JOSEPH G. CONLON
1. Construction of Brownian Motion
There are two ways in which the idea of a Markov chain on a discrete state space can be generalized: (1) The discrete time variable of the chain i.e. n = 0, 1, 2, .., can be made continuous; (2) the state space can be made into a continuum. There are many such examples, but there is one which stands out-Brownian motion- and we will study it here. Thus we have a probability space (Ω, F , P ) and a continuous set Xt, t ≥ 0, of real variables which have the following properties:
(a) Xt is a Gaussian variable with mean 0 and variance t.
(b) For any sequence 0 = t0 < t1< · · · tm, the variables Xtj − Xtj−1, j = 1, .., m,
are independent Gaussian with mean 0 and variance tj− tj−1, j = 1, .., m.
Evidently the properties (a), (b) determine the cdf of any finite set of variables Xt1, ..., Xtm. Now let Q be the non-negative rational numbers. Then the
Kol-mogorov construction allows us to create a probability space (Ω, F , P ) and vari-ables Xt, t ∈ Q, which have the properties (a), (b). The main issue in constructing
Brownian motion is to extend this set of variables to a continuous set of variables Xt, t ≥ 0. The key to doing this is the continuity result:
Proposition 1.1. Let (Ω, F , P ) be a probability space with variables Xt, t ∈ Q,
satisfying (a) and (b). Then for any a > 0 the function t → Xt(ω), t ∈ Q ∩ [0, a],
is uniformly continuous with probability 1.
Proof. We take a = 1 and use the notation Xt = X(t). For n = 0, 1, 2, .., we
associate to a dyadic interval In,k= {x ∈ [0, 1] : (k − 1)/2n< x ≤ k/2n} a random
variable Yn,k by
(1.1) Yn,k = sup{|X(t) − X((k − 1)/2n)| : t ∈ In,k∩ Q } .
Observe that the variables Yn,k, k = 1, .., 2n, are i.i.d., whence it follows that for
any δ > 0, (1.2) P ( max 1≤k≤2nYn,k > δ ) ≤ 2n X k=1 P ( Yn,k> δ ) = 2nP (Yn,1> δ ).
The key point now is to use the maximal function inequality (1.3) P (Yn,1> δ ) ≤
1
δpE[ |X(1/2 n)|p] .
This follows from Proposition 2.1 of Chapter II since any sequence X(tj), j = 1, 2, ..,
with t1 < t2 < · · · , is a Martingale. One can also prove it using the reflection
principle introduced in Chapter I. Since X(t) is Gaussian with mean 0 and variance t we have that
(1.4) E[ |X(t)|4 ] = 3t2.
Hence (1.2), (1.3) imply that
(1.5) P ( max
1≤k≤2nYn,k > δ ) ≤
3 δ42n .
Evidently (1.5) implies that (1.6) ∞ X n=1 P ( max 1≤k≤2nYn,k > δ ) < ∞,
Thus by the Borel-Cantelli lemma one has
(1.7) lim sup n→∞ max 1≤k≤2nYn,k ≤ δ with probability 1.
The inequality (1.7) implies uniform continuity of the function t → Xt(ω), t ∈
Q ∩ [0, 1], with probability 1
Corollary 1.1. There exists a probability space (Ω, F , P ) and a continuous set Xt, t ≥ 0, of real variables with the properties (a), (b). In addition the function
t → Xt, t ≥ 0, is continuous on the interval [0, ∞) with probability 1.
Proof. We define the variable Xt for non-rational t by Xt(ω) = lim{Xs(ω); s ∈
Q, s → t}. From Proposition 1.1 this limit exists with probability 1 for all t ≥ 0 and the resulting function t → Xt(ω) is continuous. It is easy to see that (a) and
(b) hold now for all variables Xt, t ≥ 0, since lims→tXs = Xt with probability
1 implies that lims→th f (Xs) i = h f (Xt) i for all bounded measurable functions
f : R → R by the dominated convergence theorem.
We have constructed Brownian motion and in the process have seen that its paths t → Xt, t ≥ 0, are continuous with probability 1. The next result shows that
they are however very irregular.
