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chapter 6

Applications of Stochastic Integration

to Brownian Motion

This chapter contains a highly incomplete selection of ways in which Ito’stheory of stochastic integration, especially his formula, has contributed toour understanding of Brownian motion. For a much more complete selection,see Revuz and Yor’s book [27].

6.1 Tanaka’s Formula for Local Time

Perhaps the single most beautiful application of Ito’s formula was madeby H. Tanaka when (as reported in [22]) he applied it to prove the existenceof local time for one-dimensional Brownian motion.

Before one can understand Tanaka’s idea, it is necessary to know whatthe preceding terminology means. Thus, let

(β(t),Ft,P

)be a 1-dimensional

Brownian motion. What we are seeking is a function ` : [0,∞)×R×Ω 7−→[0,∞) with the properties that

(a) For each y ∈ R, (t, ω) `(t, y, ω) is progressively measurable.(b) For each (y, ω) ∈ R×Ω, `(0, y, ω) = 0 and `( · , y, ω) is non-decreasing.(c) For each ω ∈ Ω, (t, y) `(t, y, ω) is continuous,(d) For P-almost every ω ∈ Ω and all t ∈ [0,∞),∫ t

0

1[a,b)

(β(τ, ω)

)dτ =

∫ b

a

`(t, y, ω) dy, for all −∞ < a < b <∞.

Alternatively, for each (t, ω) ∈ (0,∞) × Ω, define the Brownian motion,occupation time distribution µt,ω to be the Borel measure on R determinedby the rule that µt,ω(Γ) be the amount of time β( · , ω) [0, t] spends in Γ.Then `(t, · , ω) is to be the Radon–Nikodym derivative of µt,ω with respectto Lebesgue measure.

For each y ∈ R, (t, ω) `(t, y, ω) is known as the local time functionalat y of the Brownian path β( · , ω). P. Levy was the first person to discuss

151

152 6 Applications to Brownian Motion

this functional, although H. Trotter is usually credited for providing the firstrigorous proof of its existence.6.1.1. Tanaka’s Construction. The idea underlying Tanaka’s proof is tofirst write the somewhat fanciful expression

“`(t, y) =∫ t

0

δ(y − β(τ)

)dτ, ”

where δ is Dirac’s delta function. Although it seems doubtful that suchan expression should be taken seriously, Tanaka realized that Ito’s formulacan be used to rationalize it. Namely, apply Ito’s formula to the functionF (x) = x ∨ y, remember that F ′(x) = 1[y,∞)(x) and F ′′(x) = δ(y − x), andcome to the conclusion that

“12

∫ t

0

δ(y − β(τ)

)dτ = β(t) ∨ y − 0 ∨ y −

∫ t

0

1[y,∞)

(β(τ)

)dβ(τ).”

With the preceding heuristic argument in mind, we now define

(6.1.1) ˜(t, y) = 2(β(t) ∨ y − 0 ∨ y −

∫ t

0

1[y,∞)

(β(τ)

)dβ(τ)

).

When reading (6.1.1), it is important to keep in mind that it gives a definitionof (t, ω) ˜(t, y, ω) for each y ∈ R separately. However, because, for eachy, the definition has an ambiguity on a P-null set, it does not define (t, ω) ˜(t, y, ω) simultaneously for all y ∈ R, and that is one of the reasons why wehave used the notation ˜ instead of `. In any case, we already know that,for each y and P-almost every ω, ˜( · , y, ω) is a continuous function whichis 0 at time 0. We next want to show that it is, P-almost surely, a non-decreasing function. For this purpose, choose a ψ ∈ C∞c

(R; [0,∞)

)which is

supported in [−1, 1] and has total integral 1, and take Fε(x) = ψε ?F , whereψε(x) = ε−1ψ(ε−1x). By Ito’s formula,

Fε

(β(t)

)− Fε(0)−

∫ t

0

F ′ε(β(τ)

)dβ(τ) =

12

∫ t

0

F ′′ε(β(τ)

)dτ.

