ) dξ = λ2 f (λ) − 1,
(1.5)
ξ
(λ,∞)
(0,∞)
(0,∞)
1
and so −f (λ) = λ2 f (λ) + 2λf (λ), or, equivalently, f (λ) = Aλ−2 e λ for some A.
Plugging this back into (1.5), we see that A = 1. That is, we now know that
1
1
eλ
dξ
−λξ
= e λ − 1.
e
q(ξ) dξ = 2 and
e−λξ q(ξ)
(1.6)
λ
ξ
(0,∞)
(0,∞)
In particular, this proves that there is at most one q which satisfies both (1.2) and
(1.4).
As is easily verified,
∞
(1.7)
q(ξ) =
ξn
n!(n − 1)!
n=1
satisfies (1.2) and (1.4), and it is therefore the one and only function which does. In
particular, this means that
1
(1.8)
1
cosh(2ξ 2 ) − 1
≤ q(ξ) ≤ ξe2ξ 2
2
for ξ ∈ (0, ∞).
We now define q̄(x, y, t) and q(x, y, t) as in (1.3) for the q(ξ) given by (1.7). On
the basis of (1.2), it is easy to check that q(x, y, t) satisfies the Kolmogorov backward
and forward equations:
∂t q(x, y, t) = x∂x2 q(x, y, t) and ∂t q(x, y, t) = ∂y2 yq(x, y, t) .
Also, from (1.6) we know that
λx
x
(1.9)
e−λy q(x, y; t) dy = e− λt+1 − e− t
(0,∞)
1
for Reλ > − .
t
Thus, as t
0, (0,∞) q(x, y, t) dy tends to 1 uniformly for x ∈ [δ, ∞) for each δ > 0.
At the same time, from (1.8), we know that
√ xy
−1
cosh 2 t
x+y
x (√y−√x)2
− t
t
(1.10)
e
≤ q(x, y, t) ≤ 2 e−
,
2y
t
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543
THE WRIGHT–FISHER EQUATION
from which it is clear that, as t
0,
q(x, y, t) dy −→ 0
for each δ > 0
(0,∞)\(x−δ,x+δ)
uniformly fast for x in bounded subsets of (0, ∞). Hence, for each ϕ ∈ Cb ((0, ∞); R),
(1.11) uϕ (x, t) ≡
(0,∞)
ϕ(y)q(x, y; t) dy −→ ϕ(x)
uniformly on compacts in (0, ∞).
At the same time, from (1.10), it is clear that, as x
0, uϕ (x, t) −→ 0 uniformly
for t ∈ [δ, ∞) for every δ > 0. Summarizing, we have now shown that q(x, y, t) is a
fundamental solution to (1.1).
It will be important for us to know that the estimate in (1.10) can be improved
when xy ≥ t2 . Namely, there exists an δ0 ∈ (0, 1) such that
1
(1.12)
δ0
(xy) 4
1
t2
e−
√
√
( x− y)2
t
1
(xy) 4
≤ q̄(x, y, t) ≤
1
δ0 t 2
e−
√
√
( x− y)2
t
when xy ≥ t2 .
Clearly (1.12) comes down to showing that
1
1
1
1
δ0 ξ 4 e2ξ 2 ≤ q(ξ) ≤ δ0−1 ξ 4 e2ξ 2
for ξ ≥ 1.
To prove this, one can either use the relationship (cf. (7.4) and (7.6)) between q(ξ)
and Bessel functions or standard estimates for solutions to parabolic equations. The
latter approach entails using the fact that, by standard estimates for solutions to
parabolic equations (e.g., section 5.2.2 in [6]), there is an α0 ∈ (0, 1) such that
α0
t
1
2
≤ q(1, 1, t) ≤
1
1
α0 t 2
for t ∈ (0, 1],
2
which, because q(1, 1, t) = e− t q(t−2 ), gives (1.12).
We close this section with a discussion of the diffusion process associated with
(1.1) and show that q(x, y, t) is the density for its transition probability function.
Perhaps the simplest way to describe this process is to consider the Itô stochastic
integral equation:
t
2Y (τ, x) dB(τ ), (t, x) ∈ [0, ∞) × (0, ∞),
(1.13)
Y (t, x) = x +
0
where {B(τ ) : τ ≥ 0} is an R-valued Brownian motion on some probability space
(Ω, F , P). Using the Watanabe–Yamada theorem alluded to in the introduction, one
can show that, for each x ∈ [0, ∞), there is an almost surely unique solution to (1.13).
Further, if ζ0Y (x) = inf{t ≥ 0 : Y (t, x) = 0}, then t ≥ ζ0Y (x) =⇒ Y (t, x) = 0. To
describe the relationship between (1.1) and Y (t, x), we introduce the σ-algebra Ft
which is generated by {B(τ ) : [0, t]}. Then Y (t, x) is Ft -measurable. In addition,
by Itô’s formula, if w is an element of C 2,1 ((0, ∞) × (0, t)) ∩ Cb ([0, ∞) × [0, t]) and
f = ∂t w + x∂x2 w is bounded, then
τ ∧t∧ζ0Y (x)
Y
Y
w(Y τ ∧ t ∧ ζ0 (x), x), τ ∧ t ∧ ζ0 (x) −
f Y (s, x) ds, Fτ , P
0
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544
LINAN CHEN AND DANIEL W. STROOCK
is a martingale. In particular, if ϕ ∈ Cb ((0, ∞); R) and uϕ is given by (1.11), then, by
taking w(x, τ ) = uϕ (x, t − τ ) and applying Doob’s stopping time theorem, one sees
that
ϕ(y)q(x, y, t) dy = E ϕ Y (t, x) , ζ0Y (x) > t .
(1.14)
uϕ (x, t) =
(0,∞)
This proves that uϕ is the one and only solution to (1.1) and that q(x, · , t) is the
density for the distribution of Y (t, x) on {ζ0Y (x) > t}. As a consequence of uniqueness,
we know that q(x, y, t) satisfies the Chapman–Kolmogorov equation,
(1.15)
q(x, y, s + t) =
q(x, ξ, s)q(ξ, y, t) dξ,
(0,∞)
a fact which could have been also deduced from (1.9).
2. The Wright–Fisher equation. In this section we will lay out our strategy
for analyzing the Wright–Fisher equation:
(2.1)
∂t u(x, t) = x(1 − x)∂x2 u(x, t) on (0, 1) × (0, ∞)
with u(0, t) = 0 = u(1, t) and lim u( · , t) = ϕ
t0
for ϕ ∈ Cb ((0, 1); R). Specifically, we want to explain how we plan to transfer to its
fundamental solution p(x, y, t) the properties of q(x, y, t).
Because we cannot simply write it down, proving that p(x, y, t) even exists requires
some thought. Using the separation of variables and taking advantage of the facts
that the operator x(1 − x)∂x2 is formally self-adjoint in the L2 space for the measure
dy
y(1−y) and that, for each n ≥ 1, the space of polynomials of degree n is invariant for
this operator, Kimura [1] constructed a fundamental solution as an expansion in terms
of the eigenvalues and eigenfunctions. Like all such constructions, his expression for
p(x, y, t) is very useful as t → ∞, when only one term in eigenfunction expansion
dominates, but it gives very little information about p(x, y, t) for small t. In addition,
his approach relies so heavily on the particular equation that it would be difficult to
apply his methods to even slightly different equations. For this reason, we will not
use his analysis.
To get started, we will begin with the diffusion corresponding to (2.1). Thus,
for x ∈ [0, 1], let {X(t, x) : t ≥ 0} be the stochastic process determined by the Itô
stochastic integral equation:
t
2X(τ, x) 1 − X(τ, x) dB(τ ).
(2.2)
X(t, x) = x +
0
Once again, one can show that, for each x ∈ [0, 1], (2.2) has a solution which is
almost surely unique. Further, if ζξX (x) = inf{t ≥ 0 : X(t, x) = ξ} for ξ ∈ [0, 1] and
ζ X (x) = ζ0X (x) ∧ ζ1X (x), then X(t, x) = X(ζ X (x), x) for t ≥ ζ X (x).
From the preceding existence and uniqueness results, one can show (cf. Chapters
6 and 8 in [7]) that if
P (t, x, Γ) = P X(t, x) ∈ Γ and ζ X (x) > t ,
then P (t, x, · ) satisfies the Chapman–Kolmogorov equation:
(2.3)
P (s + t, x, Γ) =
P (t, y, Γ) P (s, x, dy) for x ∈ (0, 1) and Γ ∈ B(0,1) .
(0,1)
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545
THE WRIGHT–FISHER EQUATION
In addition, by the same sort of argument with which we derived (1.14), one has
t
ϕ(y)P (t, x, dy) = ϕ(x) +
0
(0,1)
(0,1)
y(1 − y)ϕ (y) P (τ, x, dy)
dτ
for any ϕ ∈ Cc2 ((0, 1); R). Hence, for each x ∈ (0, 1), P (t, x, · ) tends weakly to the
unit point mass at x as t
0, and, in the sense of Schwartz distributions P ( · , x, ∗)
is a solution to the Kolmogorov forward equation ∂t u = ∂y2 (y(1 − y)u) and therefore,
by standard hypoellicity results for parabolic operators (e.g., section 3.4.2 in [6]), is
smooth with respect to (y, t) ∈ (0, 1) × (0, ∞). In particular, we now know that, for
each x ∈ (0, 1), P (t, x, dy) = p(x, y, t) dy where (y, t) ∈ (0, 1) × (0, ∞) −→ p(x, y, t) is
smooth and satisfies
(2.4) ∂t p(x, y, t) = ∂y2 y(1 − y)p(x, y, t) in (0, 1) × (0, ∞) with lim p(x, · , t) = δx .
t0
Furthermore, another application of the uniqueness statement for solutions to (2.2)
shows that (t, x) ∈ (0, ∞)×(0, 1) −→ P (t, x, · ) is weakly continuous, and so (t, x, y) p(t, x, y) is measurable. Hence, from (2.3), we know that
(2.5) p(x, y, s + t) =
p(ξ, y, t)p(x, ξ, s) dξ, s, t ∈ (0, ∞), and x, y ∈ (0, 1).