Proposition 1.2 (Dvoretski, Erd¨os, Kakutani). Brownian paths t → Xt, t ≥ 0,
are nowhere differentiable with probability 1.
Proof. Similarly to Proposition 1.1 let us define random variable Yn,k for k =
1, .., 2n− 2, by (1.8)
Yn,k = sup{|X(k/2n)−X((k−1)/2n)|, |X((k+1)/2n)−X(k/2n)|, |X((k+2)/2n)−X((k+1)/2n)| } .
Then we have by independence of the 3 variables in the definition of Yn,k that
(1.9) P ( Yn,k ≤ a/2n ) = [P (|X(1/2n)| ≤ a/2n)]3 ≤
2a/2n
p2π/2n
!3 . The inequality (1.9) implies that
(1.10) P ( min k Yn,k ≤ a/2 n ) ≤ 2n−2 X k=1 P ( Yn,k ≤ a/2n ) ≤ a32√2 2n/2π3/2 .
Thus by the Borel-Cantelli lemma we have that
(1.11) with probability 1 there exists N (ω) such that for n ≥ N (ω),
Fix β > 0 and suppose that a function f : [0, 1] → R has a derivative at the point s ∈ [0, 1] with |f0(s)| < β. Then there exists δ > 0 such that |f (t) − f (s)| ≤ 2β|t − s| if |t − s| < δ. Assuming now that s ∈ In,k+1and δ > 2/2n, we have that
(1.12)
|f ((k + 1)/2n) − f (k/2n)| ≤ 2β/2n, |f ((k + 2)/2n) − f ((k + 1)/2n)| ≤ 6β/2n,
|f (k/2n) − f ((k − 1)/2n)| ≤ 6β/2n . Choosing a > 6β in (1.11) we conclude that with probability 1 any point s of differentiability of the function t → Xt, 0 ≤ t ≤ 1, has derivative which exceeds β
in absolute value. The result follows by letting β → ∞. It is not difficult to see that Brownian motion X(t), t ≥ 0, satisfies the Markov property i.e. for any Borel set A ⊂ R,
(1.13)
P (X(t) ∈ A | X(t1) = x1, .., X(tm) = xm) = P (X(t) ∈ A | X(tm) = xm),
0 ≤ t1< t2< · · · tm< t.
For this reason Brownian motion is a Markov process. We can seek to establish the strong Markov property for Brownian motion. First for t ≥ 0 let Ft be the
σ−field generated by the variables X(s), 0 ≤ s ≤ t. Note that by the continuity property of Brownian motion Ftis actually generated by a countable set of variables
X(s), 0 ≤ s ≤ t. In fact any countable dense set in [0, t] will suffice. A stopping time τ : Ω → [0, ∞) for Brownian motion is a measurable function such that {τ ≤ t} ∈ Ftfor all t ≥ 0.
Lemma 1.1. Let Xt(ω), t ≥ 0, ω ∈ Ω, be Brownian motion with probability space
(Ω, F , P ) and consider the function from [0, ∞) × Ω → R defined by X(t, ω) = Xt(ω). The function X(·, ·) is measurable with respect to the product σ−algebra
B × F , where B is the σ−field generated by the open sets of [0, ∞).
Proof. Consider the function Fm: [0, 1] × Rm→ R defined by linear interpolation,
so (1.14) Fm(t, x1, ..., xm) = [mt − k] xk+1+[k + 1 − mt] xk, k m ≤ t ≤ k + 1 m , k = 0, .., m−1, where we have set x0= 0. Evidently Fmis a continuous function and hence Borel
measurable on [0, 1] × Rm. For n = 1, 2, .. define X
n: [0, 1] × Ω → R by
(1.15) Xn(t, ω) = Fm(t, X(1/2n, ω), X(2/2n, ω), ..., X(1, ω)), where m = 2n.