Hence, since F ′′ε ≥ 0, the left hand side is P-almost surely non-decreasing asa function of t ≥ 0. Furthermore, Fε −→ F uniformly on compacts, while∣∣F ′ε − 1[y,∞)

∣∣ ≤ 1[y−ε,y+ε] and therefore

EP

[∥∥∥∥∫ ·0

F ′ε(β(τ)

)dβ(τ)−

∫ ·0

1[y,∞)

(β(τ)

)dβ(τ)

∥∥∥∥2

[0,t]

]

≤ 4EP[∫ t

0

∣∣F ′ε(β(τ)− 1[y,∞)(β(τ))∣∣2 dτ]

≤ 4∫ t

0

1√2πτ

(∫ y+ε

y−ε

e−ξ2

2τ dξ

)dτ ≤ 8ε

√t

π.

6.1 Tanaka’s Formula for Local Time. 153

Hence, as ε 0,∥∥∥∫ ·0 F ′′ε

(β(τ)

)dτ − ˜( · , y)

∥∥∥[0,t]

−→ 0 in P-measure, and so

we now know that, for each y ∈ R, ˜( · , y) is P-almost surely a continuous,non-decreasing function.

The next step is to use Kolmogorov’s continuity criterion (cf. Exercise2.4.17) to prove that, for each y, ˜( · , y) can be modified on a P-null set insuch a way that the resulting function `( · , y) is, P-almost surely, a continuousfunction of y. Namely, we will show that, for each T ∈ (0,∞), there existsa C(T ) <∞ for which

EP[∥∥˜( · , y2)− ˜( · , y1)

∥∥4

[0,T ]

]≤ C(T )(y2 − y1)2 when y1 ≤ y2.

To check this, first observe that∣∣β(t) ∨ y2 − β(t) ∨ y1

∣∣ ≤ |y2 − y1|. Thus,what still must be checked is that

EP

[∥∥∥∥∫ ·0

1[y2,∞)

(β(τ)

)dβ(τ)−

∫ ·0

1[y1,∞)

(β(τ)

)dβ(τ)

∥∥∥∥4

[0,t]

]

= EP

[∥∥∥∥∫ ·0

1[y1,y2)

(β(τ)

)dβ(τ)

∥∥∥∥4

[0,T ]

]≤ C(T )(y2 − y1)2.

But, by Doob’s Inequality,

EP

[∥∥∥∥∫ ·0

1[y1,y2)

(β(τ)

)dβ(τ)

∥∥∥∥4

[0,T ]

]

≤(

43

)4

EP

(∫ T

0

1[y1,y2)

(β(τ)

)dβ(τ)

)4 ,

and, by Ito’s formula,

(∫ T

0

1[y1,y2)

(β(τ)

)dβ(τ)

)4

= 4∫ T

0

1[y1,y2)

(β(t)

)(∫ t

0

1[y1,y2)

(β(τ)

)dβ(τ)

)3

dβ(t)

+ 6∫ T

0

1[y1,y2)

(β(t)

)(∫ t

0

1[y1,y2)

(β(τ)

)dβ(τ)

)2

dt.


Hence,

EP

[∥∥∥∥∫ ·0

1[y1,y2)

(β(τ)

)dβ(τ)

∥∥∥∥4

[0,T ]

]

≤ 6(

43

)4

EP

[∫ T

0

1[y1,y2)

(β(t)

)(∫ t

0

1[y1,y2)

(β(τ)

)dβ(τ)

)2

dt

]

≤ 6(

43

)4

EP

[∫ T

0

1[y1,y2)

(β(t)

)dt

∥∥∥∥∫ ·0

1[y1,y2)

(β(τ)

)dβ(τ)

∥∥∥∥2

[0,T ]

]

≤ 6(

43

)4

EP

(∫ T

0

1[y1,y2)

(β(t)

)dt

)2 1

2

× EP

[∥∥∥∥∫ ·0

1[y1,y2)

(β(τ)

)dβ(τ)

∥∥∥∥4

[0,T ]

] 12

,

and so

EP

[∥∥∥∥∫ ·0

1[y1,y2)

(β(τ)

)dβ(τ)

∥∥∥∥4

[0,T ]

]≤ KEP

(∫ T

0

1[y1,y2)

(β(t)

)dt

)2 ,

where K = 36(

43

)8. Finally, write the square of the integral over [0, T ] astwice the integral over the simplex 0 ≤ τ ≤ t ≤ T , and note that

EP

[∫ T

0

1[y1,y2)

(β(t)

)(∫ t

0

1[y1,y2)

(β(τ)

)dτ

)dt

]

=12π

∫∫0<τ<t≤T

1√τ(t− τ)

∫∫[y1,y2)2

e−ξ21

2τ e−(ξ2−ξ1)2

2(t−τ) dξ1dξ2

dτdt

≤ (y2 − y1)2

2π

∫∫0<τ<t≤T

1√τ(t− τ)

dτdt.