(0,1)
In addition, because, by uniqueness, the distribution of 1 − X(t, x) is the same as that
of X(t, 1 − x), it is clear that
p(x, y, t) = p(1 − x, 1 − y, t).
(2.6)
Finally, as we will show below,
y(1 − y)p(x, y, t) = x(1 − x)p(y, x, t).
(2.7)
Actually, if one uses Kimura’s results, (2.7) is clear. However, for the reasons given
earlier, we will give an alternative proof.
In order to focus attention on what is happening at one endpoint, let β ∈ (0, 1)
be given, and consider the fundamental solution pβ (x, ξ, t) to
u̇ = x(1 − x)u on (0, β) with u(0, t) = 0 = u(β, t) and u( · , 0) = ϕ
X
for ϕ ∈ Cc ((0, β); R). Next, given 0 < α < β < 1, define η0,[α,β]
(x) = 0,
(2.8)
X
X
(x) = inf{t ≥ η2(n−1),[α,β]
(x) : X(t, x) = β} and
η2n−1,[α,β]
X
X
η2n,[α,β]
(x) = inf{t ≥ η2n−1,[α,β]
(x) : X(t, x) = α}
for n ≥ 1. Then (cf. Theorem 4.3 below), for x ∈ (0, α] and ϕ ∈ Cc ((0, α); R),
∞
X
X
p(x, ξ, t)ϕ(ξ) dξ =
E ϕ X(t, x) , η2n,[α,β]
(x) < t < η2n+1,[α,β]
(x)
(0,1)
n=0
=
(0,∞)
pβ (x, ξ, t)ϕ(ξ) dξ
∞
E
+
n=1
(0,α)
X
X
pβ α, ξ, t − η2n,[α,β] (x) ϕ(ξ) dξ, η2n,[α,β] (x) < t .
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546
LINAN CHEN AND DANIEL W. STROOCK
Hence, for x, y ∈ (0, α],
∞
(2.9)
p(x, y, t) = pβ (x, y, t) +
X
X
E pβ α, y, t − η2n,[α.β]
(x) , η2n,[α,β]
(x) < t ,
n=1
where the convergence of the series of the right can be controlled by the fact (cf.
X
Lemma 4.2 below) that P(η2n,[α,β]
(x) < t) decreases very rapidly as n → ∞. In view
of (2.9), analysis of p(x, ξ, t) on (0, α]2 × (0, ∞) comes down to that of pβ (x, ξ, t) on
(0, α]2 × (0, ∞).
3. A change of variables. With the goal of applying the results in section 1,
we will make a change of variables, one which was used also by Feller (see [3] and [4]).
Namely, we want to choose a ψ : [0, 1] −→ [0, ∞) so that ψ(0) = 0 and the process
X ψ (t, x) ≡ ψ ◦ X(t, x) satisfies a stochastic integral equation of the form
t
t
ψ
ψ
2X (τ, x) dB(τ ) +
b X ψ (τ, x) dτ
(3.1)
X (t, x) = ψ(x) +
0
0
for t < ζ (x) = inf{τ ≥ 0 : X (τ, x) = ψ(1)}. As an application of Itô’s formula, we
see that ψ will have to be the solution to
ψ
ψ
x(1 − x)ψ (x)2 = ψ(x) with ψ(0) = 0,
(3.2)
which means that
(3.3)
√
ψ(x) = (arcsin x)2 ,
ψ(1) =
π2
,
4
√
and ψ −1 (x) = (sin x)2 .
In addition, his formula tells us that
ψψ b(x) = ψ −1 (x) 1 − ψ −1 (x) ψ ◦ ψ −1 (x) = 2 ◦ ψ −1
(ψ )
√
√
√
sin(2 x) − 2 x cos(2 x)
√
.
=
sin(2 x)
(3.4)
2
Note that, although it goes to ∞ at the right-hand endpoint π4 , b admits an extension
2
2
to (− π4 , π4 ) as a real analytic function which vanishes at 0.
Our strategy should be clear now. We want to apply the results in section 1 to analyze the fundamental solution r(x, y, t) to the heat equation ∂t u(x, t) = x∂x2 u(x, t) +
b(x)∂x u(x, t) associated with X ψ (t, x) and then transfer our conclusions to p(x, y, t)
2
via p(x, y, t) = ψ (y)r ψ(x), ψ(y), t . However, because b becomes infinite at π4 ,
2
2
x∂x + b(x)∂x is a nontrivial perturbation of x∂x . Thus it is fortunate that, by the
considerations in section 2, we have to analyze only the fundamental solution rθ (x, y, t)
to
(3.5)
∂t u(x, t) = x∂x2 u(x, t) + b(x)∂x u(x, t) on (0, θ) with u(0, t) = 0 = u(θ, t)
2
for θ ∈ (0, π4 ). Indeed, because rθ (x, y, t) does not “feel” b off of (0, θ), when analyzing
it we can replace b by any smooth, compactly supported function bθ on R which
coincides with b on (0, θ).
With the preceding in mind, we now look at equations of the form
(3.6)
∂t u(x, t) = x∂x2 u(x, t) + xc(x)∂x u(x, t) on (0, ∞) with u(0, t) = 0,
where c ∈ Cc∞ (R; R).
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547
THE WRIGHT–FISHER EQUATION
Lemma 3.1. Set
(3.7)
C(x) =
1
2
x
c(ξ) dξ
Vc (x) = −x
and
0
c (x) c(x)2
+
2
4
.
Then u is a solution to ∂t u(x, t) = x∂x2 u(x, t) + xc(x)∂x u(x, t) if and only if u(x, t) =
e−C(x) w(x, t), where w is a solution to
∂t w(x, t) = x∂x2 w(x, t) + Vc (x)w(x, t).
(3.8)
Proof. The easiest way to check it is by direct computation.
As Lemma 3.1 shows, the analysis of solutions to (3.6) reduces to that of solutions
to (3.8). Thus, we look at equations of the form
(3.9)
∂t w(x, t) = x∂x2 w(x, t) + V (x)w(x, t) on (0, ∞)2
with w(0, t) = 0,
where V is a smooth, compactly supported function. Using the time-honored perturbation equation of Duhamel, we look for the fundamental solution q V (x, y, t) to (3.9)
as the solution to the integral equation:
t
V
(3.10)
q (x, y, t) = q(x, y, t) +
q(x, ξ, t − τ )V (ξ)q V (ξ, y, τ ) dξdτ.
0
To solve (3.10), set q0V = q and
t
(3.11)
qnV (x, y, t) =
0
(0,∞)
(0,∞)
V
q(x, ξ, t − τ )V (ξ)qn−1
(ξ, y, τ ) dξdτ
for n ≥ 1.
Proceeding by induction and using the Chapman–Kolmogorov equation of (1.15) satisfied by q, one sees that
|qnV (x, y, t)| ≤
(3.12)
(V u t)n
q(x, y, t).
n!
At the same time, by the estimate in (1.12) and induction, it is clear that qnV is
continuous on (0, ∞)3 for each n ≥ 0, and from these one can easily check that
∞
q V (x, y, t) ≡
(3.13)
qnV (x, y, t)
n=0
is a continuous function of (x, y, t) ∈ (0, ∞)3 which solves (3.10). In addition, because
(by another inductive argument)
t
V
V
qn (x, y, t) =
qn−1
(x, ξ, t − τ )V (ξ)q(ξ, y, τ ) dξdτ,
(3.11 )
0
(0,∞)
one can use (1.3) and induction to see that
t
V
q̄(x, ξ, t − τ )ξ −1 V (ξ)q̄n−1
(ξ, y, τ ) dξdτ
yqnV (x, y, t) =
0
(0,∞)
0
(0,∞)
t
=
V
q̄(ξ, x, t − τ )ξ −1 V (ξ)q̄n−1
(y, ξ, τ ) dξdτ = xqnV (y, x, t).
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548
LINAN CHEN AND DANIEL W. STROOCK
Hence,
where q̄ V (x, y, t) ≡ yq V (x, y, t).
q̄ V (x, y, t) = q̄ V (y, x, t)
(3.14)
We next verify that q V is the fundamental solution to (3.9). That is, we want to
show that if ϕ ∈ Cb ([0, ∞); R) vanishes at 0 and
V
wϕ
(x, t) =
ϕ(y)q V (x, y, t) dy,
(0,∞)
V
is a smooth solution to (3.9) with initial data ϕ. Since it is clear that
then wϕ
V
V
( · , t) −→ ϕ uniformly on compacts as t
0, what remains
wϕ (0, t) = 0 and that wϕ
V
is to show that wϕ
is smooth and satisfies the differential equation in (3.9). For this
purpose, set (cf. (1.11))
V
w (x, t) ≡
q(x, y, )wϕ
(y, t) dy
(0,∞)
t
= uϕ (x, t + ) +
(0,∞)
V
q(x, ξ, t + − τ )V (ξ)wϕ
(ξ, τ ) dξdτ.
V
0, w −→ wϕ
and
Then, as (x∂x2
0
+ V − ∂t )w (x, t) =
(0,∞)
V
q(x, ξ, ) V (x) − V (ξ) wϕ
(ξ, t) dξ −→ 0
V
V
V
V
satisfies ∂t wϕ
(x, t) = x∂x2 wϕ
(x, t) + V (x)wϕ
(x, t)
uniformly on compacts. Hence, wϕ
2
on (0, ∞) in the sense of distributions and is therefore a smooth, classical solution.