The function Xn: [0, 1] × Ω → R is then measurable with respect to B1× F , where
B1is the Borel field generated by open sets of the interval [0, 1]. The measurability
of X : [0, 1] × Ω → R then follows from the fact that (1.16) lim
n→∞Xn(t, ω) = X(t, ω) for all t ∈ [0, 1] with probability 1, ω ∈ Ω.
Corollary 1.2. Let τ : Ω → [0, ∞) be a stopping time for Brownian motion. Then the function ω → X(τ (ω), ω) from Ω to R is measurable.
Proof. Since the mapping ω → (τ (ω), ω) from Ω to [0, ∞) × Ω is obviously
Proposition 1.3 (Strong Markov Property). Suppose X(t), t ≥ 0, is Brownian motion and τ : Ω → [0, ∞) a stopping time. Then the process Y (t) = X(t + τ ) − X(τ ), t ≥ 0, is also a copy of Brownian motion.
Rather than proving Proposition 1.3 we shall concentrate on a particular case. The same method can be applied to prove the proposition. The case we have in mind is the Brownian motion version of the reflection principle, which we first encountered for the standard random walk on Z in Lemma 5.1 of Chapter I. Proposition 1.4 (Reflection Principle). Let Xt, t ≥ 0, be Brownian motion and
Mt= sup0≤s≤tXs be the maximal function. Then for a > 0 there is the identity
P (Mt≥ a) = 2P (Xt≥ a).
Proof. We define a stopping time τ : Ω → [0, ∞) by (1.17) τ = inf{t ≥ 0 : X(t) > a} .
We first show that τ < ∞ with probability 1. We can see this by using a method to solve problem 6 on homework I. Thus observe that for n = 1, 2, ..,
(1.18) X(n) = ξ1+ ξ2+ · · · + ξn
where the ξj are i.i.d standard normal. Hence for any α > 0,
(1.19) P lim sup n→∞ X(n) nα < 1 = 0 or 1
since the LHS of (1.19) is a tail event. Assuming the probability is 1, it follows that if H(·) is the Heaviside function then
(1.20) lim
n→∞E[ H(1 − X(n)/n
α) ] = 1.
Since X(n) is Gaussian with mean 0 and variance n we have that (1.21) E[ H(1 − X(n)/nα) ] = √1
2π
Z nα−1/2
−∞
e−z2/2 dz .
Evidently if α < 1/2 then the RHS of (1.21) converges to 1/2 in contradiction to (1.20). We conclude that the probability in (1.19) is 0, whence τ < ∞ with probability 1.
Next observe that
(1.22) {ω ∈ Ω : τ (ω) ≤ t} = {ω ∈ Ω : Mt(ω) ≥ a} ∈ Ft,
so τ is a stopping time and X(τ (ω), ω) = a, ω ∈ Ω. We have now from (1.17) that that
(1.23) P (Mt≥ a) = P ( τ ≤ t, X(t) − X(τ ) ≥ 0 ) + P ( τ ≤ t, X(t) − X(τ ) < 0 ) .
Since it is clear that
(1.24) P ( τ ≤ t, X(t) − X(τ ) ≥ 0 ) = P (X(t) ≥ a) , we just need to establish that
(1.25) P ( τ ≤ t, X(t) − X(τ ) < 0 ) = P ( τ ≤ t, X(t) − X(τ ) ≥ 0 ) . To see this let us take t = 1 wlog and note from (1.2) that
(1.26) P ( τ ≤ 1, X(1)−X(τ ) < 0 ) = 2n X r=1 P ( (r −1)/2n< τ ≤ r/2n, X(1)−X(τ ) < 0 ) ≤ 2nP ( Y n,1> δ ) + 2n X r=1 P ( (r − 1)/2n < τ ≤ r/2n, X(1) − X(τ ) < 0, Yn,r≤ δ ) ≤ 2nP ( Y n,1> δ ) + 2n X r=1 P ( (r − 1)/2n< τ ≤ r/2n, X(1) − X(r/2n) < 2δ ) ,
where we have used the fact that
(1.27) X(1) − X(r/2n) ≤ X(1) − X(τ )+ |X(τ ) − X((r − 1)/2n)| + |X(r/2n) − X((r − 1)/2n)| ≤ 2Y n,r . Since (1.28) {ω ∈ Ω : (r − 1)/2n < τ (ω) ≤ r/2n} ∈ F r/2n ,
we have from the standard reflection principle for fixed time that (1.29) P ( (r − 1)/2n< τ ≤ r/2n, X(1) − X(r/2n) < 2δ)
= P ( (r − 1)/2n < τ ≤ r/2n, X(1) − X(r/2n) > −2δ) . We conclude then from (1.26)-(1.29) that
(1.30) P ( τ ≤ 1, X(1) − X(τ ) < 0 ) ≤ 2nP ( Yn,1> δ )+ 2n
X
r=1
P ( (r − 1)/2n< τ ≤ r/2n, X(1) − X(r/2n) > −2δ ) .
