Brownian Motion
An Introduction to Brownian Motion, Wiener Measure, and
Partial Differential Equations
Prof. Michael Mascagni
Applied and Computational Mathematics Division, Information Technology Laboratory
National Institute of Standards and Technology, Gaithersburg, MD 20899-8910 USA
AND
Department of Computer Science
Department of Mathematics
Department of Scientific Computing
Graduate Program in Molecular Biophysics
Florida State University, Tallahassee, FL 32306 USA
E-mail: mascagni@fsu.edu or mascagni@math.ethz.ch
or mascagni@nist.gov
URL: http://www.cs.fsu.edu/∼mascagni
Brownian Motion
Outline of the Lectures
Introduction to Brownian Motion as a Measure
Definitions
Donsker’s Invariance Principal
Properties of Brownian Motion
The Feynman-Kac Formula
Explicit Representation of Brownian Motion
The Karhunen-Loève Expansion
Explicit Computation of Wiener Integrals
The Schrödinger Equation
Proof of the Arcsin Law
Advanced Topics
Action Asymptotics
Brownian Scaling
Local Time
Donsker-Varadhan Asymptotics
Can One Hear the Shape of a Drum?
Probabilistic Potential Theory
Brownian Motion
Introduction to Brownian Motion as a Measure
Introduction to Brownian Motion
I
I
I
I
I
def
Let Ω = {β ∈ C[0, 1]; β(0) = 0} = C0 [0, 1], be an infinitely dimensional
space we consider for placing a probability measure
Consider (Ω, B, P), where B is the set of measurable subsets (a
σ-algebra) and P is the probability measure on Ω
hR
i
1
We would like to answer questions like P 0 β 2 (s)ds ≤ α ?
We now construct Brownian motion (BM) via some limit ideas
Central Limit Theorem (CLT): let X1 , X2 , . . . be independent, identically
P
distributed( i.i.d.) with E[Xi ] = 0, Var [Xi ] = 1 and define Sn = ni=1 Xi
1. Note if X1∗ , X2∗ , . . . are i.i.d. with E[Xi∗ ] = µ, Var [Xi∗ ] = σ 2 < ∞, then
Xi =
2. Then
Xi∗ −µ
has E[Xi ] = 0, Var [Xi ] = 1
σ
Sn
√
converges
in distribution to N(0, 1)
n
as n → ∞
Brownian Motion
Introduction to Brownian Motion as a Measure
Introduction to Brownian Motion
I
Let X1 , X2 , . . . be as before, then it follows from the CLT that
Z α
u2
1
Sn
lim P √ ≤ α = √
e− 2 du.
n→∞
n
2π −∞
I
Erdös and Kac proved (we will find the σi (·)’s):
q R
i
h
2
S
S
Sn
2 α − u2
≤
α
=
σ
(α)
=
e
1. limn→∞ P max √1n , √2n , . . . , √
du
1
π 0
n
2 2
S1 +S2 +···+Sn2
2. limn→∞ P
≤ α = σ2 (α)
n2
h
i
S +S +···+Sn
≤ α = σ3 (α)
3. limn→∞ P 1 23/2
n
I
Let Nn = #{S1 , . . . , Sn |Si > 0}, then

if α ≤ 0

0,
√
Nn
lim P
≤ α = π2 arcsin α, if 0 ≤ α ≤ 1
n→∞

n

1,
if α ≥ 1
Brownian Motion
Introduction to Brownian Motion as a Measure
Definitions
Definitions
I
X1 , X2 , . . . are as above, and ∀n ∈ N and t ∈ [0, 1] define
(S
√1 , t = 0
(n)
χ (t) = Sin
i−1
√ ,
< t ≤ ni , i = 1, 2, . . . , n
n
n
I
Let R denote the space of Riemann integrable functions on [0, 1].
I
Theorem: F : R → R and with some weak hypotheses, then
h i
lim P F χ(n) (·) ≤ α = PW [F (β) ≤ α] ,
n→∞
where PW denotes the probability called “Wiener measure,” and this
result is called Donsker’s Invariance Principal
Brownian Motion
Introduction to Brownian Motion as a Measure
Donsker’s Invariance Principal
Examples of Donsker’s Invariance Principal
R1
β 2 (s) ds, then by the theorem
" n
#
Z 1
X S2
2
i
lim P
≤ α = PW
β (s)ds ≤ α
n→∞
n2
0
1. F [β] =
0
i=1
2. F [β] = β(1), then
Z α
u2
Sn
1
e− 2 du
lim P √ ≤ α = PW [β(1) ≤ α] = √
n→∞
n
2π −∞
(
R 1 1+sgnβ(s)
1,
: x >0
3. F [β] = 0
ds, where sgn(x) =
Then
2
−1, : x ≤ 0
lim P
n→∞
Z 1
1 + sgnβ(s)
Nn
≤ α = PW
ds ≤ α
n
2
0
Brownian Motion
Introduction to Brownian Motion as a Measure
Donsker’s Invariance Principal
Defining Wiener Measure Using Cylinder Sets
I
For any integer n, any choice of 0 < τ1 < · · · < τn ≤ 1, and any
Lebesgue measurable (L-mb) set, E ∈ Rn define the “interval”
I = I(n; τ1 ; . . . ; τn ; E) := {β(·) ∈ C0 [0, 1]; (β(τ1 ), . . . , β(τn )) ∈ E}
I
Let A be the class of intervals containing all the I for all n, τ1 , . . . , τn and
all L−mb sets E ∈ Rn , then A is an algebra of sets in C0 [0, 1]
I
The I’s are the cylinder sets upon which we will define Wiener measure,
and then standard measure theoretic ideas to extend to all measurable
subsets of the infinite dimensional space, C0 [0, 1]
Brownian Motion
Introduction to Brownian Motion as a Measure
Donsker’s Invariance Principal
Defining Wiener Measure Using Cylinder Sets
I
Given I, we define its measure as
µ(I) = p
(2π)n τ1 (τ2
Z
1
− τ1 ) · · · (τn − τn−1 )
u2
Z
···
e
(u −u )2
(un −u
)2
n−1
2
1 −···−
− 2τ1 − 2(τ
2(τn −τn−1 )
1
2 −τ1 )
du1 · · · dun .
E
I
I
Let B be the smallest σ−algebra generated by A, this is the class of
Wiener measurable (W-mb) sets in C0 [0, 1]
This extension of Wiener measure, also creates a probability measure on
C0 [0, 1], and expectation w.r.t. Wiener measure will be referred to as a
1. Wiener integral or Wiener integration
2. Brownian motion expectation
Brownian Motion
Introduction to Brownian Motion as a Measure
Donsker’s Invariance Principal
Examples
I
Let A ∈ Rn×n with Aij
we have