Theorem 3.2. Set (cf. (1.14))
t
QV (t, x, Γ) = E e 0 V (Y (τ,x)) dτ , Y (t, x) ∈ Γ , Γ ∈ B(0,∞) .
Then
QV (t, x, Γ) =
q V (x, y, t) dy.
Γ
In particular,
0 ≤ q V (x, y, t) ≤ et
(3.15)
and
V+
u
q(x, y, t)
(3.16)
V
q V (x, ξ, s)q V (ξ, y, t) dξ.
q (x, y, s + t) =
(0,∞)
Proof. Given ϕ ∈ Cb ([0, ∞); R) which vanishes at 0, one can apply Itô’s formula
V
and the fact that wϕ
is a smooth solution to (3.9) which is bounded on (0, ∞) × [0, t],
to check that, for each t > 0,
s
V
e 0 V (Y (τ,x)) dτ wϕ
Y (s, x), t − s , Fs , P
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549
THE WRIGHT–FISHER EQUATION
is a martingale for s ∈ [0, t]. Hence, the first assertion follows from the equality of the
expectation values of this martingale at times s = 0 and s = t.
Given the first assertion, it is clear that q V ≥ 0. In addition, the right-hand side
of (3.15) follows from the initial expression for q V (t, x, · ), and (3.16) is a consequence
of
QV (t, ξ, Γ)QV (s, x, dξ) for Γ ∈ B(0,∞) ,
QV (s + t, x, Γ) =
(0,∞)
which is an elementary application of the Markov property for the process {Y (t, x) :
(t, x) ∈ (0, ∞)2 }.
Given θ ∈ (0, ∞), the next step is to produce from q V (x, y, t) the fundamental
solution qθV (x, y, t) to
(3.17)
∂t w(x, t) = x∂x2 w(x, t) + V (x)w(x, t)
on (0, θ) × (0, ∞)
with w(0, t) = 0 = w(θ, t) and lim w( · , t) = ϕ.
t0
Lemma 3.3. Set ζθY (x) = inf{t ≥ 0 : Y (t, x) = θ}, and define
qθV (x, y, t) = q V (x, y, t) − E e
ζθY (x)
0
V (Y (τ,x)) dτ V
q
θ, y, t − ζθY (x) , ζθY (x) ≤ t
for (x, y, t) ∈ (0, θ)2 × (0, ∞). Then qθV (x, y, t) is a nonnegative, smooth function
which is dominated by et V u,(0,θ) q(x, y, t) and satisfies
(3.18) qθV (x, y, s + t) =
qθV (x, ξ, s)qθV (ξ, y, t) dξ and yqθV (x, y, t) = xqθV (y, x, t).
(0,θ)
Finally, for each y ∈ (0, θ), (x, t) qθV (x, y, t) is a solution to (3.17) with ϕ = δy .
Proof. We begin by showing that qθV (x, y, t) is continuous, and clearly this comes
down to checking that
w0 (x, y, t) ≡ E e
ζθY (x)
0
V (Y (τ,x)) dτ V
q
θ, y, t − ζθY (x) , ζθY (x) ≤ t
is continuous. To this end, set
ws (x, y, t) = E e
ζθY (x)s
0
V (Y (τ,x)) dτ V
q
θ, y, t − ζθY (x)s , ζθY (x)s ≤ t ,
where ζθY (x)s = inf{t ≥ s : Y (t, x) = θ}. Because
|ws (x, y, t) − w0 (x, y, t)| ≤ 2 sup q V (θ, y, τ )et
τ ∈(0,t]
V
u
P ζθY (x) ≤ s ,
we know that, as s
0, ws −→ w0 uniformly on compact subsets of (0, θ)2 × (0, ∞),
and so it suffices to show that ws is continuous for each s > 0. But by the Markov
property for the process {Y (t, x) : (t, x) ∈ (0, ∞)2 },
ws (x, y, t) =
w0 (ξ, y, t − s)q V (x, ξ, s) dξ,
(0,∞)
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550
LINAN CHEN AND DANIEL W. STROOCK
and it is clear that the right-hand side is a continuous function of (x, y, t) ∈ (0, θ)2 ×
(0, ∞).
We next note that
t
(3.19)
ϕ(y)qθV (x, y, t) dy = E e 0 V (Y (τ,x)) dτ ϕ Y (t, x) , ζθY (x) > t
(0,θ)
for bounded, Borel measurable ϕ on [0, θ] which vanishes on {0, θ}. Indeed, simply
write the right-hand side of (3.19) as the difference between the expectation of the
integrand without any restriction on ζθY (x) and the one over {ζθY (x) ≤ t}. As a
consequence of (3.19) it is obvious that qθV (x, y, t) is a nonnegative function which
satisfies the asserted upper bound. In addition, the first equality in (3.18) also follows
from (3.19) combined with the Markov property. Namely,
ϕ(y)qθV (x, y, s + t) dy
(0,θ)
s+t
= E e 0 V (Y (τ,x)) dτ ϕ Y (s + t, x) , ζθY (x) > s + t
=E e
s
0
V (Y (τ,x)) dτ
ϕ(y)
=
(0,θ)
(0,θ)
V
(0,θ)
qθV
ϕ(y)qθ Y (s, x), y, t dy, ζθY (x) > s
(ξ, y, t)qθV
(x, ξ, s) dξ
dy.
To prove the second equality in (3.18), we will again apply (3.19). However, before
doing so, note that (1.3) plus the Markov property for {Y (t, x) : (t, x) ∈ (0, ∞)2 }
implies that, for any t > 0 and nonnegative, Borel measurable F on C([0, t]; [0, ∞)),
dx
E F ◦ Y ( · , x) [0, t] , ζ0Y (x) > t
x
(0,∞)
(*)
dy
,
=
E F ◦ Y̆ ( · , y) [0, t] , ζ0Y (y) > t
y
(0,∞)
where Y̆ (τ, y) = Y ((t − τ )+ , y). To see this, note that it suffices to check it for F ’s
of the form F (ω) = nm=0 ϕm (ω(τm )) for some n ≥ 1, 0 = τ0 < · · · < τn = t,
ω ∈ C([0, t]; [0, ∞)), and {ϕm : 0 ≤ m ≤ n} ⊆ Cc ((0, ∞); [0, ∞)), in which case it
is an easy application of the Markov property and (1.3). Returning to the second
equality in (3.18), take
t
F (ω) = ϕ0 ω(0) e 0 V (ω(τ )) dτ ϕ1 ω(t) 1A (ω),
where A = {ω : ω(τ ) = θ for τ ∈ [0, t]}. Then
t
F ◦ Y̆ ( · , x) [0, t] = ϕ0 Y (t, x) e 0 V (Y (τ,x)) dτ ϕ1 Y (0, x) 1(t,∞) ζθY (x) ,
and so, by (*),
dxdy
dxdy
V
=
ϕ0 (x)qθ (x, y, t)ϕ1 (y)
ϕ0 (x)qθV (y, x, t)ϕ1 (y)
x
x
(0,θ)2
(0,θ)2
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THE WRIGHT–FISHER EQUATION
551
for all nonnegative, Borel measurable ϕ0 and ϕ1 which vanish at 0. Together with the
continuity already proved, this shows that qθV (x, y, t) satisfies the second part of (3.18).
Because it is clear that, for each (y, t) ∈ (0, θ) × (0, ∞), qθV (x, y, t) −→ 0 as
x → {0, θ} and that, by (3.19), qθV ( · , y, t) −→ δy as t
0, all that remains to be
shown is that, for each y ∈ (0, θ), (x, t) ∈ (0, θ) × (0, ∞) −→ qθV (x, y, t) is smooth and
satisfies ∂t w(x, t) = x∂x2 w(x, t) + V (x)w(x, t). But, by the second equality in (3.18),
this is equivalent to showing that (y, t) qθV (x, y, t) is smooth and satisfies ∂t w̆(y, t) =
∂y2 (y w̆(y, t)) + V (y)w̆(y, t). Moreover, by standard regularity theory for solutions to
parabolic equations, we will know that it is a smooth solution to this equation as soon
as we show that it is a solution in the sense of Schwartz distributions. Finally, by an
elementary application of Itô’s formula and Doob’s stopping time theorem, we know
from (3.19) that
ϕ(y)qθV (x, y, t) dy − ϕ(x)
(0,θ)
=E
t∧ζθY (x)
exp
0
τ
V (Y (σ, x)) dσ
0
Y (τ, x)∂y2 ϕ Y (τ, x)
+ V Y (τ, x) ϕ Y (τ, x) dτ
t 2
V
=
y∂y ϕ(y) + V (y)ϕ(y) qθ (x, y, τ ) dy dτ
0
(0,θ)
for any ϕ ∈ Cc2 (0, θ); R .
4. Back to Wright–Fisher. We are now ready to return to the Wright–Fisher
equation. To be precise, let ψ and b be the functions defined in (3.3) and (3.4), and
set c(x) = b(x)
x . Given β ∈ (0, 1), let cβ be a smooth, compactly supported function
which coincides with c on (0, ψ(β)], and define
(4.1)
Vβ ψ(x), ψ(y), t
ψ(y)qψ(β)
1
for (x, y, t) ∈ (0, β)2 × (0, ∞),
pβ (x, y, t) =
y(1 − y)
ψ (x)ψ (y)
cβ (x) cβ (x)2
+
where Vβ (x) = −x
.
2
4
By (3.2) and (3.18),
pβ (x, y, s + t) =
(4.2)
(0,β)
pβ (ξ, y, t)pβ (x, ξ, s) dξ
and y(1 − y)pβ (x, y, t) = x(1 − x)pβ (y, x, t).