We can do now an exactly parallel argument as in the previous paragraph to bound the RHS of (1.30) from above in terms of P ( τ ≤ 1, X(1) − X(τ ) > 0 ) and a small error. Thus we have that
(1.31) P ( (r − 1)/2n< τ ≤ r/2n, X(1) − X(r/2n) > −2δ ) ≤
P ( (r − 1)/2n < τ ≤ r/2n, X(1) − X(τ ) > −4δ, Yn,r ≤ δ ) + P ( Yn,1> δ ) .
Evidently (1.30), (1.31) imply that (1.32)
P ( τ ≤ 1, X(1)−X(τ ) < 0 ) ≤ P ( τ ≤ 1, X(1)−X(τ ) > −4δ )+2n+1P ( Yn,1> δ ) .
Letting n → ∞ in (1.32) and using (1.3) with p = 4 we conclude that
(1.33) P ( τ ≤ 1, X(1) − X(τ ) < 0 ) ≤ P ( τ ≤ 1, X(1) − X(τ ) > −4δ ) for every δ > 0. Hence by letting δ → 0 in (1.33) we see that the LHS of (1.25) does not exceed the RHS. A symmetry argument then implies the identity (1.25). Next we show how expectations for Brownian motion can be obtained by solving differential equations. First we see that the analogue of the backward Kolmogorov
equation for the countable state space Markov chain is the heat equation. Thus let T > 0 and u(x, t), be defined by
(1.34) u(x, t) = E[ f (XT) | Xt= x] = 1 p2π(T − t) Z ∞ −∞ exp −(x − y) 2 2(T − t) f (y) dy . Then (1.34) is stating that the variable XT conditioned on Xt = x is Gaussian
with mean x and variance T − t. One easily sees that u(x, t) is the solution to the terminal value problem,
(1.35) ut +
1
2uxx = 0, x ∈ R, t < T ; t→Tlimu(x, t) = f (x), x ∈ R.
Expectations for functions of stopping times can sometimes be computed as solutions to time independent differential equations. We illustrate how this happens with an example:
Proposition 1.5. Suppose a < b and λ ≥ 0. Let uλ: (a, b) → R be the function
(1.36) uλ(x) = E[ e−λτ | X(0) = x ],
where τ is the first exit time for Brownian motion X(t), t ≥ 0, started at x ∈ (a, b) from the interval [a, b], whence X(τ ) = a or X(τ ) = b. Then uλ(·) is the unique
solution to the differential equation
(1.37) 1
2
d2uλ(x)
dx2 = λuλ(x), a < x < b,
with boundary conditions
(1.38) uλ(a) = uλ(b) = 1.
Proof. Defining uλ(x) for a < x < b by (1.36), we first prove that
(1.39) lim
x→auλ(x) = 1, x→blimuλ(x) = 1 .
This follows from the inequality
(1.40) 1 ≥ uλ(x) ≥ e−λtP (Mt> b − x), a < x < b.
From Proposition 1.4 we have that
(1.41) P (Mt> a) = 2P ( Z > a/
√
t ) , a > 0,
where Z is standard normal. Thus the limit of the RHS of (1.40) as x → b is e−λt. Now on letting t → 0 we obtain (1.39).