τ1 τ1
A =  τ1 τ2
τ1 τ2
= min(τi , τj ), i.e for the case n = 3,
τ1 < τ 2 < τ 3

τ1
τ2 
τ3
and in general we can write U = (u1 , . . . , un )> and
Z
Z
> −1
1
µ(I) = p
· · · e−U A U du1 . . . dun
(2π)n det A
E
I
Let β(·) be a BM, and 0 < τ1 < τ2 < 1, then
Z b1
2
1
− u
e 2τ1 du and
P[a1 ≤ β(τ1 ) ≤ b1 ] =
2πτ1 a1
P[a1 ≤ β(τ1 ) ≤ b1 ∩ a2 ≤ β(τ2 ) ≤ b2 ]
Z b2 Z b1
2
(u −u )2
1
− u − 2 1
e 2τ1 2(τ2 −τ1 ) du1 du2
= p
(2π)2 τ1 (τ2 − τ1 ) a2 a1
Brownian Motion
Introduction to Brownian Motion as a Measure
Properties of Brownian Motion
Useful Properties of Brownian Motion
I
I
S∞
Theorem: Let
PI∞= j=1 Ij where Ij ∩ Ik = ∅ ∀i 6= k and I, I1 , I2 , · · · ∈ A,
then µ(I) = j=1 µ(Ij )
we will see that the BM, β(t), satisfies:
1. Almost every (AE) path is non-differentiable at every point
2. AE path satisfies a Hölder condition of order α < 12 , i.e.
|β(s) − β(t)| ≤ L|s − t|α
3.
4.
5.
6.
E[β(t)] = 0
E[β 2 (t)] = t, and so β(t) ∼ N(0, t)
β(0) = 0, β(t) − β(s) ∼ N(0, t − s)
E[β(t)β(s)] = min(s, t)
Brownian Motion
Introduction to Brownian Motion as a Measure
Properties of Brownian Motion
Useful Properties of Brownian Motion
I
Let E ∈ Rn (L − mb), 0 < τ1 < · · · < τn < 1, I = I(n; τ1 ; . . . ; τn ; E), then
Z
Z
µ(I) = · · · p(τ1 , 0, u1 )p(τ2 − τ1 , u1 , u2 ) · · ·
E
p(τN − τn−1 , un , un−1 ) du1 · · · dun
where p(t, x, y ) =
I
(x−y )2
2t
Note that p(t, x, y ) = ψ(t, x, y ), the fundamental solution for the initial
value problem for the heat/diffusion equation
ψt =
I
√ 1 e−
2πt
1
ψyy ,
2
ψ(0, x, y ) = δ(y − x)
µ is finitely additive since integrals are additive set functions
Brownian Motion
Introduction to Brownian Motion as a Measure
Properties of Brownian Motion
Useful Properties of Brownian Motion
I
Theorem 1: Let a > 0, 0 < γ <
1
2
and define
Aa,γ = {β ∈ C0 [0, 1]; |β(τ2 ) − β(τ1 )| ≤ a|τ2 − τ1 |γ ∀τ1 , τ2 ∈ [0, 1]}
For any interval I ⊂ C0 [0, 1] s.t. I ∩ Aa,γ = ∅ there is a K indepedent of a
for which
m(I) < Ka
4
− 1−2γ
I
Remark: Aa,γ is a compact set in C0 [0, 1] and eventually one can prove
that AE β ∈ C0 [0, 1] satisfy some Hölder condition
I
Theorem 2: µ is countably additive on A, i.e. if In ∈ A, n ∈ N disjoint
(Ij ∩ Ik = ∅, j 6= k ) then
I=
∞
[
n=1
In ∈ A ⇒ µ(I) =
∞
X
n=1
µ(In )
Brownian Motion
Introduction to Brownian Motion as a Measure
Properties of Brownian Motion
Useful Properties of Brownian Motion
I
I
Suppose F : C0 [0, 1] → R is a measurable functional,
i.e. {β ∈ C0 [0, 1]; F [β] ≤ α} is measurable ∀α
We can consider
Z
E[F ] = EW [F [β(·)]] = F [β(·)]δW , a Wiener integral
I
Consider Cx [0, t] = {f ∈ C[0, t]; f (0) = x}, then
Z
(y −x)2
1
√
P [β(0) = x, β(t) ∈ A] =
e− 2t dy
2πt A
I
Furthermore
1
E [β(τ )] = √
2πτ
Z
∞
u2
ue− 2τ du = 0, ∀τ > 0
−∞
1
E [g (β(τ1 ), . . . , β(τn ))] = p
×
(2π)n τ1 (τ2 − τ1 ) · · · (τn − τn−1 )
Z
Z
(un −un−1 )2
u2
(u2 −u1 )2
− 2τ1 − 2(τ
−···− 2(τ −τ
−τ1 )
n
1
2
n−1 ) du · · · du
· · · g(u1 , . . . , un )e
n
1
Brownian Motion
Introduction to Brownian Motion as a Measure
Properties of Brownian Motion
Useful Properties of Brownian Motion
I
I
Let us now consider, without proof, a large deviation result for BM:
Theorem (The Law of the Iterated Logarithm for BM): Let β(s) ∈
C0 [0, ∞) be ordinary Brownian Motion, then
(1)
P
lim sup √
t→∞
β(t)
2t ln ln t
=1 =1
(2)
P
β(t)
lim inf √
= −1 = 1
t→∞
2t ln ln t
Brownian Motion
Introduction to Brownian Motion as a Measure
Properties of Brownian Motion
Dirac Delta Function
I
Let g be Borel measurable (B-mb), then
Z ∞
u2
1
E[g(β(τ ))] = √
g(u)e− 2τ du
2πτ −∞
I
Let g(u) = δ(u − x), using the Dirac delta function, then
Z ∞
u2
x2
1
1
√
E[δ(β(t) − x)] =
δ(u − x)e− 2t du = √
e− 2t
2πt −∞
2πt
thus u(x, t) = E[δ(β(t) − x)] =
the heat equation
ut =
1
uxx , u(x, 0) = δ(x)
2
x2
√ 1 e − 2t
2πt
is the fundamental solution of
Brownian Motion
The Feynman-Kac Formula
The Feynman-Kac Formula
I
Consider now V (x) ≥ 0 continuous and consider the equation
ut =
1
uxx − V (x)u, u(x, 0) = δ(x),
2
then we can write
h Rt
i
u(x, t) = E e− 0 V (β(s)) ds δ(β(t) − x)
This is the Feynman-Kac formula
I
Example:
V (x) =
x2
1
x2
, ut = uxx −
u, u(x, 0) = δ(x), then
2
2h
2
i
Rt 2
1
u(x, t) = E e− 2 0 β (s) ds δ(β(t) − x)
Brownian Motion
The Feynman-Kac Formula
The Feynman-Kac Formula
I
The following is clearly true:
P[β(τ ) ≤ x] =P ({β ∈ C0 [0, τ ]; β(τ ) ∈ E = (−∞, x]}) =
Z x
u2
1
√
e− 2τ du, and similarly
2πτ −∞
With 0 = τ0 ≤ τ1 · · · ≤ τn we have
P [β(τ1 ) ≤ x1 , . . . , β(τn ) ≤ xn ] = p
Z
xn
Z
···
−∞
I
u2
x1
e
(u −u )2
(2π)−n/2
×
(τ1 − τ0 )(τ2 − τ1 ) · · · (τn − τn−1 )
(un −u
)2
n−1
2
1 −···−
− 2τ1 − 2(τ
2(τn −τn−1 )
1
2 −τ1 )
du1 · · · dun
−∞
Hence with Aij = min(τi , τj )
1
E [g (β(τ1 ), . . . , β(τn ))] = p
×
(2π)n |A|
Z ∞
Z ∞
1 > −1
···
g(u1 , · · · , un )e− 2 U A U du1 · · · dun
−∞
−∞
Brownian Motion
The Feynman-Kac Formula
Feynman-Kac Formula: Derivation
I
Let us consider the Wiener integral below, where expectation is taken
over all of C0 [0, t]
o
n R
E e−
I
I
I
t
0
V (β(τ )) dτ
We will show that this is equal to the solution of the Bloch equation using
an elementary proof of Kac
We assume that 0 ≤ V (x) < M is bounded from above and
non-negative; however, the upper bound will be relaxed
We know
Z t
k
∞
Rt
X
(−1)k
V (β(τ )) dτ /k !
e− 0 V (β(τ )) dτ =
k =0
0
I
Since V (·) is bounded we also have
Z t
0<
V (β(τ )) dτ < Mt
I
This allows us to use Fubini’s theorem as follows
(Z
k )
∞
n Rt
o X
t
− 0 V (β(τ )) dτ
k
E e
=
(−1) E
V (β(τ )) dτ
/k !
0
k =0
0
Brownian Motion
The Feynman-Kac Formula
Feynman-Kac Formula: Derivation
I
Now let us consider the moments
(Z
k )
t
µk (t) = E
V (β(τ )) dτ
0
I
Consider first k = 1
Z t
Z tZ
Z t
Fubini
E {V (β(τ ))} dτ =
E
V (β(τ )) dτ
=
0
0
I
0
∞
V (ξ) √
−∞
ξ2
1
e− 2τ dξ dτ
2πτ
The case k = 2 is a bit more complicated
(Z
2 )
Z t Z τ2
t
Fubini
E
V (β(τ )) dτ
= 2!E
V (β(τ1 ))V (β(τ2 )) dτ1 dτ2
=
0
0
Z tZ
0
τ2
E {V (β(τ1 ))V (β(τ2 ))} dτ1 dτ2 =
2!
0
0
ξ2
Z tZ
τ2
Z
∞
Z
∞
2!
0
0
−∞
−∞
− 2τ1
e
V (ξ1 )V (ξ2 ) √
−
(ξ2 −ξ1 )2
e 2(τ2 −τ1 )
p
dξ1 dξ2 dτ1 dτ2
2πτ1 2π(τ2 − τ1 )
1
Brownian Motion
The Feynman-Kac Formula
Feynman-Kac Formula: Derivation
I
For general k we will proceed by defining the function Qn (x, t) as follows
1. Q0 (x, t) =
x2
√ 1 e− 2t
2πt
2. Qn+1 (x, t) =
Rt R∞
0
−∞
(x−ξ)2
√
−
1
e 2(τ −t)
2π(τ −t)
Rt
V (ξ)Qn (ξ, τ ) dξ dτ
Qk (x, t) dx
I
We have that µk (t) = k !
I
By the boundedness of V (·) we also have, by induction, that
n
0 ≤ Qn (x, t) ≤ (Mt)
Q0 (x, t)
n!
P
k
Now define Q(x, t) = ∞
k =0 (−1) Qk (x, t)
I
0
I
This series converges for all x and t 6= 0 and |Q(x, t)| < eMt Q0 (x, t)
I
One can easily check that the definitions of the Qk (x, t)’s ensures that
Q(x, t) satisfies the following integral equation
Z tZ ∞
(x−ξ)2
1
1
−
p
Q(x, t) + √
e 2(t−τ ) V (ξ)Q(ξ, τ ) dξ dτ = Q0 (x, t)
2π 0 −∞ (t − τ )
Brownian Motion
The Feynman-Kac Formula
Feynman-Kac Formula: Derivation
I
It also follows that
n Rt
o Z
E e− 0 V (β(τ )) dτ =
∞
Q(x, t) dx
−∞
I
Recall that his Wiener integral is over all of C0 [0, t], let us restrict this
only to a < β(t) < b, thus
n Rt
o Z b
E e− 0 V (β(τ )) dτ ; a < β(t) < b =
Q(x, t) dx
a
I
This tell us immediately that Q(x, t) ≥ 0
I
Now we will relax the upper bound on V (·) by considering the function
(
V (x), if V (x) ≤ M
VM (x) =
M,
if V (x) ≥ M
and we denote Q (M) (x, t) as the respective “Q” function
Brownian Motion
The Feynman-Kac Formula
Feynman-Kac Formula: Derivation
I
By the additivity of Wiener measure we have that
n Rt
o
n Rt
o
lim E e− 0 VM (β(τ )) dτ ; a < β(t) < b = E e− 0 V (β(τ )) dτ ; a < β(t) < b
M→∞
I
Furthermore, as M → ∞ the functions Q (M) (x, t) form a decreasing
sequence with limM→∞ Q (M) (x, t) = Q(x, t) existing with the resulting
limiting function, Q(x, t) satisfying the (Bloch) equation
1 ∂2Q
∂Q
=
− V (x)Q
∂t
2 ∂x 2
with the initial condition Q(x, t) → δ(x) as t → 0
Brownian Motion
The Feynman-Kac Formula
Feynman-Kac Formula: Derivation Variation
I
I
I
I
Recall the integral equation solved by Q(x, t)
Z tZ ∞
(x−ξ)2
x2
1
1
1
−
p
e 2(t−τ ) V (ξ)Q(ξ, τ ) dξ dτ = √
e− 2t
Q(x, t)+ √
2πt
2π 0 −∞ (t − τ )
R∞
−st
Let us define Ψ(x) = −∞ Q(x, t)e dt with s > 0, this is the Laplace
transform of Q(x, t)
Now multiply the integral equation by e−st and integrate out t to get the
equation satisfied by the Laplace transform of Q(x, t)
Z ∞
√
√
1
1
√
e− 2s|x−ξ| V (ξ)Ψ(ξ) dξ = √ e− 2s|x|
Ψ(x) +
2s −∞
2s
It is easy to verify that Ψ(x) also satisfies the following differential
equation
1 00
Ψ − (s + V (x))Ψ = 0, with the following conditions
2
1. Ψ → 0 as |x| → ∞
2. Ψ0 is continuous except at x = 0
3. Ψ0 (−0) − Ψ0 (−0) = 2
Brownian Motion
The Feynman-Kac Formula
Explicit Representation of Brownian Motion
Explicit Representation of Brownian Motion
I
Suppose that F [β] =
t
Z
E
Rt
0
β 2 (s) ds, then it follows
β 2 (s) ds
t
Z
=
0
h
Fubini
0
Rt
β(s) ds
i
Z t
h
i
t2
E β 2 (s) ds =
s ds =
2
0
I
To compute E e
I
Consider the eigenvalue problem for this integral equation
Z t
ρ
u(s) min(τ, s) ds = u(τ )
0
, we need to do some classical analysis
0
I
Find eigenvalues ρ0 , ρ1 , . . . and corresponding
orthonormalized
Rt
eigenfunctions u0 (τ ), u1 (τ ), . . . with 0 uj (τ )uk (τ ) dτ = δjk , ∀j, k ≥ 0
Brownian Motion
The Feynman-Kac Formula
Explicit Representation of Brownian Motion
Explicit Representation of Brownian Motion
I
For t > τ we have
Z t
Z τ
su(s) ds + ρ
τ u(s) ds = u(τ )
ρ
0
τ
d
dτ
t
Z
=⇒ ρτ u(τ ) − ρτ u(τ ) + ρ
u(s) ds = u 0 (τ )
τ
d
dτ
=⇒
−ρu(τ ) = u 00 (τ )
Thus u 00 (τ ) + ρu(τ ) = 0 and with u(0) = 0, u 0 (t) = 0 we get

1 2 π2

ρk = (k +
)
q2 t 2
k = 0, 1, 2, . . .