Lemma 4.1. For each β ∈ (0, 1) and ϕ ∈ Cc ((0, β); R),
X
(4.3) E ϕ X(t, x) , ζβ (x) > t =
ϕ(y)pβ (x, y, t) dy for (x, t) ∈ (0, β) × (0, ∞).
(0,β)
X
(x) : n ≥ 0} is defined as in (2.8),
Furthermore, if 0 < α < β < γ < 1 and {ηn,[α,β]
then
∞
(4.4) pγ (x, y, t) = pβ (x, y, t)+
X
X
E pβ α, y, t−η2n,[α,β]
(x) , η2n,[α,β]
(x) < t∧ζγX (x)
n=1
2
for (x, y, t) ∈ (0, α) × (0, ∞).
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552
LINAN CHEN AND DANIEL W. STROOCK
Proof. To prove (4.3), let u(x, t) denote the right-hand side, and set w(x, t) =
u(ψ −1 (x), t). Then, using the second equality in (3.4), one can easily check that
Vβ
w(x, t) = e−Cβ (x)
qψ(β)
(x, ξ, t)ϕ̃(ξ) dξ,
(0,ψ(β))
1 x
where Cβ (x) = 2 0 cβ (ξ) dξ and ϕ̃ = eCβ ϕ ◦ ψ −1 . Hence, by Lemmas 3.1 and 3.3, w
is smooth and satisfies ∂t w(x, t) = x∂x2 w(x, t) + xc(x)∂x w(x, t) in (0, ψ(β)) × (0, ∞)
with boundary conditions w(0, t) = 0 = w(ψ(β), t) and limt0 w( · , t) = ϕ ◦ ψ −1 .
Thus, using (3.2), one can check that u is smooth on (0, β) × (0, ∞) and satisfies
∂t u(x, t) = x(1 − x)∂x2 u(x, t) there with boundary conditions u(0, t) = 0 = u(β, t) and
limt0 u( · , t) = ϕ. In particular, one can use Itô’s formula and Doob’s stopping time
theorem to conclude from this that u(x, t) equals the left-hand side of (4.3).
Given (4.3), the proof of (4.4) is essentially the same as that of (2.9). The only
change is that one has to take into account the condition ζγX (x) > t, but this causes
no serious difficulty.
Before going further, we will need the estimates contained in the following lemma.
Lemma 4.2. Let 0 < x < α < β < 1 and t ∈ (0, 1]. Then
Y ψ(x) − (ψ(β)−ψ(α))2
4ψ(β)t
P ζψ(β)
e
ψ(x) ≤ t ≤
ψ(α)
and
2
X
x
2 (β−α)
t
P η2n,[α,β]
(x) ≤ t ≤ e−4n
α
for n ≥ 1.
Proof. Because Y ( · , ψ(x)) and X( · , x) get absorbed at 0 and, up to the time
they hit {0, 1}, are random time changes of Brownian motion, one has
ψ(x)
x
Y and P ζαX (x) < ∞ = .
ψ(x) < ∞ =
P ζψ(α)
ψ(α)
α
Y
Y
(ψ(x)) < ∞, ζψ(β)
(ψ(x)) −
Further, because, by the Markov property, given that ζψ(α)
Y
Y
Y
(ψ(α)):
ζψ(α) (ψ(x)) is independent of ζψ(α) (ψ(x)) and has the same distribution as ζψ(β)
ψ(x) Y Y P ζψ(β) ψ(α) ≤ t .
ψ(x) ≤ t ≤
P ζψ(β)
ψ(α)
Similarly,
X
x X
P η2n,[α,β]
(x) ≤ t ≤ P η2n,[α,β]
(α) ≤ t .
α
To complete the first estimate, use Itô’s formula, Doob’s stopping time theorem,
2
Y
and Fatou’s lemma to see that eλψ(β) E[e−λ ψ(β)ζψ(β) (ψ(α)) ] is dominated by
Y
t∧ζψ(β)
(ψ(α)) Y
2
lim E exp λY t ∧ ζψ(β) ψ(α) , ψ(α) − λ
Y τ, ψ(α) dτ
= eλψ(α)
t→∞
0
∀ λ ∈ R. Hence,
Y P ζψ(β)
ψ(α) ≤ t ≤ exp λ2 ψ(β)t − λ ψ(β) − ψ(α) ∀ λ ∈ R,
and so the asserted estimate follows when one takes λ =
ψ(β)−ψ(α)
2ψ(β)t .
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553
THE WRIGHT–FISHER EQUATION
The argument for the second estimate is similar. Namely, for any n ≥ 0, given
X
X
X
X
that ηn,[α,β]
(α) < ∞, ηn+1,[α,β]
(α) − ηn,[α,β]
(α) is independent of ηn,[α,β]
(α) and has
X
the same distribution as, depending on whether n is even or odd, ζβ (α) or ζαX (β).
Hence, for n ≥ 1 and λ ∈ R,
n n
2 X
2 X
2 X
E e−λ η2n,[α,β] (α) = E e−λ ζα (β) E e−λ ζβ (α) .
Finally, once one takes into account the fact that x(1 − x) ≤
reasoning which we used before shows that
λ2 X λ2 X E e− 4 ζα (β) ∨ E e− 4 ζβ (α) ≤ e−λ(β−α)
1
4
for x ∈ [0, 1], the same
for all λ > 0
and therefore that
X
4n2 (β − α)2
P η2n,[α,β] (α) ≤ t ≤ exp −
t
.
Theorem 4.3. There is a unique continuous function (x, y, t) ∈ (0, 1)2 × (0, ∞)
−→ p(x, y, t) ∈ (0, ∞) such that
(4.5)
ϕ(y)p(x, y, t) dy = E ϕ X(t, x)
for (x, t) ∈ (0, 1) × (0, ∞)
(0,1)
and all bounded, Borel measurable functions ϕ on [0, 1] which vanish on {0, 1}. Moreover, p(x, y, t) satisfies (2.5), (2.6), (2.7), and, for each 0 < α < β < 1,
∞
(4.6)
p(x, y, t) = pβ (x, y, t) +
X
X
E pβ α, y, t − η2n,[α,β]
(x) , η2n,[α,β]
(x) < t
n=1
∀ (x, y, t) ∈ (0, α]2 × (0, ∞). Finally, p(x, y, t) is smooth and, for each y ∈ (0, 1),
satisfies
(4.7)
∂t p(x, y, t) = x(1 − x)∂x2 p(x, y, t) on (0, 1) × (0, ∞)
with p(0, y, t) = 0 = p(1, y, t) and lim p( · , y, t) = δy .
t0
Proof. Let 0 < α < β < 1 and (x, t) ∈ (0, α] × (0, ∞) be given. From (4.4) and
Lemma 4.2, it is clear that the family {pγ (x, · , t) (0, α] : γ ∈ (β, 1)} is equicontinuous and, by (4.4), γ pγ (x, · , t) is nondecreasing. Hence, there is a unique continuous
function y ∈ (0, 1) −→ p(x, y, t) ∈ (0, ∞) such that p(x, y, t) = limγ 1 pγ (x, y, t). Furthermore, from (4.3) and (4.4), one knows that (4.5) and (4.6) hold. In addition, (2.5)
and (2.7) follow from (4.2), and (2.6) follows from (4.5) together with the observation
that 1 − X(t, x) has the same distribution as X(t, 1 − x).
Knowing (2.5), (2.7), and (4.5) and that p(x, · , t) is continuous, one can easily
check that (x, y, t) p(x, y, t) is continuous. Finally, the proof that p(x, y, t) is
smooth and satisfies (4.6) is a repeat of the sort of reasoning which we used in the
final part of the proof of Lemma 3.3.
Corollary 4.4. Set p̄(x, y, t) = y(1 − y)p(x, y, t) and, for β ∈ (0, 1),
q̄ Vβ (ξ, η, t) = ηq Vβ (ξ, η, t).
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554
LINAN CHEN AND DANIEL W. STROOCK
Then, for each 0 < α < β and θ ∈ (0, 1),
ψ (x)ψ (y)p̄(x, y, t) − q̄ Vβ ψ(x), ψ(y), t ≤
K(α, β, θ)etUβ
μβ (β − α)2
ψ(x)ψ(y) exp −
β−α
t
∀ (x, y, t) ∈ (0, θα]2 × (0, 1], where (cf. (4.1))
Uβ =
sup
Vβ (ξ), μβ =
ξ∈(0,ψ(β))
1 + 4ψ(β)
4ψ(β)
1
∧ 4, and K(α, β, θ) =
3
2 2 π 2 ψ(β)
.
ψ(α)α4 (1 − θ)4
In particular, if
Oβ = sup{Vβ (η) − Vβ (ξ) : (ξ, η) ∈ 0, ψ(β) }
and
νβ =
1
∧ 3,
4ψ(β)
then (cf. (1.12))
ψ (x)ψ (y)p̄(x, y, t)
K(α, β, θ)etOβ νβ (β − α)2
− 1 ≤
t2 ∨ ψ(x)ψ(y) exp −
Vβ q̄ ψ(x), ψ(y), t
δ0 (β − α)
t
∀ (x, y, t) ∈ (0, θα] × (0, 1] with Proof. Because
∞
p̄(x, y, t) = p̄β (x, y, t) +
ψ(y) −
ψ(x) ≤ β − α.
X
X
E p̄β α, y, t − η2n,[α,β]
(x) , η2n,[α,β]
(x) < t ,
n=1
ψ (x)ψ (y)p̄(x, y, t) equals
∞
Vβ X
ψ (x)
Vβ
X
q̄ψ(β)
E q̄ψ(β)
(x) , η2n,[α,β]
(x) < t .