The key to proving that the equation (1.37) holds is to observe that
(1.42) E[ e−λτ H(τ − t) | X(0) = x ] = e−λtE[ uλ(X(t)) H(τ − t) | X(0) = x ],
where H(·) is the Heaviside function. This follows since the function H(τ − t) is Ft
measurable. Thus (1.43)
E[ e−λτ H(τ − t) | Ft] = e−λtH(τ − t)E[ e−λ(τ −t)|X(t) ] = H(τ − t)uλ(X(t)) .
Hence we have that (1.44) uλ(x) = e−λt √ 2πt Z ∞ −∞ exp −(x − y) 2 2t uλ(y) dy + Error(t).
In (1.44) we have extended the function uλ(·) by 1 outside of the interval (a, b).
The error term is bounded by
(1.45) |Error(t)| ≤ 2P ( τ ≤ t | X(0) = x ) ≤ 4P (Z > min{x − a, b − x}/√t ) , where Z is standard normal.
We obtain the equation (1.37) by dividing (1.44) by t and letting t → 0. It is easy to see that limt→0Error(t)/t = 0, whence we have that
(1.46) lim
t→0uλ(x) − e
−λtE[ u
λ(X(t)) | X(0 = x ] /t = 0.
The limit in (1.46) yields the equation (1.37) if we assume uλ(y) is C2 for y in a
neighborhood of x, by expanding uλ(X(t)) in a Taylor series about x to second
order. Of course the definition (1.36) of uλ(·) allows us to conclude only that it is
a continuous function from the arguments we have been using. There is a gap here in our argument, which can really only be filled by going further into the theory of differential equations. We therefore omit it and leave the proof at this point. It is easy to see that the solution to the boundary value problem (1.37), (1.38) is given by (1.47) uλ(x) = cosh√2λ(x − c) cosh√2λR , c = a + b 2 , R = b − a 2 ,
so c is the center and R is the radius of the interval (a, b). Suppose now we consider the semi-infinite interval (−∞, b) with b > 0. In that case it is easy to see that
(1.48) uλ(x) = exp[−
√
2λ(b − x) ], x < b.
Assuming the exit time τ from the interval (−∞, b) conditioned on X(0) = x, has a density ρx(s), s > 0, with respect to Lebesgue measure, then (1.48) yields the
formula (1.49) Z ∞ 0 e−λtρx(s) ds = exp[− √ 2λ(b − x) ] .
This is simply an explicit formula for the Laplace transform of the function ρx(s), s >
0. We compare this with the formula given by the reflection principle Proposition 1.4. Thus P ( τ ≤ t | X(0) = 0 ) = 2P ( X(t) > b | X(0) = 0 ) (1.50) Z t 0 ρ0(s) ds = 2 √ 2π Z ∞ b/√t e−z2/2 dz .
Differentiating the second equation in (1.50) with respect to t, we see that
(1.51) ρ0(t) = b √ 2πt3/2exp −b 2 2t .
On comparing (1.49) and (1.51) we conclude on setting b = 1/√2 that (1.52) Laplace transform of 1 2√πt3/2exp −1 4t = exp[−√λ] .
One can of course verify (1.52) directly, but it is of some interest that the identity is a consequence of a symmetry property of Brownian motion i.e. the reflection principle.
2. Gaussian Processes
We have so far in the study of Brownian motion noted that it is a Markov process. We shall see in this section that it is also a Gaussian Process. To see what this means we need to review some basic facts about Gaussian random variables. Let (Ω, F , P ) be a probability space and Y : Ω → Rn be a random variable with mean
h Y (·) i = 0. We shall think of Y (ω) ∈ Rn as a column vector. Then Y is Gaussian
if its pdf is determined by its covariant matrix Γ, where Γ is a positive definite symmetric n × n matrix defined by
(2.1) v∗Γv = h [v∗Y (·)]2 i , v ∈ Rn.
The characteristic function χY(·) for Y is then given by the formula
(2.2) χY(σ) = E[ eiσ ∗Y (·) ] = exp −1 2σ ∗Γσ, σ ∈ Rn.