uk (s) = 2 sin (k + 1 ) πs
t
I
2
t
By the spectral theorem the integral equation kernel can be represented
as:
min(s, τ ) =
∞
X
uk (s)uk (τ )
ρk
k =0
Brownian Motion
The Feynman-Kac Formula
Explicit Representation of Brownian Motion
Explicit Representation of Brownian Motion
I
Let α0 (ω), α1 (ω), . . . be i.i.d. N(0, 1), then we claim that the following is
an explicit representation of BM
∞
X
αk (ω)uk (τ )
= β(τ )
√
ρk
(2.1)
k =0
I
This is a Fourier series with random coefficients and we will prove that
this converges for AE path ω with the following properties
1. We use ω to denote an individual sample of i.i.d. N(0, 1) αi (ω)’s
2. E[αi (ω)] = 0, ∀i ≥ 0
3. E[αi (ω)αj (ω)] = δij , ∀i, j ≥ 0
I
This is the simplest version of the Karhunen-Loève expansion of
stochastic processes
Brownian Motion
The Feynman-Kac Formula
Explicit Representation of Brownian Motion
Explicit Representation of Brownian Motion (Proof)
I
We now use the representation (2.1) to compute some expectations
w.r.t. the αi ’s ∼ N(0, 1)
#
"∞
X αk (ω)
i.i.d. N(0,1) & Fubini
E
=
√ uk (τ )
ρk
k =0
∞
∞
X
X
E[αk (ω)]uk (τ )
0 × uk (τ )
=
= 0 = E[β(τ )]
√
√
ρk
ρk
k =0
k =0
I
We now use the representation (2.1) to compute some expectations
"∞
#
∞
X αk (ω)
X
i.i.d.N(0,1)
αl (ω)
E
=
√ uk (τ )
√ ul (τ )
ρk
ρl
k =0
l=0
∞
h
i
X
uk2 (τ )
= min(τ, τ ) = τ = E β 2 (τ )
ρk
k =0
Brownian Motion
The Feynman-Kac Formula
Explicit Representation of Brownian Motion
Explicit Representation of Brownian Motion (Proof)
I
Similarly we compute
#
"∞
∞
X
X αk (ω)
i.i.d.N(0,1)
αl (ω)
E
=
√ uk (τ )
√ ul (s)
ρk
ρl
k =0
l=0
∞
X
uk (τ )uk (s)
= min(τ, s) = E [β(τ )β(s)]
ρk
k =0
I
We have computed the mean, variance, and correlation of the process
defined in (2.1), and it is clear that it is ∼ N(0, τ ) and hence Brownian
motion, β(τ )
Brownian Motion
The Feynman-Kac Formula
The Karhunen-Loève Expansion
An Introduction to the Karhunen-Loève Expansion
I
Karhunen-Loève (KL) expansion writes the stochastic processes Y (ω, t)
as a stochastic linear combination of a set of orthonormal, deterministic
functions in L2 , {ei (t)}∞
i=0
Y (ω, t) =
∞
X
Zi (ω)ei (t)
i=0
1. Given the covariance function of the random process Y (ω, t) as CYY (s, τ )
the KL expansion is
Y (ω, t) =
∞ p
X
λi ξi (ω)φi (t)
i=0
2. Here λi and φi (t) are the eigenvalues and L2 -orthonormal eigenfunctions of
the covariance function and ξi (ω)φi (t) are i.i.d.
√ random variables whose
distribution depends on Y (ω, t), i.e. Zi (ω) = λi ξi (ω), and ei (t) = φi (t)
3. It can be shown that such an expansion converges to the stochastic process
in L2 (in distribution)
Brownian Motion
The Feynman-Kac Formula
The Karhunen-Loève Expansion
An Introduction to the Karhunen-Loève Expansion
4. By the spectral theorem, we can expand the covariance, thought of as
an integral equation kernel, as follows
∞
X
CYY (s, τ ) =
λi φi (s)φi (τ )
i=0
5. Here λi and φi (t) are the eigenvalues and eigenfunctions of the following
integral equation
Z ∞
CYY (s, τ )φj (τ ) dτ = λj φj (s)
0
I
For ordinary BM, Y (ω, t) = β(t), we have from above
1. CYY (s, τ ) = Cββ (s, τ ) = min(s, τ )
2
, where ρj = (j + 12 )2 πs2
q
3. φj (t) = uj (t) = 2s sin((j + 21 ) πts )
2. λj =
1
ρj
4. ξj (ω) = αj (ω) ∼ N(0, 1)
P αj (ω)uj (t)
5. Y (ω, t) = ∞
= β(t)
√
j=0
ρ
j
Brownian Motion
The Feynman-Kac Formula
Explicit Computation of Wiener Integrals
Explicit Computation of Wiener Integrals
I
We are now in position to compute
R P
i
h Rt
t
∞
=
E e 0 k =0
E e 0 β(s) ds
P R
∞
t
E e k =0 0
α
√k
ρk
uk (s) ds
indep.
=
∞
Y
α
√k
E e ρk
αk uk (s)
√
ρk
Rt
0
ds
=
uk (s) ds
=
k =0
∞
Y
1
e 2ρk
Rt
(
0
2
uk (s) ds)
1
=
e2
R t R t P∞
0
0
k =0
uk (s)uk (τ )
ρk
ds dτ
=
k =0
1
I
Rt Rt
1
e 2 0 0 min(s,τ ) ds dτ
=
e2
We have used the following results
u2
τ2
0 [( 2
Rt
+(τ (t−τ ))] dτ
t3
=e6
1. E[eαu ] = e 2 , with α ∼ N(0, 1) via moment generating function
R
R
R
2
2. 0t min(s, τ ) ds = 0τ s ds + τt τ ds = τ2 + (τ (t − τ ))
Brownian Motion
The Feynman-Kac Formula
Explicit Computation of Wiener Integrals
Explicit Computation of Wiener Integrals
I
Moreover
λ2
E e− 2
Rt
indep.
=
0
β 2 (s) ds
∞
Y
"
"
E
2
=E e
2 α2
k
−λ
e 2 ρk
#
=
k =0
=
=
∞
Y
k =0
∞
Y
k =0
1
√
2π
− λ2
∞
Y
k =0
Z
1
q
1+
2
∞
e
− α2
α2
k
k =0 ρk
#
1
√
2π
Z
P∞
2
1+ λ
ρ
k
∞
e
2 2
− λ2 α
ρ
k
−∞
dα
−∞
λ2
ρk
1
= p
cosh(λt)
= s
Q∞
k =0
1
1+
λ2 t 2
(k + 12 )2 +π 2
e−
α2
2
dα
Brownian Motion
The Feynman-Kac Formula
The Schrödinger Equation
The Schrödinger Equation
I
Let us review the Schrödinger equation from quantum mechanics
1. The “standard," time-dependent Schrödinger equation
"
#
−~2
∂
∆ + V (x, t) Ψ(x, t) = Ĥ(x, t)Ψ
i~ Ψ(x, t) =
∂t
2m
2. We can make the equation dimensionless as
1
∂
−i ψ(x, t) =
∆ − V (r, t) ψ(x, t) = H(x, t)ψ
∂t
2
3. We also are interested in the spectral properties of the time-independent
problem
1
∆ − V (x, t) ψ(x, t) = H(x, t)ψ = λψ
2
Brownian Motion
The Feynman-Kac Formula
The Schrödinger Equation
The Schrödinger and Bloch Equations
I
We now arrive at the Bloch equation
1. Consider transformation (analytic continuation) of the Schrödinger to
imaginary time, τ = it, this gives us the Bloch equation, but is sometimes
also called the Schrödinger equation (going back to u(x, t))
1
∂u(x, t)
= ∆u(x, t) − V (x, t)u(x, t)
∂τ
2
2. The time dependent Bloch equation can be solved via separation of
variables as
u(x, t) = U(x)T (t), and so we apply this to the Bloch equation
∂u(x, t)
= U(x)T 0 (t) =
∂t
1
∆U(x) − V (x, t)U(x) T (t)
2
Brownian Motion
The Feynman-Kac Formula
The Schrödinger Equation
The Schrödinger and Bloch Equations
3. Placing the time and space dependent on different sides of the equation
gives
1
∆ − V (x, t) U(x)
T 0 (t)
=λ= 2
, where λ is constant
T (t)
U(x)
4. Thus we have that T (t) and U(x) satisfy the following equations
T 0 (t) − λT (t) = 0,
1
∆ − V (x, t) U(x) = λU(x)
2
5. Thus the λj ’s and ψj (x, t)’s are eigenvalues and eigenfunctions of the
above eigenvalue problem, and the solution by separation variables is
u(x, t) =
∞
X
j=1
cj e−λj t ψj (x), where, cj =
Z
∞
u0 (x)ψj (x) dx
−∞
Brownian Motion
The Feynman-Kac Formula
The Schrödinger Equation
The Schrödinger and Bloch Equations
I
i
h 1 Rt 2
Let λ = 1, as t → ∞, E e− 2 0 β (s) ds = √
lim
t→∞
I
1
cosh(t)
∼
√ −t
2e 2 and
i
h 1 Rt 2
1
1
ln E e− 2 0 β (s) ds = − .
t
2
Theorem: If V (y ) → ∞ as |y | → ∞, then
lim
t→∞
i
h Rt
1
ln E e− 0 V (β(s)) ds = −λ1 ,
t
where λ1 is the lowest eigenvalue of the Bloch equation
1 00
ψ (y ) − V (y )ψ(y ) = λψ(y )
2
Brownian Motion
The Feynman-Kac Formula
The Schrödinger Equation
The Schrödinger and Bloch Equations
I
Feynmann-Kac: Let V be measurable and bounded below, then the
solution of the Bloch equation
1
uxx − V (x)u, u(x, 0) = u0 (x)
2
h Rt
i
is u(x, t) = Ex e− 0 V (β(s)) ds u0 (β(t))
ut =
I
This equation is the imaginary time analog of the Schrödinger
1 00
ψ (y ) − V (y )ψ(y ) = λψ(y )
2
Equation
1. Special case: V ≡ 0:
Ex [u0 (β(t))] = √
1
2πt
Z
∞
−∞
u0 (y )e−
(x−y )2
2t
dy = u(x, t)
Brownian Motion
The Feynman-Kac Formula
The Schrödinger Equation
Another special case
x2
,
2
u0 ≡ 1:
h 1 Rt 2
i
h 1 Rt
i
2
u(x, t) = = Ex e− 2 0 β (s) ds · 1 = E0 e− 2 0 (β(s)+x) ds
i
h
Rt
x2t
1 Rt 2
=e− 2 E e−x 0 β(s) ds− 2 0 β (s) ds
"
#
2
2. For V (x) =
=e−
x2t
2
=e−
x2t
2
∞
Y
2
− x2 t
k =0
∞
Y
=e
E e
k =0
=e−
x2t
2
−x
αk
k =0 √ρk
P∞
"
E e
α
−x √ρk
1
√
2π
0
Rt
0
k
Z
Rt
∞
e
uk (s) ds− 12
α2
uk (s) ds− 12 ρ k
k
−x √α
ρ
Rt
k
0
α
k
k =0 ρk
P∞
#
2
uk (s) ds− α2 (1+ ρ1 )
k
−∞
x2
1
p
e2
cosh(t)
R t R t P∞
0
0
k =0
uk (s)uk (τ )
ρk +1
ds dτ
dα
Brownian Motion
The Feynman-Kac Formula
The Schrödinger Equation
I
Define R(s, τ ; −λ2 ) such that
Z t
min(s, τ ) = λ2
min(s, ξ)R(ξ, τ ; −λ2 ) dξ
0
Note that R(s, τ ; −1) = −
I
P∞
k =0
uk (s)uk (τ )
.
ρk +1
Consider
−
∞
∞
X
uk (s)uk (τ ) X uk (s)uk (τ )
+
ρk + λ 2
ρk
k =0
k =0
Z tX
∞
∞
uk (s)uk (ξ) X ul (ξ)ul (τ )
= λ2
dξ
ρk
ρk + λ 2
0
k =0
l=0
Brownian Motion
The Feynman-Kac Formula
The Schrödinger Equation
I
I
For 0 ≤ s ≤ t we have
( cosh(λ(t−τ )) sinh(λs)
−
2
λ cosh(λt)
R(s, τ ; −λ ) =
sinh(λτ )
− cosh(λ(t−s))
λ cosh(λt)
s≥τ
Thus
u(x, t) = p
I
s≤τ
Rt Rt
x2
x 2 tanh t
1
1
e− 2 (t+ 0 0 R(s,τ ;−1) ds dτ ) = p
e− 2
cosh(t)
cosh(t)
2
Exercise: compute
t) for Vi (x) = x2 , u0 (x) = x. Hint: the solution is
h 1 R tu(x,
2
u(x, t) = Ex e− 2 0 β (s) ds β(t) . Calculate
h
i
1 Rt 2
ũ(x, t, λ) =Ex eλβ(t)− 2 0 β (s) ds ,
u(x, t) =
d
ũ(x, t, λ) .
dλ
λ=0
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Proof of the Arcsin Law
I
Theorem: Let X1 , X2 , . . . be i.i.d. r.v.’s with
P E [Xi ] = 0, Var (Xi ) = 1, and
Nn is the number of partial sums Sj = ji=1 Xi out of S1 , . . . , Sn which
are ≥ 0:


α<0
0
√
Nn
lim P
< α = Σ(α) = π2 arcsin α 0 ≤ α ≤ 1
n→∞

n

1
α≥1
I
Proof: (Using the Feynman-Kac formula and Donsker’s Invariance
Principal) Define the random step function
(S
√1
τ =0
(n)
X (τ ) = Sin i−1
√
< τ ≤ ni
n
n
The invariance principle states that for a large class of functionals F and
F ∈F
h h
i
i
lim P F X (n) (·) ≤ α = PBM [F [β(·)] ≤ α]
(2.2)
n→∞
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Proof of the Arcsin Law
I
For example, let
t
Z
F [β] =
0
1 + sgn[β(s)]
ds, where sgn(x) =
2
(
1
−1
I
Then (2.2) says that
Z 1
1 + sgn[β(s)]
Nn
≤ α = PBM
ds ≤ α
lim P
n→∞
n
2
0
I
We drop the BM from the probabilities as it is understood
of the Brownian motion that is positive
x ≥0
x <0
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Proof of the Arcsin Law
I
Let
t
Z
σ(α, t) = P
0
I
1 + sgn[β(s)]
ds ≤ α
2
Then for λ > 0 we can define the Laplace Transform/Moment Generating
Function of σ(α, t)
Z ∞
R t 1+sgn[β(s)]
ds
2
=
e−λα dσ(α, t)
E e−λ 0
0
I
Now define
Rt
u(x, t; λ) = E e−λ 0
1+sgn[β(s)]
2
ds
δ(β(t) − x)
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Proof of the Arcsin Law
I
By Feynman-Kac this is a solution to the following PDE
u(x, t; λ)t =
1
u(x, t; λ)xx − λV (x)u(x, t; λ),
2
(
where V (x) =
I
We also realize that
Z ∞
Z
u(x, t; λ) dx =
−∞
Z
∞
Rt
E e−λ 0
u(x, 0; λ) = δ(x)
1 x ≥0
0 x <0
1+sgn[β(s)]
2
ds
Fubini
δ(β(t) − x) dx =
−∞
∞
E
e−λ
R t 1+sgn[β(s)]
0
2
ds
Rt
δ(β(t) − x) dx = E e−λ 0
−∞
∞
Z
0
e−λα dσ(α, t)
1+sgn[β(s)]
2
ds
=
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Proof of the Arcsin Law
I
It is know that u(x, t; λ) also solves the following integral equation
−x 2
1
e 2t −
u(x, t; λ) = √
2πt
Z t
Z ∞
−(x−ξ)2
1
e 2(t−τ )
λ
dτ
dξV (ξ)u(ξ, τ ; λ) p
2π(t − τ )
0
−∞
I
Now we apply the heat equation operator,
∂
∂t
−
1 ∂2
2 ∂x 2
∂u
1 ∂2u
−
= 0 − λV (x)u(x, t; λ)
∂t
2 ∂x 2
I
And we the Laplace transform of u(x, t; λ)
Z ∞
Ψ(x, s; λ) =
e−st u(x, t; λ)dt
−∞
to this
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Proof of the Arcsin Law
I
If we take the Laplace transform of the integral equation we get
√
1
Ψ(x, s; λ) = √ e− 2s|x|
2s
Z ∞
√
1
dξV (ξ)Ψ(ξ, s; λ) √ e− 2s|x−ξ|
−λ
2s
−∞
I
This is equivalent to the following ordinary differential equation (ODE)
1 00
Ψ (x) − (s + λV (x))Ψ(x) = 0, Ψ → 0 as |x| → ∞
2
Ψ(x) and Ψ0 (x) is continuous at x 6= 0, and Ψ0 (0− ) − Ψ0 (0+ ) = 2
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Proof of the Arcsin Law
I
The solution to the above ODE is
( √
Ψ(x, s; λ) =
I
x ≥0
x <0
Thus we have that
Z
∞
−∞
I
√
2 √
√
e− 2(s+λ)x
s+λ+
s
√
√
2 √
√
e− 2sx
s+λ+ s
1
Ψ(x, s; λ) dx = p
s(s + λ)
So we have the following
Z ∞
Z ∞
Z ∞
Ψ(x, s; λ) dx =
e−st
u(x, t; λ) dx ds =
−∞
−∞
0
Z ∞
Z ∞
1
e−st
e−λα dσ(α, t) ds = p
s(s
+ λ)
0
0
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Proof of the Arcsin Law
I
The last line test us that we know the Laplace transform of
Z ∞
F (t) =
e−λα dσ(α, t)
0
I
The inverse Laplace transform of √
λt
F (t) = e− 2 Io (
I
1
s(s+λ)
λt
)=
2
∞
Z
tells us that
e−λα σ 0 (α, t) dα
0
Which is itself the Laplace transform of σ 0 (α, t), so we have

 √1
0<α<t
0
σ (α, t) = π α(t−α)
0
α>t
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Proof of the Arcsin Law
I
We now integrate the previous result
Z
α
−∞
I
σ 0 (ᾱ, t) d ᾱ = σ(α, t) =