ψ(x), ψ(y), t +
ψ(α), ψ(y), t−η2n,[α,β]
ψ (α) n=1
we know that
ψ (x)ψ (y)p̄(x, y, t) is dominated by
∞ X
ψ (x)
Vβ Vβ
sup
ψ(α),
ψ(y),
τ
q̄ ψ(x), ψ(y), t +
q̄
P η2n,[α,β] (x) < t
ψ(β)
ψ (α) τ ∈(0,t]
n=1
Thus,
and
ψ (x)ψ (y)p̄(x, y, t) dominates
which, by Lemma 3.3, is equal to
q̄ Vβ ψ(x), ψ(y), t
Y
−E e
ζψ(β) (ψ(x))
0
V
β
ψ (x)ψ (y)p̄β (x, y, t) = q̄ψ(β)
(ψ(x), ψ(y), t),
V (Y (τ,ψ(x)))dτ Vβ
Y
Y
ψ(x) , ζψ(β)
ψ(x) < t .
q̄ ψ(β), ψ(y), t − ζψ(β)
1
Since (1 − ξ) 2 ψ (ξ) ∈ [1, π2 ] for ξ ∈ [0, 1] and, by the second part of Lemma 4.2,
X
x
P η2n,[α,β]
(x) < t ≤
(4.8)
α
n=1
√
πt
1+
4(β − α)
e−
4(β−α)2
t
≤
4(β−α)2
2x
e− t ,
α(β − α)
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555
THE WRIGHT–FISHER EQUATION
the right-hand side of the upper bound can be replaced by
4(β−α)2
π
2x
Vβ q̄ Vβ ψ(x), ψ(y), t +
sup q̄ψ(β)
e− t .
ψ(α), ψ(y), τ
2 τ ∈(0,t]
α(β − α)
√
At the same time, because the derivative of ψ dominates 1 and y ≤ θα, the upper
bound in (1.10) leads to
Vβ q̄ψ(β)
ψ(α), ψ(y), τ
√
√
2
ψ(α)ψ(y) − ( ψ(y)− ψ(α))
τ
≤e
e
τ2
ψ(α)ψ(y) − α2 (1−θ)2
τ
≤ etUβ sup
e
τ2
τ ∈(0,t]
tUβ
≤ etUβ
ψ(α)ψ(y)
α4 (1 − θ)4
for τ ∈ (0, t]. Thus, after combining this with the preceding, we find that
√
2
2πxψ(α)ψ(y)etUβ − 4(β−α)
Vβ
t
e
ψ (x)ψ (y)p̄(x, y, t) ≤ q̄ ψ(x), ψ(y), t + 5
,
α (1 − θ)4 (β − α)
which, because ξ ≤ ψ(ξ) ≤ πξ, means that
(4.9)
ψ (x)ψ (y)p̄(x, y, t) ≤ q̄
Vβ
ψ(x), ψ(y), t +
√
2π 3 ψ(x)ψ(y)etUβ − 4(β−α)2
t
e
.
α4 (1 − θ)4 (β − α)
In view of the lower bound in the first paragraph of the proof, to prove the
complementary lower bound, we need to estimate the term on the right which is
preceded by a minus sign. But by the first part of Lemma 4.2 and (1.10), it is clear
that this term is dominated by
Y Y
Y
ψ(x) , ζψ(β)
ψ(x) < t
E eζψ(β) (ψ(x))Uβ q̄ Vβ ψ(β), ψ(y), t − ζψ(β)
≤
2
ψ(x)etUβ − (ψ(β)−ψ(α))
4ψ(β)t
e
sup q̄ ψ(β), ψ(y), τ
ψ(α)
τ ∈(0,t]
√
−
2
− (ψ(β)−ψ(α))
4ψ(β)t
(
ψ(β)−
√
ψ(y)
τ
ψ(x)ψ(y)ψ(β)
e
≤ etUβ e
sup
2
ψ(α)
τ
τ ∈(0,t]
tUβ
1 + 4ψ(β) (β − α)2
ψ(x)ψ(y)ψ(β)e
≤
exp −
,
ψ(α)α4 (1 − θ)4
4ψ(β)t
since (
ψ(β) −
)2
ψ(y))2 ≥ α2 (1 − θ)2 + (β − α)2 . Hence, we now know that
(4.10)
ψ (x)ψ (y)p̄(x, y, t)
2
Uβ t
1
+
4ψ(β)
(β
−
α)
ψ(x)ψ(y)ψ(β)e
.
exp −
≥ q̄ Vβ ψ(x), ψ(y), t −
ψ(α)α4 (1 − θ)4
4ψ(β)t
Given (4.9) and (4.10), the initial assertion follows immediately. To prove the
second estimate, note that
⎧ ψ(x)ψ(y) ψ(x)+ψ(y)
−
t
if ψ(x)ψ(y) ≤ t2 ,
⎨ t2 e
√
√
2
1
1
q̄ ψ(x), ψ(y), t ≥
ψ(y)−
ψ(x)
)
⎩δ ψ(x) 4 1ψ(y) 4 e− (
t
if ψ(x)ψ(y) ≥ t2 .
0
t2
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556
LINAN CHEN AND DANIEL W. STROOCK
Hence, since ψ(x) + ψ(y) ≥ ( ψ(y) − ψ(x))2 , the second estimate is an easy consequence of the first when one takes into account the condition that | ψ(y)− ψ(x)| ≤
β − α.
Corollary 4.5. For each 0 < α < β < 1 and θ ∈ (0, 1),
ψ (x)ψ (y)p̄(x, y, t) − q̄ ψ(x), ψ(y), t ≤ tet
Vβ
u
π 2 K(α, β, θ)
(β−α)2
ψ(x)ψ(y)e− t
Vβ u q̄ ψ(x), ψ(y), t +
3
e(β − α)
∀ (x, y, t) ∈ (0, θα]2 × (0, 1] and
ψ (x)ψ (y)p̄(x, y, t)
− 1
q̄ ψ(x), ψ(y), t
≤ tet
Vβ
Vβ u +
u
π 2 K(α, β, θ) 2 t ∨ ψ(x)ψ(y)
3
eδ0 (β − α)
∀ (x, y, t) ∈ (0, θα]2 × (0, 1] satisfying | ψ(y) − ψ(x)| ≤ β − α.
Proof. Just as in the proof of Corollary 4.4, the second inequality is a consequence
of the first. Furthermore, given the first inequality in Corollary 4.4, the first inequality
here comes down to the estimate
n V
tV
β
u
β
q̄ (ξ, η, t) − q̄(ξ, η, t) ≤
q̄(ξ, η, t),
n!
n=1
which is a trivial consequence of (3.12) and (3.13).
5. Derivatives. In order to get regularity estimates, it will be important to
know how to represent derivatives of the solutions to (1.1) in terms to derivatives of
the initial data. For this purpose, observe that if u is a smooth solution to (1.1) and
u(m) is its mth derivative with respect to x, then u(m) satisfies
∂t w(x, t) = x∂x2 w(x, t) + m∂x w(x, t).
(5.1)
Thus, we should expect that
u
(m)
ϕ(m) (y)q (m) (x, y, t) dy,
(x, t) =
(0,∞)
where ϕ(m) is the mth derivative of the initial data ϕ and q (m) (x, y, t) is the fundamental solution to (5.1).
In order to verify this suspicion, we need to know something about q (m) (x, y, t)
when m ≥ 1, and there are two ways of going about this. The first is to notice that
if w satisfies (5.1) and v(x, t) = w(x2 , t), then v satisfies
1
2m − 1
∂x v(x, t) .
∂t v(x, t) =
∂x2 v(x, t) +
4
x
Since, apart from the factor of 14 , the operator on the right-hand side is the radial
part of the Laplacian in R2m , one sees that
(5.2)
q (m) (x, y, t) =
y m−1 e−
tm
x+y
t
q (m)
xy t2
, where q (m) (ξ) =
1
2π m
1
e2ξ 2 (ω,ω0 ) dω,
S2m−1
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557
THE WRIGHT–FISHER EQUATION
ω0 being any reference point on the 2m− 1 unit sphere S2m−1 in R2m and dω denoting
integration with respect to the surface measure for S2m−1 . Starting from here, one
can use elementary facts about Bessel functions to see that q (m) is given by the series
q (m) (ξ) =
(5.3)
∞
ξn
.
n!(n + m − 1)!
n=0
Alternatively, knowing that (5.2) holds, one can proceed as in section 1 to show q (m)
must solve
ξ∂ξ2 q (m) (ξ) + m∂ξ q (m) (ξ) − q (m) (ξ) = 0
and therefore that
q (m) (ξ) = q (m) (0)(m − 1)!
∞
ξn
.
n!(n + m − 1)!
n=0
Hence, since, by the expression for q (m) in (5.2), q (m) (0) must be the area of S2m−1
divided by 2π m , (5.3) follows.
Observe that, as distinguished from q(x, y, t), q (m) (x, · , t) has a total integral 1
for all m ≥ 1 and (x, t) ∈ (0, ∞)2 . To understand the behavior of q (m) (x, y, t) for
small t > 0, one can use (cf. (7.4) and (7.6)) the expression in (5.2). Alternatively,
proceeding as in the second derivation of (1.12), ones knows that there is an δm ∈ (0, 1)
1
−1
) for t ∈ (0, 1]. Thus,
such that t 2 q (m) (1, 1, t) ∈ (δm , δm
1
1
m
1
m
1
−1 4 − 2 2ξ 2
ξ
e
δm ξ 4 − 2 e2ξ 2 ≤ q (m) (ξ) ≤ δm
for ξ ≥ 1,
and so
1
(5.4) δm
m
y m−1 (xy) 4 − 2
1
t2
e−
1
1
(y 2 −x 2 )2
t
1
−1
≤ q (m) (x, y, t) ≤ δm
m
y m−1 (xy) 4 − 2
1
t2
e−
1
1
(y 2 −x 2 )2
t
for xy ≥ t2 . On the other hand, by (5.3), after readjusting δm , one has
(5.5)
δm
m−1
x+y
y m−1 − x+y
−1 y
e t ≤ q (m) (x, y, t) ≤ δm
e− t
m
m
t
t
when xy ≤ t2 .