The probability measure associated with (2.2) is given by
(2.3) exp −1 2φ ∗Γ−1φ n Y j=1 dφ(j)normalization .
The normalization in (2.3) can be computed explicitly as (2.4) normalization = (2π)n/2(det Γ)1/2 ,
but we wish to de-emphasize this since the normalization in (2.3) is determined by the fact that the measure (2.3) is a probability measure.
Since the Gaussian variable Y is determined by the symmetric positive definite covariance matrix Γ, its structure can be understood from the structure of the matrix Γ, in particular the fact that Γ has a basis of orthogonal eigenvectors, all with positive eigenvalues. Suppose the eigenvectors are an orthonormal set v1, ..., vn ∈ Rn, i.e. orthogonal and with Euclidean norm |vj| = 1, j = 1, .., n.
Then the n × n matrix O = [v1, .., vn] is orthogonal so that
(2.5) OO∗ = O∗O = identity .
If the eigenvalues of the vj are λj, j = 1, .., n, then we can form the diagonal n × n
matrix Λ, (2.6) Λ = λ1 0 · · · 0 λ2 · · · · · · and one has the identity
(2.7) Γ = OΛO∗.
Observe now that since the vj, j = 1, ., n, are an orthonormal basis for Rn that
(2.8) Y (ω) = n X j=1 (Y (ω), vj)vj = n X j=1 pλj ξj(ω)vj , ω ∈ Ω,
where (·, ·) denotes the Euclidean inner product on Rn. Thus (2.8) defines the set
standard normal. This follows from (2.2). Thus let ξ(ω) = (ξ1(ω), .., ξn(ω)) ∈ Rn
and observe from (2.8) that for any σ ∈ Rn,
(2.9) σ∗ξ(ω) = n X j=1 σj pλj (Y (ω), vj) = (Y (ω), w) , where (2.10) w = n X j=1 σj pλj vj .
Noting now that
(2.11) Γw = n X j=1 σjpλj vj , we conclude that (2.12) χξ(σ) = E[ eiσ ∗ξ(·) ] = exp −1 2w ∗Γw = exp −1 2|σ| 2 , σ ∈ Rn, whence the ξj, j = 1, .., n, are i.i.d. standard normal.
We wish now to generalize the above considerations to Gaussian processes, by which we mean a continuous set of random variables Y (t), t ∈ R, such that any finite set of them (Y (t1), .., Y (tn)) have joint distribution which is Gaussian. Its
covariance “matrix” is now a function Γ(t, s) defined by (2.13) Γ(t, s) = h Y (t)Y (s) i s, t ∈ R .
We have already seen that Brownian motion is a Gaussian process, and its covari-ance is given by
(2.14) Γ(t, s) = h X(t)X(s) i = min[s, t] .
We would like to obtain an infinite dimensional version of (2.8) and write Brownian motion as a sum of i.i.d. standard normal variables. Thus we wish to write for some suitable functions aj(·), j = 0, 1, ..,
(2.15) X(t) =
∞
X
j=0
aj(t)ξj
where the ξj, j = 0, 1, 2, .., are i.i.d. standard normal.
Let us try to find such a representation at least for t in a finite interval, say the interval [0, π]. If we were to follow the method above for Gaussian variables Y ∈ Rn, we would look to find the eigenfunctions of the self-adjoint operator Γ
defined by
(2.16) Γf (t) =
Z π
0
Γ(t, s)f (s) ds , 0 ≤ t ≤ π.
It is known from the theory of integral equations that the operator (2.16) is compact on L2([0, π]), and this implies that there is an orthonormal basis for L2([0, π])
by vj(t), 0 ≤ t ≤ π, j = 0, 1, 2, ..., with corresponding non-negative eigenvalues
λj, j = 0, 1, 2, .. we expect following (2.8) the identity
(2.17) X(t, ω) =
∞
X
j=0
pλj ξj(ω)vj(t), ω ∈ Ω, 0 ≤ t ≤ π.