0
2
π

1
Setting t = 1 we get the Arcsin Law


0
√
σ(α, 1) = Σ(α) = π2 arcsin t


1
arcsin
pα
t
0<α
0<α<t
α>t
0<α
0 < α < 1 Q. E. D.
α>1
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Another Wiener Integral
I
We wish to compute the probability of
P max β(s) ≤ α
0≤s≤t
I
By Donsker’s Invariance Principal this is equal to
S1 S2
Sn
lim max √ , √ , · · · , √
≤ α = P max β(s) ≤ α = H(α, t)
n→∞
0≤s≤t
n
n
n
I
Consider the step-function potential
(
1
Vα (x) =
0
I
x ≥α
x <α
Since β(·) is a continuous function AE, if max0≤s≤t β(s) ≤ α then
Vα (β(s)) = 0 on a set of positive measure
Brownian Motion
The Feynman-Kac Formula
Proof of the Arcsin Law
Another Wiener Integral
I
Consider the following Wiener integral
h
i
Rt
lim E e−λ 0 Vα (β(s)) ds = H(α, t)
λ→∞
I
I
This is because the λ limit kills walks that exceed α and only count the
walks that satisfy the condition
for a fixed λ this is, by Feynman-Kac, the solution to
u(x, t; λ)t =
I
1
u(x, t; λ)xx − λV (x)u(x, t; λ),
2
(
1 x ≥α
where V (x) =
0 x <α
u(x, 0; λ) = 1
The solution of the PDE is very similar to the solution of the PDE from
the Arcsin Law, and is left to the reader
r Z √α
u2
t
2
H(α, t) =
e− 2 du
π 0
Brownian Motion
Advanced Topics
Action Asymptotics
Action Asympotics: A Heuristic for Wiener Integrals
I
Von Neumann proved that there is no translationally invariant Haar
measure in function space; Wiener measure is not translationally
invariant
I
Consider the following problem where we write our heuristic via a “flat”
integral
1
F [β]e− 2
E {F [β]} “ = ”
I
0
[β 0 (τ )]2 dτ δβ
Here we define the Action as
A[β] = −
I
Rt
1
2
Z
t
0 2
β (τ ) dτ
0
This is obviously a heuristic, as BM is nondifferentiable AE
Brownian Motion
Advanced Topics
Action Asymptotics
Action Asympotics: A Heuristic for Wiener Integrals
I
Now consider computing the following with Action Asymptotics
h √1 R t
i
β(s) ds
E e 0
I
We first compute this using our standard techniques
i
h √1 R t
1 R t P∞
√
β(s) ds
=
E e 0 k =0
E e 0
1 P∞
√
E e k =0
α
√k
ρk
Rt
0
uk (s) ds
indep.
=
∞
Y
k =0
I
And thus
1
√
2π
αk uk (s)
√
ρk
∞
Z
e
−∞
h √1 R t
i
t3
β(s) ds
lim ln E e 0
=
→0
6
√α
ρk
ds
Rt
0
=
2
uk (s) ds − α
e 2
dα =
Brownian Motion
Advanced Topics
Action Asymptotics
Action Asympotics: A Heuristic for Wiener Integrals
I
Let’s “derive” the action asymptotics heuristic with a construction due to
Kac and Feynman by considering
n Rt
o
F (t) = E e− 0 V (β(τ )) dτ
where β(·) ∈ C0 [0, t], and the expectation is taken w.r.t. Wiener measure
I
Since we assume that V (·) is continuous and non-negative, and
Rt
β(·) ∈ C0 [0, t]) is continuous, F (t) exists as 0 V (β(τ )) dτ is measurable
I
Now let us consider a discrete approximation of this Wiener integral by
breaking it up into N sized time intervals of size t/N, which gives us F (t)
from bounded convergence and the Riemann summability
o
n t PN
tk
F (t) = lim E e− N k =1 V (β( N ))
N→∞
Brownian Motion
Advanced Topics
Action Asymptotics
Action Asympotics: A Heuristic for Wiener Integrals
I
If we consider the expectation in the limit we can rewrite it as follows
Z ∞
Z ∞
o
n t PN
PN
tk
···
e−h k =1 V (βk ) ×
lim E e− N k =1 V (β( N )) = lim
N→∞
N→∞
−∞
−∞
P(0, β1 ; h)P(β1 , β2 ; h) · · · P(βN−1 , βN ; h) dβ1 dβ2 · · · dβN
where we have
1. h = Nt
2. βk = β(kh)
3. P(βk −1 , βk ; h) =
√ 1 e−
2πh
(βk −βk −1 )2
2h
I
This limit exists and is equal to the Wiener integral
I
However, Feynman chose to rewrite the above as (suppressing the limit)
with β0 = 0
(
1
(2πh)N/2
Z
∞
Z
∞
···
−∞
−h
e
−∞
PN
k =1
V (βk )+ 12
PN
k =1
βk −βk −1
h
2 )
dβ1 dβ2 · · · dβN
Brownian Motion
Advanced Topics
Action Asymptotics
Action Asympotics: A Heuristic for Wiener Integrals
I
If we look at the exponent in Feynman’s we notice that
( N
)
2 )
2
Z t( N X
1 X βk − βk −1
1 dβ
h→0
V (βk ) +
h→
+ V (β(τ )) dτ
2
h
2 dτ
0
k =1
k =1
I
This is the Hamiltonian the along the path, β(τ ), and with the classical
action along the path is
)
2
Z t( 1 dβ
− V (β(τ )) dτ
2 dτ
0
I
thus Feynman writes the above integral instead as
n Rt
o Z −
F (t) = E e− 0 V (β(τ )) dτ = e
Rt
0
1
2
( dβ
dτ )
2
+V (β(τ )) dτ
d(path)
Brownian Motion
Advanced Topics
Action Asymptotics
Action Asympotics: A Heuristic for Wiener Integral
I
h 1 √ i
How does E e F [ β] behave as → 0?
I
We can approach this with Action Asymptotics
h 1 √ i
E e F [ β] “ = ”
I
Now let
1
1
e [F [ω]− 2
Rt
0 [ω
0
(s)]2 ds]
δβ
Using Laplace asymptotics the above will behave like
1
e
I
R
2
β] − 12 0t [β 0 (s)] ds
δβ
e
√
β = ω
“=”
I
√
1
e F[
R
supω∈C ∗ [0,t] [F [ω]− 12 0t [ω 0 (s)]2 ds]
0
Where the space C0∗ [0, t] is made up functions, ω(t), with
1. ω(t) continuous in [0, t]
2. ω(0) = 0
3. ω 0 (t) ∈ L2 [0, t]
Brownian Motion
Advanced Topics
Action Asymptotics
Action Asympotics: Examples
I
A conjecture using Action Asymptotics
h 1 √ i
lim ln E e F [ β] =
→0
I
Consider F [β] =
Rt
0
Z
1 t 0
F [ω] −
[ω (s)]2 ds
2 0
ω∈C0∗ [0,t]
sup
β(s) ds
h 1 √ i
h √1 R t
i
β(s) ds
E e F [ β] = E e 0
I
From the conjecture we have that
h
lim ln E e
→0
1
√
Rt
0
β(s) ds
i
t
Z
=
sup
ω∈C0∗ [0,t]
0
1
ω(s) ds −
2
Z
t
0
2
[ω (s)] ds
0
Brownian Motion
Advanced Topics
Action Asymptotics
Action Asympotics: Examples
I
From the calculus of variations we have that the Euler equation for
following maximum principle is
Z t
Z
1 t 0
2
sup
[ω (s)] ds =⇒
ω(s) ds −
2 0
ω∈C0∗ [0,t]
0
1. 1 + ω 00 (s) = 0
2. ω(0) = 0
3. ω 0 (t) = 0
I
2
The solution is ω(s) = − s2 + ts and ω 0 (s) = −s + t so
Z t 2
Z
s
1 t
t3
− + ts ds −
[s − t]2 ds =
2
2 0
6
0
Brownian Motion
Advanced Topics
Brownian Scaling
Brownian Scaling
I
Recall some basic properties of the BM, β(·) and constant, c:
1.
2.
3.
4.
5.
6.
I
β(τ ) ∼ N(0, τ )
β(cτ ) ∼ N(0, cτ )
√
cβ(τ ) ∼ N(0, cτ )
E[β(τ )β(s)] = min(τ, s)
E[β(cτ )β(cs)] = c min(τ,
s)√
√
E[β(cτ )β(cs)] = E[ cβ(τ ) cβ(s)] = cE[β(τ )β(s)] = c min(τ, s)
Now consider the following
h
i
E esup0≤s≤t β(s)
=
i
h
√
E esup0≤τ ≤1 tβ(τ )
=
h 1
i
√
E e sup0≤τ ≤1 β(τ )
h
i
E esup0≤τ ≤1 β(tτ ) =
h
1 β(τ ) i
√
t sup
E e 0≤τ ≤1 t
=
using the substitution t =
1
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics: Examples
I
So we now have that
i
h
i
h 1
√
1
lim ln E esup0≤s≤t β(s) = lim ln E e sup0≤τ ≤1 β(τ )
t→∞ t
→0
By Action Asymptotics we have
h 1
i
√
lim ln E e sup0≤τ ≤1 β(τ )
=
I
"
#
Z
1 1 0
2
[ω (τ )] dτ
sup
sup ω(τ ) −
→0
2 0
ω∈C0∗ [0,1] 0≤τ ≤1
a2
1
= max a −
=
a>0
2
2
I The supremum comes on straight lines, that minimize arc-length i.e. the
second term, so consider ω(τ ) = aτ , and a = 1 is the maximizer
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics: Examples
I
Consider a more complicated problem for Action Asymptotics is
h √
i
√
1
E G ( β(·)) e F ( β(·))
h 1 √
i
lim
“=”
→0
E e F ( β(·))
i
h √
√
2
0
1
1 Rt
E G ( β(·)) e F ( β(·))− 2 0 [β (s)] ds δβ
h
E e
1F
(
i
R
√
β(·))− 12 0t [β 0 (s)]2 ds
=
δβ
√
We now change variables with x(·) = β(·)
h
i
2
1 Rt 0
1
E G (x(·)) e [F (x(·))− 2 0 [x (s)] ds] δx
h 1
i
1 Rt 0
2
E e [F (x(·))− 2 0 [x (s)] ds] δx
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics: Examples
I
As → 0 the exponential term goes to something like a “delta" function
in function space and we get
= G [ω ∗ (·)] where ω ∗ (·) = argsup [F [ω] − A[ω]]
ω∈C0∗ [0,t]
I
We now apply this to some PDE problems: Burger’s Equation
ut + uux =
uxx ,
−∞ ≤ x ≤ ∞, t > 0
2
Z
∞
u(x, 0)
=
u0 (η) dη = o(x 2 ) as |x| → ∞
u0 (x),
0
I
We now apply the Hopf-Cole transformation, if we define the solution to
(x,t)
Burger’s equation u(x, t) = − vvx(x,t)
= −∂x [ln v (x, t)] then v (x, t)
satisfies
1 Rx
vt = vxx , v (x, 0) = e− 0 u0 (η) dη
2
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics: Examples
I
So by Feynman-Kac we can write the solution as
Z ∞
(x−y )2
1 Ry
1
v (x, t; ) = √
e− 0 u0 (η) dη e− 2t dy
2πt −∞
I
We now apply the Hopf-Cole transformation (taking the logarithmic
derivative)
(y −x)2
1 Ry
dy
(x−y ) − 0 u0 (η) dη+ 2t
e
t
−∞
2
R ∞ − 1 R0y u0 (η) dη+ (y −x)
dy
2t
R∞
u(x, t; ) =
−∞
I
I
I
e
Ry
2
Now let F (y ) = 0 u0 (η) dη + (y −x)
, this is the function that Action
2t
Asymptotics tells us to minimize (due to the negative sign)
Note that lim|y |→∞ Fy(y2 ) = 2t1 by the assumptions, and so there is a
minimum, y (x, t) = argmin F (y )
Hopf showed that if at (x, t) there is a single minimizer to F (y ) then
lim u(x, t; ) =
→0
x − y (x, t)
= u0 (y (x, t))
t
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics: Examples
I
Consider the related equation
ut + uux =
uxx − V 0 (x),
2
Z
−∞ ≤ x ≤ ∞,
t >0
∞
u(x, 0)
=
u0 (x),
u0 (η) dη = o(x 2 ) as |x| → ∞
0
I
Again we use the Hopf-Cole transformation to get
vt =
I
1
vxx − V 0 (x)v ,
2
1
v (x, 0) = e− Rx
0
u0 (η) dη
And so we can write down the solution to the transformed equation via
Feynman-Kac
√
√
1 R β(t) u (η) dη
1 Rt
0
v (x, t; ) = Ex e− 0 V ( β(s)) ds− 0
hR
i
R √β(t)+x
√
− 1 0t V ( β(s)+x) ds 0
u0 (η) dη
= E0 e
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics: Examples
I
We now take apply the Hopf-Cole transformation and get
i
h √
√
1
E G[ β(·)]e− F [ β(·)]
h 1 √
i
where we define
u(x, t; ) =
E e− F [ β(·)]
t
Z
F [β(·)] =
0
t
Z
G[β(·)] =
0
√
V ( β(s)) ds −
√
Z
β(t)
u0 (η) dη
0
√
√
V 0 ( β(s) + x) ds + u0 ( β(t) + x)
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics: Examples
I
By Action Asymptotics we have that
lim u(x, t; ) = G[ω ∗ (·)] where ω ∗ (·) = arginf [F [ω] + A[ω]]
→0