Finally, set q (0) (x, y, t) = q(x, y, t). Then, because q (m) (ξ) = ∂ξm q(ξ), it is easy to
check that
1
∂x q (m) (x, y, t) = q (m+1) (x, y, t) − q (m) (x, y, t) ∀ m ≥ 0
t
and therefore that
(5.6)
∂xm q (k) (x, y, t)
1
= m
t
m
(−1)
=0
m (m+k−)
(x, y, t) ∀ k, m ∈ N.
q
Theorem 5.1. Suppose that ϕ ∈ C m ((0, ∞); R), set ϕ() = ∂x ϕ, and assume
that ϕ() is bounded on (0, 1) and has subexponential growth at ∞ for each 0 ≤ ≤ m.
If ϕ tends to 0 at 0, then
m
ϕ(y)q(x, y, t) dy =
ϕ(m) (y)q (m) (x, y, t) dy.
(5.7)
∂x
(0,∞)
(0,∞)
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558
LINAN CHEN AND DANIEL W. STROOCK
Moreover, for any ≥ 1,
m
()
(5.8)
∂x
ϕ(y)q (x, y, t) dy =
(0,∞)
ϕ(m) (y)q (m+) (x, y, t) dy
(0,∞)
even if ϕ does not vanish at 0.
Proof. Since every ϕ satisfying our hypotheses can be written as the th derivative
of one which vanishes at 0, it is clear that (5.8) is an immediate corollary of (5.7).
In proving (5.7), first observe that, by an obvious cutoff argument and the estimates in (5.4) and (5.5), it suffices to handle ϕ’s which vanish off of a compact subset
of [0, ∞). Second, suppose that we knew (5.7) when ϕ ∈ Cc∞ ((0, ∞); R). Given a
ϕ ∈ C m ((0, ∞); R) which vanishes at 0 and to the right of some point in (0, ∞),
set ϕ (y) = 1[2 ,∞) (y)ϕ(y − 2) for > 0. Next, choose ρ ∈ Cc∞ ((−1, 1); [0, ∞))
with a total integral 1, and set ϕ̃ (y) = ρ ϕ , where ρ (y) = −1 ρ(−1 y). Then
ϕ̃ ∈ Cc∞ ((0, ∞); R), and so we would know that
∂xm
ϕ̃ (y)q(x, y, t) dy =
ϕ̃(m) (y)q (m) (x, y, t) dy.
(0,∞)
(0,∞)
For each (t, x) ∈ (0, ∞)2 , it is clear that the left-hand side of the preceding tends to
the left-hand side of (5.7) as 0. To handle the right-hand side, write it as the
sum of
ρ(m) ϕ (y)q (m) (x, y, t) dy and
ρ ϕ(m) (y)q (m) (x, y, t) dy.
(0,3 ]
(3 ,∞)
As 0, the second of these tends to the right-hand side of (5.7). At the same time,
by (5.6), we know that, for each (t, x) ∈ (0, ∞)2 , for sufficiently small > 0, the first
of these is dominated by a constant times
(m) m−1
ρ (ξ)ϕ (y − ξ) dξ dy
y
(0,3 ]
(0,∞)
=
[2 ,3 ]
(m)
ρ (ξ − 2)
≤ (3)
m
y
m−1
[ξ,3 ]
(m) ρ (ξ) dξ
|ϕ(y − ξ)| dy
(0,∞)
(0,3 )
dξ
|ϕ(y)|
dy −→ 0 as y
0.
Hence, it suffices to treat ϕ ∈ Cc∞ ((0, ∞); R).
To prove (5.7) for ϕ ∈ Cc∞ ((0, ∞); R), assume that ϕ vanishes off of (a, b), and
set
ϕ(y)q(x, y, t) dy.
u(x, t) =
(0,∞)
Because
(x∂x2 )
ϕ(y)q(x, y, t) dy =
(0,∞)
(0,∞)
(y∂y2 ) ϕ(y)q(x, y, t) dy,
it is obvious that ∂x u(x, t) is bounded on [ a2 , ∞) × (0, ∞) ∀ ≥ 0. At the same time,
from (5.6) and the estimates in (5.4) and (5.5), we know that, for each ≥ 0 and T > 0,
Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.
THE WRIGHT–FISHER EQUATION
559
∂x q(x, y, t) is uniformly bounded on (0, a2 ] × (a, ∞) × (0, T ]. Hence, for all ≥ 0 and
T > 0, ∂x u(x, t) is uniformly bounded on (0, ∞)×(0, T ], and so it is now easy to check
that u(m) (x, t) ≡ ∂xm u(x, t) is the unique solution to ∂t w(x, t) = x∂x2 w + m∂x w(x, t)
which is bounded on finite time intervals and has an initial value ϕ(m) . Since this
means that u(m) (x, t) is given by the right-hand side of (5.7), we are done.
Corollary 5.2. For 0 ≤ k ≤ m and (s, x), (t, y) ∈ (0, ∞)2 , set
(m,k)
(x, y, s, t) =
q (m) (x, ξ, s)q (k) (ξ, y, t) dξ.
q
(0,∞)
Then
(5.9)
q (m,k) (x, y, s, t) =
m−k 1
(s + t)m−k
j=0
m − k j m−k−j (k+j)
q
(x, y, s + t).
s t
j
Proof. By the Chapman–Kolmogorov equation, (5.9) is obvious when k = m. To
prove it in general, assume that it holds for some m and 0 ≤ k ≤ m. Then, by (5.6)
and the induction hypothesis, for 0 ≤ k ≤ m,
1 (m+1,k)
1
q
(x, y, s, t) − q (m,k) (x, y, s, t)
s
s
(m,k)
(x, y, s, t)
= ∂x q
m−k m − k j m−k−j (k+j+1)
1
q
(x, y, s + t)
=
s t
m+1−k
j
(s + t)
j=0
−
1
(s + t)m+1−k
m−k j=0
m − k j m−k−j (k+j)
q
(x, y, s + t),
s t
j
and so, again by the induction hypothesis, q (m+1,k) (x, y, s, t) equals
⎛
m+1−k m − k j m+1−k−j (k+j)
1
⎝
q
(x, y, s + t)
s t
j−1
(s + t)m+1−k
j=1
⎞
m−k m − k j m+1−k−j (k+j)
s t
+
q
(x, y, s + t)⎠
j
j=0
=
1
(s + t)m+1−k
m+1−k j=0
m + 1 − k j m+1−k−j (k+j)
q
(x, y, s + t).
s t
j
The reason for our interest in the preceding results is that they allow us to produce
rather sharp estimates on the derivatives of q V (x, y, t).
Lemma 5.3. Set
Cm (V ) =
5m
max{∂ V u : 0 ≤ ≤ m}
2
and
m
Sm (x, y, t) =
=0
m ()
q (x, y, t).
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560
LINAN CHEN AND DANIEL W. STROOCK
Then
m t2m
m V
e
∂ q (x, y, t) ≤ 1 + tCm (V )
x
1 − tCm (V )
V
u
− 1 Sm (x, y, t)
tm
for (x, y, t) ∈ (0, ∞)2 × (0, 1].
Proof. Clearly (cf. (3.11) and (3.13)) it suffices to prove that, for each n ∈ N,
+
m V
∂x qn (x, y, t) ≤ Cm (V )m∧n 2m V u (n−m)
tn
Sm (x, y, t)
(n − m)+ !
tm
for (x, y, t) ∈ (0, ∞)2 × (0, 1]. When n = 0 this is obvious from (5.6) with k = 0, and
when m = 0 there is nothing to do. Since (5.6) already explains how to proceed for
general m ≥ 1, we will concentrate on the case when m = 1.
Using (3.11) and Theorem 5.1, one can write ∂x q1V (x, y, t) as the sum of
t 2
0
and
0
t
2
(0,∞)
∂x q(x, ξ, t − τ )V (ξ)q(ξ, y, τ ) dξ
(0,∞)
q (1) (x, ξ, τ )∂ξ V (ξ)q(ξ, y, t − τ ) dξ
dτ
dτ.
By (5.6) and (5.9), the first of these is dominated by V u times
2t
−1
(1)
(t − τ )
q(x, ξ, t − τ ) + q (x, ξ, t − τ ) q(ξ, y, τ ) dξ dτ
0
(0,∞)
= 12 q (1) (x, y, t) +
q(x, y, t)
t
t
2
0
t+τ
dτ ≤ 12 q (1) (x, y, t) + 54 q(x, y, t).
t−τ
As for the second, it can be dominated by
0
t
2
V u + V u (t − τ )−1 q (1,0) (x, y, τ, t − τ ) dτ + log 2V u q (1) (x, y, t)
1
t 3t 3
V u + V u q(x, y, t) +
V u + V u q (1) (x, y, t).
≤
2
8
8
2
After combining these, one gets that ∂x q1V (x, y, t) ≤ C1 (V )S1 (x, y, t) for t ∈ (0, 1].
Given the preceding, the argument for n ≥ 2 is easy. Namely, by (3.11 ),
V
∂x qn+1
(x, y, t) =
t
0
(0,∞)
∂x qnV (x, ξ, t − τ )V (ξ)q(ξ, y, τ ) dξ
dτ.