We cannot however find the eigenfunctions in (2.17) explicitly. To get an explicit representation we need to use the fact that Brownian motion is the derivative of the white noise Gaussian process W (t), t ∈ R. This process is defined as the distributional derivative of Brownian motion, so
(2.18) Z ∞ −∞ f (s)W (s) ds = Z ∞ −∞ f0(s)X(s) ds
for all C∞functions f : R → R with compact support. Formally W (t) = dX(t)/dt or in other words white noise paths are the infinitesimal time increments for Brow-nian motion. We have already seen that BrowBrow-nian paths are differentiable nowhere with probability 1, which is why we must define white noise as a distribution. The covariance Γ of (2.13) for white noise is Γ(t, s) = δ(t − s), where δ(·) is the Dirac delta function. Thus the corresponding integral operator (2.16) is simply the identity. To see this observe from (2.13) that if X(·) denotes Brownian motion then (2.19) Z ∞ −∞ g(t)Γ(t, s)f (s) dt ds = h [X(s2) − X(s1)][X(t2) − X(t1)] i = Z ∞ −∞ g(t)f (t) dt if g(·) is the characteristic function of the interval [s1, s2] and f (·) is the
character-istic function of the interval [t1, t2].
Now any orthonormal basis of L2([0, π]) is an orthonormal basis of eigenfunctions for the identity operator Γ on L2([0, π]) with corresponding eigenvalues 1. Choosing
the basis consisting of the functions (2.20) v0(t) = 1 √ π, vj(t) = r 2 πcos jt, j = 1, 2, ..., 0 ≤ t ≤ π, we conclude from (2.17) that
(2.21) W (t) = √ξ0 π+ r 2 π ∞ X j=1 ξjcos jt, 0 ≤ t ≤ π,
where the ξj, j = 0, 1, .., are i.i.d. standard normal. We have not rigorously
proven (2.21) of course, and in fact we have been rather vague about what we mean about the white noise process W (·). Let us continue however and formally integrate equation (2.21) from 0 to t, observing now that Brownian motion is the integral of white noise. Thus we obtain the representation
(2.22) X(t) = √tξ0 π+ r 2 π ∞ X j=1 ξj sin jt j , 0 ≤ t ≤ π,
for Brownian motion. The formula (2.22) is known as the Polya-Wiener represen-tation for Brownian motion. One might ask what is the advantage of (2.22) over
the representation
(2.23) X(t) = √∆t [ξ0+ ξ1+ · · · + ξm] , t = (m + 1)∆t, m = 0, 1, ...,
where the ξj, j = 0, 1, .., are i.i.d. standard normal. One advantage is that the
partial sums in (2.22) converge in L2(Ω) uniformly in t for 0 ≤ t ≤ 1 since
(2.24) h ∞ X j=N +1 ξj sin jt j 2 i = ∞ X j=N +1 sin jt j 2 ≤ C N ,
for some constant C. Hence representations like (2.22) can be efficient ways of generating Brownian motion starting with i.i.d. standard normal variables.
We have so far defined two Gaussian processes-Brownian motion and its deriva-tive the white noise process. We shall define other Gaussian processes by considering the infinite dimensional version of the formula (2.3) for the probability density. Con-sider Brownian motion φ(t), t ≥ 0, for which we know that if 0 < t1< t2< · · · tn,
the joint pdf for the variables φ(tj), j = 1, .., n, is given by
(2.25) exp − n X j=1 {φ(tj) − φ(tj−1)}2 2(tj− tj−1) n Y j=1 dφ(tj)normalization,
where t0= 0. We can rewrite the sum in (2.25) as
(2.26) n X j=1 {φ(tj) − φ(tj−1)}2 2(tj− tj−1) = Z tn 0 dφ(t) dt 2 dt
where [t, φ(t)], 0 ≤ t ≤ tn, is the graph obtained by linear interpolation of the
points [0, 0] and [tj, φ(tj)], j = 1, .., n. Thus we might be tempted to write the
Brownian motion measure as
(2.27) exp " −1 2 Z ∞ 0 dφ(t) dt 2 dt # Y t>0 dφ(t)normalization .