I
ω∈C0∗ [0,t]
If for (x, t) ∃! minimizer, ω ∗ , then the limit exists and is
Z t
G[ω ∗ (t)] = u(x, t) =
V 0 (ω ∗ (s) + x) ds + u0 (ω ∗ (t) + x)
0
I
Now consider the related variational problem
"Z
Z
t
inf
∗
ω∈C0 [0,t]
I
ω(t)+x
V (ω(s) + x) ds
0
u0 (η) dη +
0
1
2
We refer to the functional to be minimized as H[ω(·)]
t
Z
0
#
[ω 0 (s)]2 ds
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics: Examples
I
To arrive derive an equivalent system via the Calculus of Variations we
need to form the Frechet derivative, in the direction of the arbitrary
function, Ψ, as follows
Z t
dH[ω + hΨ] =
V 0 (ω(s) + x)Ψ(s) ds + u0 (ω(t) + x)Ψ(t)
δH|Ψ =
dh
0
h=0
Z t
ω 00 (s)Ψ(s) ds
+ ω 0 (t)Ψ(t) −
0
I
Note that the last two terms come from the following computation
Z t
dJ[ω + hΨ] def 1
J[ω(·)] =
[ω 0 (s)]2 ds =⇒
2 0
dh
h=0
Z t
Z t
1
=
[ω 0 (s) + hΨ0 (s)]2 ds =
[ω 0 (s) + hΨ0 (s)]2 ds
2 0
0
Z t
Z t
=
ω 0 (s)Ψ0 (s) ds =
ω 0 (s) dΨ0 (s)
0
0
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics: Examples
I
We now integrate by parts using the natural boundary conditions
1. ω(0) = 0
2. ω 0 (0) = 0
Z
t
ω 0 (s) dΨ0 (s) = ω 0 (t)Ψ0 (s) −
0
I
t
Z
ω 00 (s)Ψ ds
0
So the solution to this problem is
1. V 0 (ω(s) + x) = ω 00 (s) for 0 ≤ s ≤ t
2. ω(0) = 0
3. ω 0 (t) = −u0 (ω(s) + x)
I
We can now apply this Hpof’s result with V ≡ 0
1. ω 00 (s) = 0 for 0 ≤ s ≤ t
2. ω(0) = 0
3. ω 0 (t) = −u0 (ω(s) + x)
I
The solution is then very simply
1. ω(s) = cs for some constant, c
2. ω 0 (s) = c = −u0 (ct + x)
y (x,t)−x
3. Let c =
= −u0 (y (x, t)) or u0 (t(x, t)) =
t
I
With a unique y (x, t) we get a unique ω ∗ (s) =
x−y (x,t)
t
x−y (x,t)
t
s
Brownian Motion
Advanced Topics
Brownian Scaling
Action Asympotics
I
I
We now consider some tools with the “flat integral"
The Cameron-Martin Translation Formula
E {F [β + y ]} , with y ∈ C0 [0, t]
I
We now use the “flat integral"
1
F [β + y ]e− 2
E {F [β + y ]} “ = ”
1
F [ω]e− 2
“=”
1
“ = ”e− 2
Rt
0
Rt
0 [ω
(s)]2 ds
0
Rt
0 [β
0
(s)]2 ds
δβ, and let ω = β + y
(s)−y 0 (s)]2 ds
Rt
δω
0
0
1
F [ω]e+ 0 [ω (s)y (s)] ds− 2
n
o
Rt 0
2
1 Rt 0
“ = ”e− 2 0 [y (s)] ds E F [β]e 0 y (s) dβ(s)
I
0 [y
Rt
0 [ω
0
(s)]2 ds
And so our result is that
1
E {F [β + y ]} = e− 2
Rt
0 [y
0
(s)]2 ds
n
o
Rt 0
E F [β]e 0 y (s) dβ(s) , with y ∈ C0 [0, t]
δω
Brownian Motion
Advanced Topics
Local Time
Local Time
I
Spectral Theory:
I
If V (x) ≥ 0 and V (x) → 0 as |x| → ∞ then the eigenvalue problem
1 00
Ψ (x) − V (x)Ψ(x) = −λΨ(x)
2
1. Has discrete spectrum: λ1 , λ2 , · · ·
2. With corresponding eigenfunctions: Ψ1 , Ψ2 , · · ·
I
Theorem (1949):
lim
t→∞
1 h − 12 R0t V (β(s)) ds i
E e
= −λ1
t
Note: The expectation can start at any x due to ergodicity
I
Proof We will first prove this using Feynman-Kac
h 1 Rt
i
u(x, t) = Ex e− 2 0 V (β(s)) ds
Brownian Motion
Advanced Topics
Local Time
Local Time
I
Satisfies the following PDE
ut =
I
1
uxx − V (x)u,
2
u(x, 0) = 1
By separation of variables we have
u(x, t) =
∞
X
cj e−λj t ψj (x), where, cj =
Z
u(x, 0)ψj (y ) dy
−∞
j=1
I
∞
But since u(x, 0) = 1 we have that cj =
two representations must be equal
R∞
−∞
ψj (y ) dy , ∀j ≥ 0, and so the
Z
∞
h 1 Rt
i X
u(x, t) = Ex e− 2 0 V (β(s)) ds =
e−λj t ψj (x)
j=1
I
∞
ψj (y ) dy
−∞
And so the largest eigenvalue, λ1 , controls the behavior
i
1 h 1 Rt
lim E e− 2 0 V (β(s)) ds = −λ1
t→∞ t
Brownian Motion
Advanced Topics
Local Time
Local Time
I
We also have a variational representation of λ1
Z ∞
Z
1 ∞ 0
λ1 = inf
V (y )Ψ2 (y ) dy +
[Ψ (y )]2 dy
2 −∞
Ψ∈L2
−∞
||Ψ||=1
I
Which has a corresponding Euler equation
1 00
Ψ (x) − V (x)Ψ(x) = −λΨ(x)
2
I
I
i
h 1 Rt
We notice that in the Wiener integral representation, E e− 2 0 V (β(s)) ds ,
since the internal integral is in an negative exponential, the main
contribution comes for paths that remain close to where V (·) is smallest,
which leads us to dissect this problem as follows
Let β(s), 0 ≤ s < ∞; β(0) = x be BM for t > 0 and consider the
proportion of time that β(·) spends in a set A ⊂ R
Z
1 t
`t (β(·), ·) =
χ (β(s)) ds
t 0 A
Brownian Motion
Advanced Topics
Local Time
Local Time
I
Some properties of Lt (β(·), ·) with t > 0, x fixed, and β(·) a particular,
fixed, path
1. Lt (β(·), ·) is a countable additive, non-negative function
2. Lt (β(·), R) = 1
3. Lt (β(·), ·) : Cx [0, t] → M, the space of probability measures on R
I
As a set function, Lt (β(·), ·) for fixed x ∈ R and t > 0 and for almost all
β(·) has a density function which we call the normalized local time
Z
1 t
`t (β(·), y ) =
δ(β(s) − y ) dy and
t 0
Z ∞
Lt (β(·), A) =
χA (y )`t (β(·), y ) dy
I
`t (β(·), ·) → 0 as Table → ∞ for compact A and almost every β(·)
I
Now consider the following representation
h Rt
i
h
i
R∞
Ex e− 0 V (β(s)) ds = Ex e−t −∞ V (y )`t (β(·),y ) dy
−∞
Brownian Motion
Advanced Topics
Local Time
Local Time
I
For fixed x ∈ R and t > 0 we define a probability measure on M,
Qx,t = PL−1
t , as follows
I
If C ⊂ M then we can write
Qx,t (C) = P {β(·) ∈ Cx [0, ∞] : Lt (β(·), ·) ∈ C}
I
Lt (β(·), ·) is an occupation measure so we can write
i
i
i
h Rt
h
h
R∞
R∞
Ex e− 0 V (β(s)) ds = Ex e−t −∞ V (y )`t (β(·),y ) dy = Ex e−t −∞ V (y ) dLt (β(·),y )
h
i
h
i
R∞
R∞
Q
Q
Ex x,t e−t −∞ V (y ) µ(dy ) = Ex x,t e−t −∞ V (y )f (y ) dy
I
We define F as the space of probability density functions on R, then this
an expected value on F
I
To understand how the expected value on F behaves as t → ∞, we
need to understand how Qx,t and therefore also how Lt (β(·), A) behaves
as t → ∞
Brownian Motion
Advanced Topics
Local Time
Local Time
I
Long time behavior of local time measures
1. Lt (β(·), A) → 0 as t → ∞ for A ⊂ R, compact, and AE β(·)
2. `t (β(·), A) → 0 as t → ∞ for A ⊂ R, compact, and AE β(·) by the ergodic
theorem for BM, if β(·) were not BM, then this would converge AE to the
invariant measure
3. Qx,t (C) → 0 as t → ∞ if C ⊂ M, C 6= M, i.e. C is a reasonable set
I
Theorem on Speed of Convergence: We first need to put the Levý
topology on F
1. If C ∈ F is closed, then
lim sup
t→∞
1
ln Qx,t (C) ≤ inf I(f )
f ∈C
t
2. If G ∈ F is open, then
lim inf
t→∞
3. Where
I(f ) =
1
8
1
ln Qx,t (C) ≥ inf I(f )
f ∈G
t
Z
∞
−∞
n
o
[f 0 (y )]2 /f (y ) dy
Brownian Motion
Advanced Topics
Donsker-Varadhan Asymptotics
Donsker-Varadhan Asympotics
I
This is a simple case of what is referred to as “Donsker-Varadhan
Asymptotics" and are a large deviation result
I
An example, suppose f (y ) ∼ N(0, σ 2 ), i.e. f (y ) =
y
f 0 (y ) = − σ3 √
e
2π
I(f ) =
1
8
Z
∞
−∞
n
2
− y 2
2σ
and f 0 (y )2 =
[f 0 (y )]2 /f (y )
o
dy =
−2
y2
e
σ 6 2π
1 1
8 σ4
Z
y2
2σ 2
∞
−∞
y2
σ
−
√1 e 2σ2
2π
, then
and finally we have
2
y2
σ2
1
− y
√ e 2σ2 dy =
=
8σ 4
8σ 2
σ 2π
Note: the last integral is the variance, σ 2 , of a N(0, σ 2 ) random variable
I
We refer to the functional I : F → [0, ∞] as the entropy, and roughly
speaking
Qx,t (f ) ∼ e−t inff ∈A I(f ) for “nice" A
Brownian Motion
Advanced Topics
Donsker-Varadhan Asymptotics
Donsker-Varadhan Asympotics
I
I
I
Now let us apply the “Entropy Asymptotics" with the “Flat Integral"
i
h 1 Rt
h
i
R∞
Q
Ex e− 2 0 V (β(s)) ds = Ex x,t e−t −∞ V (y )f (y ) dy for t large
R∞
V (y )f (y ) dy
R∞
V (y )f (y ) dy +I(f )]
“=”
e−t
“=”
e−t [
−∞
−∞
e−tI(f ) δf
δf
As t → ∞ we use Laplace asymptotics to get
Z ∞
Z
h 1 Rt
i
1 ∞ [f 0 (y )]2
1
V (y )f (y ) dy +
dy
lim ln Ex e− 2 0 V (β(s)) ds = − inf
t→∞ t
f ∈y
8 −∞ f (y )
−∞
p
R∞ 2
R∞
Let f (y ) = Ψ(y ), then −∞ Ψ (y ) dy = −∞ f (y ) dy = 1 since f (y ) is a
p.d.f., and so Ψ(·) ∈ L2 [−∞, ∞] and ||Ψ|| = 1
Brownian Motion
Advanced Topics
Donsker-Varadhan Asymptotics
Donsker-Varadhan Asympotics
I
We now transform the “Entropy Asymptotics" expression with some
substitutions
p
R
R
∞
∞
1. Let f (y ) = Ψ(y ), then −∞
Ψ2 (y ) dy = −∞
f (y ) dy = 1 since f (y ) is a
2
p.d.f., and so Ψ(·) ∈ L [−∞, ∞] and ||Ψ|| = 1 2. Also Ψ0 (y ) = √1
2
f (y )
f 0 (y ), and so [Ψ0 (y )]2 =
1
4
f 0 (y )2
f (y )
I
These allow us to write
Z ∞
Z
1 ∞ [f 0 (y )]2
− inf
V (y )f (y ) dy +
dy =
f ∈y
8 −∞ f (y )
−∞
Z ∞
Z
1 ∞ 0
[Ψ (y )]2 dy = −λ1
− inf
V (y )Ψ2 (y ) dy +
2 −∞
Ψ∈L2
−∞
I
Theorem:Let Φ : F → R be bounded and continuous then, by the
“general structure theorem"
h
i
h
i
1
1
Q
lim ln Ex x,t e−tΦ(f ) = lim ln Ex e−tΦ(`t (β(·),·)) = − inf [Φ(f ) + I(f )]
t→∞ t
t→∞ t
f ∈F
||Ψ||=1
Brownian Motion
Advanced Topics
Donsker-Varadhan Asymptotics
Donsker-Varadhan Asympotics
I
This is more subtle than action asymptotics, for example consider
h
i
1
Q
lim ln Ex x,t e+tΦ(f ) = sup [Φ(f ) − I(f )]
t→∞ t
f ∈F
1. There is always a fight between the two terms in the supremum
2. In statistical mechanics we often consider αΦ(f ) and want to compute
supf ∈F [αΦ(f ) − I(f )] = g(α), where α is a convex function of α
3. Them may be a critical value of α, call it α0 , where there is a phase
transition, this is due to nonuniqueness in the f that maximized the functional
Brownian Motion
Advanced Topics
Donsker-Varadhan Asymptotics
An Example Using Action and Entropy Asymptotics
I
Now we will use “Entropy Asymptotics" to revisit a topic we have already
considered
I
Recall that
(
P
)
sup β(s) ≤ α
r
=
0≤s≤t
h
E e
sup0≤s≤t β(s)
i
2
πt
α
Z
0
∞
Z
e dP{ sup β(s) ≤ α} =
= h(t) =
eα
0
2
e
πt
2
− α2t
r
dα =
with the substitution u =
I
2
πt
e
0≤s≤t
∞
Z
e−
(α−t)2
2t
t
e+ 2 dα =
0
2 2t
e
π
Then we have
r
2 2t
e
π
Z
∞
√
t
α
0
r
α−t
√
t
1
lim ln h(t) =
t→∞ t
r
∞
Z
α
0
r
∞