Thus, one can use the preceding and induction on n ≥ 1 to get the asserted estimate.
For general m ≥ 1, the strategy is the same. One has to apply the argument just
given to estimate ∂x q1V (x, y, t) not only once but m times. At the end of the mth
repetition, ones arrives at the estimate |∂xm qnV (x, y, t)| ≤ Cm (V )n tn−m Sm (x, y, t) for
0 ≤ n ≤ m. After the mth repetition, one switches from (3.11) to (3.11 ) and proceeds
inductively as above.
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561
THE WRIGHT–FISHER EQUATION
Theorem 5.4. Set p̄(x, y, t) = y(1 − y)p(x, y, t). For each 0 < α < β < 1,
θ ∈ (0, 1), and m ∈ Z+ , there exists a Km (α, β, θ) < ∞ such that
(m)
m
4(β−α)2
q
ψ(x),
ψ(y),
t
∨
q
(ψ(x),
ψ(y),
t
∂x p̄(x, y, t) ≤ Km (α, β, θ)ψ(y)
+ e− t
tm
∀ (x, y, t) ∈ (0, θα] × (0, 1]. In particular, for each > 0, p̄(x, y, t) has bounded derivatives of all orders on (0, 1)2 × [, ∞).
Proof. Given the first assertion, the second assertion is an easy application of the
symmetry properties of p̄(x, y, t), the equation ∂t p̄(x, y, t) = x(1 − x)∂x2 p̄(x, y, t), and
the fact that
p(x, ξ, s)p̄(y, ξ, t) dξ,
p̄(x, y, s + t) =
(0,1)
which is a consequence of symmetry and the Chapman–Kolmogorov equation for
p(x, y, t).
Next set p̄β (x, y, t) = y(1−y)pβ (x, y, t), note that, by (4.2), p̄β (x, y, t) = p̄β (y, x, t),
and observe that, by symmetry,
∞
∂xm p̄(x, y, t) = ∂xm p̄β (x, y, t) +
X
X
E ∂xm p̄β x, α, t − η2n,[α,β]
(y) , η2n,[α,β]
(y) < t
n=1
and therefore, by (4.8), that
m
∂ p̄(x, y, t) ≤ ∂ m p̄β (x, y, t) + y
x
x
α
1+
√
πt
4(β − α)
4(β−α)2
sup ∂xm p̄β (x, α, τ )e− t .
τ ∈(0,t]
In addition, if q̄ Vβ (ξ, η, t) = ηq Vβ (ξ, η, t), because both p̄β and q̄ Vβ are symmetric,
(4.1) implies that p̄β (x, y, t) equals
q̄ Vβ ψ(x), ψ(y), t
ψ (x)ψ (y)
Y
−E e
ζψ(β) (ψ(y))
0
V (Y (τ,ψ(y))) dτ
Y
q̄ Vβ ψ(x), ψ(β), t − ζψ(β)
(ψ(y))
ψ (x)ψ (y)
,
Y
ζψ(β)
ψ(y) < t .
Thus, there is a Cm (β) < ∞ such that
m
∂x p̄β (x, y, t) ≤ Cm (β)
max ∂x q̄ Vβ ψ(x), ψ(y), t 0≤≤m
+
2
ψ(y) − (ψ(y)−ψ(β))
4ψ(β)t
e
sup max ∂x q̄ Vβ ψ(x), ψ(β), τ .
ψ(α)
τ ∈(0,t] 0≤≤m
Finally, by combining these with the estimates in (5.4), (5.5), and Lemma 5.3, using
(ψ(β) − ψ(α))2 ≥ ψ(β)(β − α)2 , one gets the desired result.
6. Two concluding results. In this section we will present two results which
might be of computational interest. The first answers one of the questions which Nick
Patterson originally asked. Namely, he wanted to know whether one can justify a
Taylor’s approximation for solutions to the Wright–Fisher equation (1.1).
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562
LINAN CHEN AND DANIEL W. STROOCK
In the following, L = x(1 − x)∂x2 is the Wright–Fisher operator, X(t, x) is the
Wright–Fisher diffusion given by (2.2), and ζ X (x) = inf{t ≥ 0 : X(t, x) ∈
/ (0, 1)}. By
Itô’s formula,
t
E Lϕ X(τ, x) dτ
E ϕ X(t, x) = ϕ(x) +
0
for any ϕ ∈ C 2 ([0, 1]; R). In addition, since ξ(1 − ξ) ≤
1
4
for ξ ∈ [0, 1],
2(x∧(1−x))2
t
.
P ζ X (x) ≤ t ≤ 2e−
(6.1)
Lemma 6.1. Assume that ϕ ∈ C 2(n+1) ((0, 1); R) for some n ≥ 0 and that Lm ϕ
admits a continuous extension to [0, 1] for each 0 ≤ m ≤ n + 1. Then, for each
(x, t) ∈ (0, 1) × (0, ∞),
E ϕ X(t, x) , ζ X (x) > t −
−
1
n!
0
t
n
tm m
L ϕ(x)
m!
m=0
(t − τ )n E Ln+1 ϕ X(τ, x) , ζ X (x) > τ dτ ≤ 2e−
2(x∧(1−x))2
t
tm m
|L ϕ(0)| ∨ |Lm ϕ(1)| .
m!
m=0
n
Proof. Set ζrX (x) = inf{t ≥ 0 : |X(t, x)| ∧ |1 − X(t, x)| ≤ r} for r ∈ (0, 1). By
Itô’s formula and Doob’s stopping time theorem,
t∧ζrX (x)
X
E ϕ X(t ∧ ζr (x), x) = ϕ(x) + E
Lϕ X(τ, x) dτ
0
for each r ∈ (0, 1). Thus, after letting r
E ϕ X(t ∧ ζ X (x), x) = ϕ(x) +
0, we know first that
0
t
E Lϕ X(τ, x) , ζ X (x) > τ dτ
and then, by (6.1), that the result holds for n = 0.
Next, assume the result for n. By applying the result for n = 0 to Ln+1 ϕ, we see
that
t
n+1
(t − τ )n E Ln+1 ϕ X(τ, x) , ζ X (x) > τ dτ − t
Ln+1 ϕ(x)
n
+
1
0
t
n+2 X
1
n+1
−
(t − τ )
E L
ϕ X(τ, x) , ζ (x) > τ dτ n+1 0
t
2(x∧(1−x))2
−
n+1
n+1
t
≤e
|L
ϕ(0)| ∨ |L
ϕ(1)|
(t − τ )n dτ
0
≤ 2e
2
− 2(x∧(1−x))
t
|Ln+1 ϕ(0)| ∨ |Ln+1 ϕ(1)|
tn+1
.
n+1
Thus, by induction, the result follows for general n’s.
Given f ∈ C((0, 1); R), set f¯(x) = x(1 − x)f (x).
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563
THE WRIGHT–FISHER EQUATION
Theorem 6.2. Assume that f ∈ C ∞ ((0, 1); R) and that, for each n ≥ 0, Ln f¯
extends to [0, 1] as a continuous function. Then, for each n ≥ 0 and (y, t) ∈ (0, 1) ×
(0, ∞),
n
f (x)p(x, y, t) dx −
(0,1)
1
−
y(1 − y)n!
≤e
m
t
0
tm
fm (y)
m!
m=0
X
n+1 ¯
f X(τ, y) , ζ (y) > τ dτ (t − τ ) E L
n
2
2
− 2(x ∧(1−x))
t
tm |fm (0)| ∨ |fm (1)| ,
m!
m=0
n
¯
L f
where fm (y) ≡ y(1−y)
for m ≥ 0.
Proof. Because y(1 − y)p(x, y, t) = x(1 − x)p(y, x, t),
f (x)p(x, y, t) dx =
f¯(x)p(y, x, t) dx = E f¯ X(t, y) , ζ X (y) > t .
y(1 − y)
(0,1)
(0,1)
Hence, the desired result follows from Lemma 6.1 with ϕ = f¯.
The second topic deals with possible improving of the first estimate in Corollary 4.5. A look at the argument there reveals that the weak link is the replacement
of q̄ Vβ by q̄ in the first estimate in Corollary 4.4. Of course, the problem with q̄ Vβ is
that we can only estimate it but cannot give an explicit expression for it in terms of
familiar quantities. Nonetheless, if one is interested in p̄(x, y, t) when all the variables
are small, one can make a modest improvement by using a Taylor approximation for
Vβ .
In order to carry this out, we will need the following computation, which is of
some independent interest on its own.
(m)
Lemma 6.3. For each m ≥ 1, define {Ck,j : k ∈ N & 0 ≤ j ≤ k} so that
(m)
Ck,0 =
(m + k − 1)!
(m)
(m)
(m)
(m)
, Ck,k = 1, and Ck+1,j = (m + k + j)Ck,j + Ck,j−1 , 1 ≤ j ≤ k.
(m − 1)!
(2)
Next, set C1,1 = 1 and Ck,j = Ck−1,j−1 for k ≥ 2 and 1 ≤ j ≤ k. Then
q
(m)
k
k
(x, y, t)y =
(m)
Ck,j tk−j xj q (m+k+j) (x, y, t)
for m ≥ 1 and k ≥ 1,
j=0
and
k
q(x, y, t)y k =
Ck,j tk−j xj q (k+j) (x, y, t)
for k ≥ 1.
j=1
In particular, for k ≥ 1,
t 0
q(x, ξ, t − τ )ξ q(ξ, y, τ ) dξ
k
(0,∞)
k
dτ = tx
Q(k,j) (x, y, t),
j=1
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564
LINAN CHEN AND DANIEL W. STROOCK
where
Q(k,j) (x, y, t) =
(k + j)!Ck,j k−j j−1
t x
(2k + 1)!
k+j
i=0
(k − j + i)! (i)
q (x, y, t).
i!