Comparing (2.27) with (2.3) we expect that if Γ is the covariance operator for Brownian motion given by (2.14), then
(2.28) [φ(·), Γ−1φ(·)] = Z ∞ 0 dφ(t) dt 2 dt, where (2.29) [f (·), g(·)] = Z ∞ 0 f (t)g(t) dt ,
for functions f, g ∈ L2([0, ∞)). We conclude that for Brownian motion
(2.30) Γ−1g(t) = −d
2
dt2g(t) , 0 ≤ t < ∞.
We can verify this intuition directly by observing from (2.14) that
(2.31) Γf (t) = Z t 0 sf (s) ds + t Z ∞ t f (s) ds . Differentiating (2.31) twice we see that
(2.32) −d
2
and hence Γ−1 is given by (2.30).
It is clear now how we can naturally generalize Brownian motion by considering Gaussian processes for which Γ−1 is a second order positive definite differential operator. Differential operators come together with boundary conditions so we need to be a little careful here. Thus Γ−1 for Brownian motion is actually defined on functions f (t), t ≥ 0, with the Dirichlet boundary condition f (0) = 0. Thus (2.33) [f (·), Γ−1g(·)] = − Z ∞ 0 f (t)g00(t) dt = Z ∞ 0 f0(t)g0(t) dt ,
where we have used the boundary condition f (0) = 0 to integrate by parts. It follows from the second integral formula on the RHS of (2.33) that Γ−1 is positive
definite. Consider now Γ−1 defined on functions f : [0, 1] → R with Dirichlet
boundary conditions f (0) = f (1) = 0 by (2.34) [f (·), Γ−1g(·)] = − Z 1 0 f (t)g00(t) dt = Z 1 0 f0(t)g0(t) dt ,
whence Γ−1 is positive definite. The Gaussian process associated with (2.34) is called the Brownian bridge. One can easily see from (2.34) that
Γ(t, s) = s(1 − t) if s < t, (2.35)
Γ(t, s) = t(1 − s) if s > t .
A realization of the Brownian bridge process α(t), 0 ≤ t ≤ 1, can be given in terms of Brownian motion X(t), t ≥ 0, by the formula α(t) = X(t) − tX(1). To see this all we need to do is to verify from (2.14) that the covariance of α(t), 0 ≤ t ≤ 1, is given by (2.35). Note that α(t), 0 ≤ t ≤ 1, is not Markovian.
Next consider Γ−1 defined on functions f : (−∞, ∞) → R by (2.36) [f (·), Γ−1g(·)] = Z ∞ −∞ f (t){−g00(t) + g(t)} dt = Z ∞ −∞ {f0(t)g0(t) + f (t)g(t)} dt , whence Γ−1 is positive definite. It seems that we are not imposing any boundary
conditions on the functions f (·), g(·), but in the integration by parts we are im-plicitly assuming Dirichlet boundary condition at t = ±∞ i.e. limt→±∞f (t) = 0.
The Gaussian process Y (t), t ∈ R, associated with (2.36) is called the Ornstein-Uhlenbeck process. One sees from (2.36) that
(2.37) Γ(t, s) = 1
2e
−|t−s| .
The process Y (t), t ∈ R, is Markovian and can be represented by Brownian motion X(s), s ≥ 0, as (2.38) Y (t) = Y (0)e−t/2+√1 2X(t) − 1 2√2 Z t 0 e(s−t)/2X(s) ds, t > 0. In (2.38) the variable Y (0) is independent of the Brownian motion. Taking Y (0) to be Gaussian with mean 0 and variance 1/2, we see from (2.14) that h Y (t)Y (s) i is given by the RHS of (2.37) for s, t ≥ 0. In particular Y (t) is Gaussian with mean 0 and variance 1/2 for all t ≥ 0. Hence the Markov process Y (t) has an invariant measure which is the Gaussian variable with mean zero and variance 1/2.
University of Michigan, Department of Mathematics, Ann Arbor, MI 48109-1109 E-mail address: [email protected]