Z
u2
e− 2t du, so that we also have
e−
u2
2
du =
1
2
Z
∞
√
t
2 − α2t2
e
dα
πt
e−
u2
2
du
Brownian Motion
Advanced Topics
Donsker-Varadhan Asymptotics
An Example Using Action and Entropy Asymptotics
I
First we turn the t → ∞ limit into an → 0 limit
i
h
i
h 1
√
1
lim ln E esup0≤s≤t β(s) = lim ln E e sup0≤τ ≤1 β(τ )
t→∞ t
→0
Recall that by Action Asymptotics we have
"
#
Z
h 1
i
√
1 1 0
sup0≤τ ≤1 β(τ )
2
lim ln E e
[ω (τ )] dτ
=
sup
sup ω(τ ) −
→0
2 0
ω∈C0∗ [0,1] 0≤τ ≤1
a2
1
= max a −
=
a>0
2
2
I The supremum comes on straight lines, that minimize arc-length i.e. the
second term, so consider ω(τ ) = aτ , and a = 1 is the maximizer
I
Brownian Motion
Advanced Topics
Donsker-Varadhan Asymptotics
An Example Using Action and Entropy Asymptotics
I
Now we solve the same problem using Entropy Asymptotics by using a
result of Paul Levý that the following have the same probability
distributions
(
)
P
sup β(s) ≤ α
= P {t`t (β(·), 0)}
0≤s≤t
I
Thus we have that
h
i
h
i
h
i
h(t) = E esup0≤s≤t β(s) = E et`t (β(·),0) = E etΦ[`t (β(·),0)] , where Φ[f ] = f (0)
I
So from Entropy Asymptotics we get
lim
t→∞
I
Z
i
1 ∞ [f 0 (y )]2
1
1 h
ln h(t) = lim E etΦ[`t (β(·),0)] = sup f (0) −
dy
t→∞ t
t
8 −∞ f (y )
f ∈F
Recall that f ∈ F is a probability distribution, and so the maximizing
family of functions (proven below) is fa (y ) = ae−2a|y |
Brownian Motion
Advanced Topics
Donsker-Varadhan Asymptotics
An Example Using Action and Entropy Asymptotics
I
Recall that f ∈ F is a probability distribution, and so the maximizing
family of functions (proven below) is fa (y ) = ae−2a|y |
I
We can write
fa (y ) = ae−2a|y | =
(
ae−2ay
ae2ay
[fa0 (y )]2 =
I
y ≥0
, so fa0 (y ) =
y <0
(
4a4 e−4ay
4a4 e4ay
(
−2a2 e−2ay
2a2 e2ay
y ≥0
, and so
y <0
y ≥0
= 4a4 e−4a|y |
y <0
This gives us
Z
Z
1 ∞ [f 0 (y )]2
1 ∞
sup f (0) −
dy = sup a −
4a3 e−2a|y | dy
8 −∞ f (y )
8 −∞
a>0
a>0
Z
a2 ∞
1
a2
= sup a −
ae−2a|y | dy = sup a −
= , which occurs at a = 1
2 −∞
2
2
a>0
a>0
Brownian Motion
Advanced Topics
Donsker-Varadhan Asymptotics
An Example Using Action and Entropy Asymptotics
I
Now we find the maximizing family of functions by the same
transformation
as before
p
1.
f (y ) = Ψ(y ) or f (y ) = Ψ2 (y ), and so
2
2. f (0)
= Ψ (0)
3.
I
1
4
f 0 (y )2
f (y )
= [Ψ0 (y )]2
And so we obtain
Z
Z
1 ∞ [f 0 (y )]2
1 ∞ 0
dy = sup Ψ2 (0) −
[Ψ (y )]2 dy
sup f (0) −
8 −∞ f (y )
2 −∞
f ∈F
Ψ∈L2
||Ψ||=1
I
Let Ψ(0) = a we get the following constrained Euler-Lagrange equation
Ψ00 (y ) − 2λΨ(y ),
I
Ψ(0) = a
This is maximized with a stretched version of Ψ(y ) = e−2|y |
Brownian Motion
Advanced Topics
Can One Hear the Shape of a Drum?
Kac’s Drum
I
I
Let Ω ⊂ R2 be an open domain with sufficiently smooth boundary, ∂Ω,
so that the following problem has a unique solution
1
∆u + λu = 0, with u = 0 on ∂Ω
2
Under these circumstances we know that
1. ∃0 < λ1 < λ2 < · · · a discrete spectrum
2. ∃u1 (x, y ) < u1 (x, y ) < · · · corresponding normalized eigenfunctions
I
Consider
C(λ) =
X
1 = # of eigenvalues < λ
λj <λ
I
C(λ) is an increasing function in λ, and Hermen Weyl proved that
I
|Ω|λ
as λ → ∞
2π
Additionally, Carlemann proved that
X
λ
u(x, y ) ∼
, ∀ (x, y ) ∈ Ω as λ → ∞
2π
C(λ) ∼
λj <λ
Brownian Motion
Advanced Topics
Can One Hear the Shape of a Drum?
Kac’s Drum
I
I
I
Now consider starting a BM at (x0 , y0 ) ∈ Ω
Let p(x0 , y0 , x, y , t) be the probability density function of a 2D BM
starting at (x0 , y0 ) reaching (x, y ) at time t without hitting ∂Ω
Einstein-Smoluchowski: Then p(x0 , y0 , x, y , t) is the solution to
1
∂p
= ∆p in Ω
∂t
2
p = 0 on ∂Ω, ∀t > 0
I
We note that as t → 0
Z
g(x, y )p(x0 , y0 , x, y , t) dx dy → g(x0 , y0 )
I
Assume we can find p using separation of variables:
p(x0 , y0 , x, y , t) = T (t)U(x, y ), then
Ω
T
∆U, U = 0 on ∂Ω, ∀t > 0
2
0
∆U
T
=
= −λ yields
T
2
−λt
T (t) = e , and U = the eigenfunction corresponding to λ
T 0U =
Brownian Motion
Advanced Topics
Can One Hear the Shape of a Drum?
Kac’s Drum
I
So this means that we can write explicitly
p(x0 , y0 , x, y , t) =
∞
X
e−λj t uj (x0 , y0 )uj (x, y ), and so we know
j=1
p(x0 , y0 , x0 , y0 , t) =
∞
X
e−λj t uj2 (x0 , y0 )
j=1
I
Let p∗ (x0 , y0 , x, y , t) be the probability density function of unrestricted 2D
BM starting at (x0 , y0 ) reaching (x, y ) at time t
p∗ (x0 , y0 , x, y , t) =
I
1 − (x−x2t 0 )2 − (y −y2t 0 )2
e
2πt
Thus we conclude that
∞
X
j=1
e−λj t uj2 (x0 , y0 ) ∼ p∗ (x0 , y0 , x, y , t) ∼
1
as t → 0
2πt
Brownian Motion
Advanced Topics
Can One Hear the Shape of a Drum?
Kac’s Drum
I
Karamata Tauberian Theorem: Consider
Z ∞
f (t) =
e−λt dα(λ), and assume
0
1. The above Laplace-Stiltje’s transform exists
2. α(λ) is non-decreasing on (0, ∞)
I
If f (t) ∼ At −γ as t → 0 for A and γ constants then
α(λ) ∼
I
Aλγ
as λ → ∞(λ → 0)
Γ(γ + 1)
We now apply the Karamata Tauberian Theorem to
Z ∞
∞
X
X 2
f (t) =
e−λt dα(λ) =
e−λj t uj2 (x0 , y0 ), where α(λ) =
uj (x0 , y0 )
0
λj <λ
j=1
1
2πt
as t → 0, and so α(λ) ∼
λ
2π
I
We know f (t) ∼
I
By integrating this over Ω we get Weyl’s theorem
as λ → ∞
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
1. Let Ω ∈ R3 be a bounded closed domain
2. Let r(t) ∈ C be a continuous function starting at the origin
3. Let χΩ (·) be the indicator function of Ω
I
Consider the following functional on C
Z ∞
TΩ (y, r(·)) =
χΩ (y + r(τ )) dτ,
y ∈ R3
0
I
I
This functional is the total occupations time of r(·), a 3D BM, in Ω
translated by y
Now impose Wiener measure on C and consider the following Wiener
integral
Z
∞
E {TΩ (y, r(·))} =
P {y + r(τ ) ∈ Ω} dτ
0
I
Note that because we are using Wiener measure we know
Z ∞
|r−y|2
1
P {y + r(τ ) ∈ Ω} =
e− 2τ dr
3/2
(2πτ )
0
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
We now use Fubini’s theorem to exchange the order of integration
Z
Z ∞
|r−y|2
1
e− 2τ dτ
E {TΩ (y, r(·))} =
dr
3/2
(2πτ )
Ω
Z 0
1
dr
=
< ∞ in R3
2π Ω |r − y|
I
We see that in R3 AE BM path starting at y spends a finite amount of
time in Ω
I
Now consider the k th moment of the occupation time
Z
Z
n
o
[k ]
drk
k!
dr1
dr2
E TΩk (y, r(·)) =
·
·
·
···
(2π)k Ω
|r
−
y|
|r
−
r
|
|r
−
rk −1 |
1
2
1
k
Ω
I
k = 1, 2, · · ·
We focus on the second moment, k = 2
n
o Z ∞Z ∞
2
E TΩ (y, r(·)) =
P {y + r(τ1 ) ∈ Ω} P {y + r(τ2 ) ∈ Ω} dτ1 dτ2
0
0
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
We focus on the second moment, k = 2
n
o Z ∞Z ∞
P {y + r(τ1 ) ∈ Ω} P {y + r(τ2 ) ∈ Ω} dτ1 dτ2
E TΩ2 (y, r(·)) =
0
ZZ
=2
0≤τ1 <τ2 <∞
0
dτ1 dτ2
Ω
=
I
2
Z Z
Ω
|r −r |
|r1 −y|2
1
1
− 2 1
e− 2τ
e 2(τ2 −τ1 ) dr1 dr2
3/2
3/2
(2πτ1 )
[2π(τ2 − τ1 )]
2
(2π)2
Z Z
Ω
Ω
dr1
dr2
|r1 − y| |r2 − r1 |
The formula for the k th moment suggests that we should consider the
following eigenvalue problem
Z
φ(ρ)
1
dρ = λφ(r), r ∈ Ω
2π Ω |r − ρ|
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
The integral kernel in the eigenvalue problem is Hilbert-Schmidt
1. Since the single integral is convergent, we have
Z Z
1
dr dρ < ∞
2
Ω Ω |r − ρ|
2. We also need to show that the kernel is positive definite:
Z Z
φ(r)φ(ρ)
dr dρ > 0 ∀φ(ρ) 6= 0 in L2 (Ω)
Ω Ω |r − ρ|
Note that:
∞
Z
0
∞
|r−y|2
1
e− 2τ dτ =
3/2
(2πτ )
0
Z
−|ζ|2 τ
1
τ 3/2
dτ
dζ =
eiζ·(r−ρ) e 2
3/2
3/2
(2πτ )
(2π)
R3
Z ∞
Z
−|ζ|2 τ
1
dτ
dζeiζ·(r−ρ) e 2
3
(2π) 0
R3
1
1
=
2π |r − ρ|
Z
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
I
So
Z Z
φ(r)φ(ρ)
dr dρ =
Ω Ω |r − ρ|
2
Z
−|ζ|2 τ dζe 2 φ(ρ)eiζ·ρ dρ > 0, ∀φ(ρ) 6= 0 in L2 (Ω)
Z ∞
Z
1
dτ
(2π)3 0
Ω
R3
With the kernel being Hilbert-Schmidt, we know that the integral
equation has
1. Discrete spectrum: λ1 , λ2 , · · ·
2. With corresponding eigenfunctions that form a complete, orthonormal basis
for L2 (Ω)
I
Lemma:
∞
o X
1 n k
λkj −1
E TΩ (y, r(·)) =
k!
Z
j=1
φj (r) dr
Ω
1
2π
1. This holds for all y ∈ R3
2. If y ∈ Ω, then we note that
1
2π
φj (ρ)
Z
Ω
|ρ − y|
dρ = λj φj (y)
Z
Ω
φj (ρ)
dρ
|ρ − y|
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
Proof: Recall that
Z
Z
o
[k ]
1 n k
1
dr1
dr2
drk
E TΩ (y, r(·)) =
·
·
·
···
k!
(2π)k Ω
|rk − rk −1 |
Ω |r1 − y| |r2 − r1 |
I
We recognize this as an iterated integral equation of the form
a(y, r1 )a(r1 , r2 ) · · · a(rk −1 , rk )
I
We can then rewrite this using Mercer’s theorem representation of the
kernel of the integral operator
∞
X
1
=
λj φj (ρ)φj (y)
|ρ − y|
j=1
I
Next we apply Mercer’s theorem only to the terms not involving y to get
Z
Z X
∞
o
1
1 n k
1
E TΩ (y, r(·)) =
λkj −1 φj (r1 )φj (rk ) dr1 drk
k!
2π Ω |r1 − y| Ω
j=1
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
To review we have that
o
1 n k
E TΩ (y, r(·)) =
k!
I
(P
∞
R
R
1
λkj −1 Ω φj (r) dr 2π
Ω
R
k
λ
φ
(r)φ
(y)
dr,
j
j
j
j=1
Ω
j=1
P∞
φj (ρ)
|ρ−y|
dρ, y ∈ R3
y∈Ω
Now let us consider the moment generation function (Laplace transform)
with z ∈ C
∞
n
o X
o
zk n k
E ezTΩ (y,r(·)) =
E TΩ (y, r(·))
k!
k =0
I
Now we use the above lemma to get
Z
Z
∞ φj (ρ)
z X
1
dρ
=1+
φj (r) dr
2π
1 − λj z
Ω
Ω |ρ − y|
j=1
1. This series converges if |z| < λ 1
max
2. The moment generating function is analytic if <{z} < 0 since TΩ ≥ 0
3. The last series is analytic for <{z} < 0, so by analytic continuation this
identity holds with <{z} < 0
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
Let u > 0 and define
Z
Z
∞ n
o
φj (ρ)
u X
1
h(y, u) = E e−uTΩ (y,r(·)) = 1−
φj (r) dr
dρ
2π
1 + λj u
Ω
Ω |ρ − y|
j=1
(*)
I
This series converges on compact sets in C because
1.
1
<1
1 + λj u
2.