(m)
Proof. Starting from the expression for q (x, y, t) in (5.2) and remembering
that q (m) (x, y, t) satisfies (5.1), one sees that
x+y
m (m)
2x
x∂x2 q (m) (x, y, t) + m∂x q (m) (x, y, t) =
−
q (x, y, t) − 2 q (m+1) (x, y, t).
t2
t
t
At the same time, by (5.6), x∂x2 q (m) (x, y, t) + m∂x q (m) (x, y, t) equals
x
m 2x
m (m)
x (m+2)
(m+1)
−
q (x, y, t).
q
(x,
y,
t)
+
(x,
y,
t)
+
−
q
t2
t
t2
t2
t
Hence,
yq (m) (x, y, t) = xq (m+2) (x, y, t) + mtq (m+1) (x, y, t)
(6.2)
∀ m ≥ 0.
Given (6.2), an easy inductive argument proves the first equation, and given the
first equation, the second one follows immediately from q(x, y, t)y = xq (2) (x, y, t),
which is (6.2) when m = 0 and k = 1.
Finally, from the second equation, we know that (cf. Corollary 5.2 and (5.9))
t k
q(x, ξ, t − τ )ξ q(ξ, y, τ ) dξ dτ
0
(0,∞)
k
j
=
Ck,j x
j=1
k
=
Ck,j t
0
t
(t − τ )k−j q (k+j,0) (x, y, t − τ, τ ) dτ
k+j t
k+j
k−j+i k+j−i
(i)
x
(t − τ )
τ
dτ q (x, y, t)
i
0
i=0
−k−j j
j=1
k+j
k
Ck,j tk−j xj
=t
j=1
i=0
(k + j)!(k − j + i)! (i)
q (x, y, t),
i!(2k + 1)!
which is easy to recognize as the expression we want.
It is amusing to note that the numbers {Ck,j : 1 ≤ j ≤ k} are the coefficients for
the polynomial which gives the kth moment of q(x, · , t). That is,
k
y k q(x, y, t) dy =
Ck,j tk−j xj .
(6.3)
(0,∞)
j=1
To see this, simply start with the equation for y m q(x, y, t) in Lemma 6.3 and, remembering that the integral of q (m) (x, · , t) is 1 for all m ≥ 1, integrate both sides with
respect to y ∈ (0, ∞).
With Lemma 6.3 we can implement the idea mentioned above. To this end, first
use (3.12) to check that
t V
34
q(x, ξ, t − τ )V 34 (ξ)q(ξ, y, τ ) dξ dτ q (x, y, t) − q(x, y, t) −
0
(0,∞)
≤
(tV 34 u )2 e
2
t V3
4
u
q(x, y, t).
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565
THE WRIGHT–FISHER EQUATION
Second, use Taylor’s theorem and the fact that V 34 (0) = 0 to see that
t 0
(0,∞)
− V 3 (0)
4
≤
q(x, ξ, t − τ )V 34 (ξ)q(ξ, y, τ ) dξ dτ
t 0
(0,∞)
q(x, ξ, t − τ )ξq(ξ, y, τ ) dξ
V 3 u t 4
2
0
(0,∞)
dτ q(x, ξ, t − τ )ξ 2 q(ξ, y, τ ) dξ
dτ.
28
, and combine this with the
Third, use elementary calculus to show that V 3 (0) = − 45
4
preceding and the second equation in Lemma 6.3 to conclude that
(6.4)
3
V
28tx
34
(i)
q (x, y, t)
q (x, y, t) − q(x, y, t) +
135 i=0
⎡
⎤
3
4
V 3 u
V 34 2u t V 3 u 2
4
≤
e 4 t q(x, y, t) +
tx ⎣t
jq (j) (x, y, t) + 2x
q (j) (x, y, t)⎦ .
2
10
j=0
j=0
After putting (6.4) together with the first estimate in Corollary 4.4, we arrive at
the following theorem.
Theorem 6.4. For (x, y, t) ∈ (0, 14 ]2 × (0, 1],
28tψ(x)ψ(y) 3 (j) ψ (x)ψ (y)p̄(x, y, t) − q̄ ψ(x), ψ(y), t +
q
ψ(x), ψ(y), t 135
j=0
V 43 2u t V 3 u 2 √ μ
e 4 t q̄ ψ(x), ψ(y), t
≤KetU 1 + πt ψ(x)ψ(y)e− 16t +
2
⎡
⎤
3
4
V 3 u
4
tψ(x)ψ(y) ⎣t
+
jq (j) ψ(x), ψ(y), t + 2ψ(x)
q (j) ψ(x), ψ(y), t ⎦ ,
10
j=0
j=0
where U = U 43 , K = K( 12 , 34 , 12 ), and μ = μ 34 are the constants determined by the
prescription given in Corollary 4.4.
It should be evident that the preceding line of reasoning can be iterated to obtain better and better approximations to p(x, y, t). However, even the next step is
dauntingly messy, since it requires one to deal with integrals of the form
⎛
⎞
⎜
⎟
q(x, ξ2 , t − τ2 )ξ2 q(ξ2 , ξ1 , τ2 − τ1 )ξ1 q(ξ1 , y, τ1 ) dξ1 dξ2 ⎠ dτ1 dτ2 ,
⎝
0≤τ1 <τ2 ≤t
(0,∞)2
a task which is possible but probably not worth the effort.
7. Appendix. In this appendix we discuss a few important facts about the
functions q (m) (ξ) which appear in this article. The function q (0) (ξ) is the same as the
function q(ξ) which was introduced as the solution to (1.2) satisfying the boundary
conditions in (1.4). As we saw, one representation of q(ξ) is as the power series in
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566
LINAN CHEN AND DANIEL W. STROOCK
(1.7). By comparing this expression with the power series in (5.3) for q (2) (ξ), one
finds that
q(ξ) = ξq (2) (ξ).
(7.1)
We now want to show how to estimate the numbers δm which occur in (1.12) and
(5.4). Isolating the polar angle corresponding to ω0 and using the fact that the 2m− 2
1
sphere has an area
in (5.2) to
(7.2)
q
(m)
2π m− 2
,
Γ(m− 12 )
one can pass from the integral representation of q (m) (ξ)
1
(ξ) = √ πΓ m − 12
π
√
ξ cos θ
e2
sin2m−1 θ dθ
0
for m ≥ 1.
Aficionados of Bessel functions will recognize that (7.2) says that, when m ≥ 1,
√
1−m
q (m) (ξ) = ξ 2 Im−1 (2 ξ), where Im−1 is the (m − 1)st Bessel function with
√
√ a purely
imaginary argument. In conjunction with (7.1), this means that q(ξ) = ξI1 (2 ξ),
which is the route which Elkies took when he inverted Fefferman’s Fourier expression.
Turning to the estimation of the δm , set
π
gm (η) = e−η
eη cos θ sin2m−2 θ dθ for η ∈ [0, ∞).
0
By (7.2),
√
1
2 ξ gm 2 ξ .
q (m) (ξ) = √ 1 e
πΓ m − 2
1
1
Because 1 − cos θ ≥ 8− 2 θ2 for θ ∈ 0, π4 , cos θ ≤ 2− 2 for θ ∈ π4 , π , and sin θ ≤ θ for
θ ∈ [0, π], gm (η) is bounded above by
π4
1
2m−1
−1
2
− 1−2− 2 η π
e−8 2 ηθ θ2m−2 dθ + e
2m − 1
0
∞
1
2m−1
3m
7
1
3
− 1−2− 2 η π
≤ 2 2 − 4 η 2 −m
e−t tm− 2 dt + e
2m − 1
0
⎡
1⎤
⎛
2 ⎞m− 2
1
m− 2 π
1
⎥ 12 −m
⎢ 3m 5 ⎝ ⎠
= ⎣2 2 − 4 Γ m − 12 +
.
⎦η
2m − 1 e 1 − 2− 12
Hence, for m ≥ 1,
(7.3)
⎡
1
⎢ 2m−5
q (m) (ξ) ≤ ⎣2 4 π − 2
and so
(7.4)
1⎤
⎛
2 ⎞m− 2
1
m− 2 π
1
⎥ 14 − m2 2√ξ
⎠
+
e ,
⎝ ⎦ξ
1
1
1
(2m − 1)π 2 Γ m − 2
e 1 − 2− 2
⎡
1
⎢ 1
q(ξ) ≤ ⎣2− 4 π − 2 +
⎤
1
3
3
1
2
⎥ 14 2√ξ
.
3 ⎦ξ e
2
1
2 2 e 2 1 − 2− 2
∀ ξ ∈ (0, ∞), although neither of these estimates is close to optimal unless ξ ∈ [1, ∞).
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567
THE WRIGHT–FISHER EQUATION
The corresponding lower bound is easy, although it holds only when η ≥ 1.
Namely, when η ≥ 1,
gm (η) ≥
≥
π
4
e
−(1−cos θ)η
0
23m−3 1 −m
η2
π 2m−2
2m−2
sin
π
4
e−
θ2
2
0
θ dθ ≥
π
4
2
e
− ηθ2
0
θ2m−2 dθ ≥
3
22 θ
π
2m−2
dθ
π
2m+1 (2m
1
− 1)e
π2
32
η 2 −m ,
and so
(7.5)
q (m) (ξ) ≥
√
π
(2m − 1)2
2m+ 12
e
1
π2
32
Γ(m − 12 )
√
ξ
ξ 2 −m e2
and
(7.6)
7 π2 −1 1 2√ξ
q(ξ) ≥ 2 2 3e 32
ξ4e
for ξ ∈ [1, ∞).
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New York, 1970.
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Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.