∞ Z
X
φj (r) dr

j=1
φj (ρ)
Z
Ω
Ω
|ρ − y|
2
dρ ≤
j=1
Z
|Ω|
Ω
I
∞ Z
X
2 X
∞ Z
φj (r) dr
Ω
j=1
φj (ρ)
Ω
|ρ − y|
2
dρ
=
dρ
<∞
|ρ − y|
This gives uniform convergence via the Weierstrass M-test and thus this
is also analytic
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
If y ∈ Ω then we get
h(y, u) = 1 −
∞ X
j=1
I
And so we can multiply both sides by
1
2π
I
λj u
1 + λj u
Z
Ω
h(y, u) dy
1
=
|y − r|
2π
Z
Ω
Z
1
2π|y−r|
φj (r) dr φj (y)
Ω
and integrate over Ω
Z
Z
∞ X
λj u
φj (y) dy
dy
1
−
φj (ρ) dρ
|y − r|
1 + λj u
2π Ω |y − r|
Ω
j=1
But we know that
Z
Z
∞ Z
X
φj (y) dy
1
dy
1
=
φj (ρ) dρ
2π Ω |y − r|
2π Ω |y − r|
Ω
j=1
I
Thus we cane write that
Z
Z
Z
∞ X
φj (y) dy
h(y, u) dy
1
1
1
=
φj (ρ) dρ
2π Ω |y − r|
1 + λj u
2π
Ω
Ω |y − r|
j=1
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
We recognize the left hand side of the previous equation from (*), and so
we use this ro rewrite this as
Z
h(y, u) dy
1
1
= (1 − h(r, u)) , ∀r ∈ R3
2π Ω |y − r|
u
I
Moreover, if we rename variables we get
Z
h(ρ, u) dρ
1
1
= (1 − h(y, u)) ,
2π Ω |y − ρ|
u
I
∀y ∈ R3
(**)
We now make some important observations
1. From (*) we see that if y ∈
/ Ω then h(y, u) is harmonic in y, and the series in
(*) converges uniformly on compact Ω’s
2. Again from (*) we get




2 1/2 X
2 1/2
∞ Z
∞ Z
φj (ρ)
u X
h(y, u) > 1 −
dρ
φj (ρ) dρ



2π 
Ω
Ω |ρ − y|
j=1
>1−
j=1
u
|Ω|1/2
2π
Z
Ω
dρ
|ρ − y|
1/2
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
3. So we now know that 0 ≤ h(y, u) ≤ 1, and so
lim h(y, u) = 1
(***)
u%∞
4. And for from Courant-Hilbert II, pp. 245–246
Z
h(y, u) dy
∆
= −4πh(y, u)
|y − r|
Ω
I
Now apply the Laplacian to both sides of (**) to get
1
−2h(y, u) = − ∆h(y, u)
u
or we get
I
1
∆h(y, u) − uh(y, u) = 0, y ∈ Ω
2
Now consider U(y) = limu%∞ (1 − h(y, u)) = P {TΩ (y, r(·)) > 0}, this is
the capacitory potential (capacitance) and follows easily from the
definition of the moment generating function
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
Example: Let Ω be a sphere of radius 1 centered at the origin
1. h(y, u) is clearly spherically symmetric
2. h(y, u) is harmonic outside Ω, so we have
h(y, u) =
α(u)
+ β(u),
|y|
y∈
/Ω
α(u)
3. From (***) we see that β(u) = 1 and so h(y, u) = |y| + 1 for y ∈ Ω
4. We also know that for y ∈ Ω we have
√
sinh( 2u |y|)
h(y, u) = γ(u)
|y|
5. If we substitute this into the equation (**) we get that γ(u) =
1 √
√
2u cosh(a 2u)
6. h(y, u) is continuous ∀y so from the uniform convergence of the series, and
so
√
α(u)
1 sinh( 2ua) 1
√
+1= √
a
2u cosh( 2ua) a
to finally give us
√

2u)
1 − 1 1 − tanh(a
√
, y∈
/Ω
|y| √
a 2u
h(y, u) =
 √ sinh( 2u|y|)
√
,
y∈Ω
2u cosh( 2ua)|y|
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
Recall that
(
U(y) = lim (1 − h(y, u)) = P TS(0,a) (y, r(·)) > 0 =
u%∞
a
,
|y|
y∈
/Ω
1,
y∈Ω
I
This is the capacitory potential of S(0, a)
I
Now back to the general case, ∀y ∈ R3 we have
!Z
Z
∞
n
o X
φj (ρ) dρ
1
1
−uTΩ (y,r(·))
φj (r) dr
1−E e
=
1
2π
λ
+
j
Ω
Ω |ρ − y|
u
j=1
1. We note that 0 ≤ 1 − h(y, u) ≤ 1
2. The function 1 − h(y, u) is non-decreasing in u: 1 − h(y, u1 ) ≤ 1 − h(y, u2 )
if u1 < u2
3. This is true due to the following
3.1 0 ≤ e−uTΩ (y,r(·)) ≤ 1 and
3.2
lim e
u%∞
−uTΩ (y,r(·))
=
(
0, TΩ > 0
1, TΩ = 0
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
From the previous results and the bounded convergence theorem we
have
U(y) = lim (1 − h(y, u)) = P {TΩ (y, r(·)) > 0}
u%∞
and hence also
U(y) = lim
∞
X
u%∞
j=1
1
u
1
+ λj
!Z
φj (r) dr
Ω
1
2π
Z
Ω
φj (ρ) dρ
|ρ − y|
and this holds ∀y ∈ R3
Case 1. Let y ∈ Ωo (the interior), clearly the continuity of r(·) immediately implies
U (y) = P {TΩ (y, r(·)) > 0} = 1
Remark: with y ∈
summability result
Ωo
1 = lim
u%∞
we have U (y) = 1 and so we have the following
∞
X
j=1
!Z
λj
λj +
1
u
φj (r) dr
Ω
1
2π
φj (ρ) dρ
Z
Ω
|ρ − y|
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
Case 2. Let y ∈
/ Ω, we already know that 1 − h(y, u) is harmonic in y, and it is
nondecreasing in u, and the previous limit in u exists and equals
P {TΩ (y, r(·)) > 0}, thus by Harnack’s theorem, U(y) is harmonic with
y∈
/ Ω. Assume that Ω ⊂ S(0, a), then
P {TΩ (y, r(·)) > 0} ≤ P TS(0,a) (y, r(·)) > 0
From the last problem this means
P {TΩ (y, r(·)) > 0} ≤
a
,
|y|
y∈
/ S(0, a)
and so lim|y|→∞ U(y) = 0
Case 3. Let yo ∈ ∂Ω, and assume that it is regular in the sense of Poincaré: ∃ a
sphere S(y∗ , ) lying completely in Ω so that yo ∈ S(y∗ , ) Consider now
y∈
/Ω
U(y) = P {TΩ (y, r(·)) > 0} ≥ P TS(0,a) (y, r(·)) > 0 =
|y − y∗ |
As y → yo with y ∈
/ Ω we have |y−y
→ |y −y
, and since U(y) ≤ 1 we
∗|
o
∗|
have finally that
lim U(y) = 1
y→yo
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
Probabilistic Potential Theory
I
Thus if Ω is a closed and bounded region, each point on the boundary
that is regular in the Poincaré sense has U(y) as the capacitory potential
of Ω
I
Recall that
U(y) = lim
δ→0
I
∞ X
j=1
1
λj + δ
Z
Ω
Z
Ω
φj (ρ) dρ
|ρ − y|
We note that this implies that
lim |y|(1 − h(|y|, u)) =
|y|→∞
I
1
2π
φj (r) dr
1
2π
Z
uh(ρ, u) dρ
Ω
Again assume that Ω ∈ S(0, a), then h(y, u) = E e−uTΩ ≥ e−uTS(0,a) ,
a
a
there for y ∈
/ S(0, a) we have h(y, u) ≥ 1 − |y|
or 1 − h(y, u) ≤ |y|
and so
u
2π
Z
h(ρ, u) dρ ≤ a
Ω
Brownian Motion
Advanced Topics
Probabilistic Potential Theory
c
Michael
Mascagni, 2016