Hence, by Kolmogorov’s criterion, there exists each ω ∈ Ω a continuousy ∈ R 7−→ `( · , y, ω) ∈ C

([0,∞); R

)such that `( · , y) = ˜( · , y) P-almost

surely for each y ∈ R. In fact, after further adjustment on a P-null set, wemay and will assume that, for all ω ∈ Ω, (t, y) `(t, y, ω) is continuous,and, for all (y, ω), `(0, y, ω) = 0 and `( · , y, ω) is non-decreasing.

It remains to show that ` satisfies (d). To this end, let ϕ ∈ C∞c (R; R)be given, and set Φ(x) =

∫R x ∨ yϕ(y) dy. Then Φ′′ = ϕ, and so, by the


Mean Value Theorem, we can choose ξNm ∈

(m2−N , (m + 1)2−N

)so that

Φ′((m+ 1)2−N

)− Φ′(m2−N ) = 2−Nϕ(ξN

m). Thus

12

∫Rϕ(y)`(t, y) dy = lim

N→∞2−N

∑m∈Z

ϕ(ξNm)`(t,m2−N )

= Φ(β(t)

)− Φ(0)− lim

N→∞

∑m∈Z

∫ t

0

Φ′(m2−N )1[(m−1)2−N ,m2−N )

(β(τ)

)dβ(τ)

= Φ(β(t)

)− Φ(0)−

∫ t

0

Φ′(β(τ)

)dβ(τ)

since∫ t

0

Φ′(β(τ)

)dβ(τ)−

∑m∈Z

∫ t

0

Φ′(m2−N )1[(m−1)2−N ,m2−N )

(β(τ)

)dβ(τ)

=∫ t

0

(Φ′(β(τ)

)− Φ′

([β(τ)

]N

))dβ(τ)

and

EP

[(∫ t

0

(Φ′(β(τ)

)− Φ′

([β(τ)]N

))dβ(τ)

)2]

= EP[∫ t

0

(Φ′(β(τ)

)− Φ′

([β(τ)]N

))2

dτ

]−→ 0

as N →∞. Thus, we now know that

12

∫Rϕ(y)`(t, y) dy = Φ

(β(t)

)− Φ(0)−

∫ t

0

Φ′(β(τ)

)dβ(τ)

P-almost surely. At the same time, by Ito’s formula, the right hand sideof the preceding is also P-almost surely equal to 1

2

∫ t

0ϕ(β(τ)

)dτ , and so we

also know that, for each ϕ ∈ C∞c (R; R)∫Rϕ(y)`(t, y, ω) dy =

12

∫ t

0

ϕ(β(τ, ω)

)dτ =

12

∫Rϕ(y)µt,ω(dy), t ≥ 0,

for all P-almost every ω. Starting from this, it an elementary exercise inmeasure theory to see that ` has the property in (d).

We summarize our progress in the following theorem, in which the con-cluding equality is Tanaka’s formula.


6.1.2 Theorem. Given a one dimensional Brownian motion(β(t),Ft,P

)there exists a P-almost surely unique function ` : [0,∞)× R× Ω −→ [0,∞)with the properties described in (a)–(d) above. Moreover, for each y ∈ R,

(6.1.3)`(t, y)

2= β(t) ∨ y − 0 ∨ y −

∫ t

0

1[y,∞)

(β(τ)

)dβ(τ), t ∈ [0,∞)

P-almost surely.

Remark 6.1.4. It is important to realize that the very existence of `(t, · ) isa reflection of just how fuzzy the graph of a Brownian path must be. Indeed,if p : [0, t] −→ R is a smooth path, then its occupation time distribution µt

will usually have a non-trivial part which is singular to Lebesgue measure.For example, if the derivative p of p vanishes at some τ ∈ [0, t], then it is easyto see that this prevents µt from being absolutely continuous with respect toLebesgue measure. Moreover, even if p never vanishes on [0, t], and thereforeµt is absolutely continuous, the Radon–Nikodym of µt will have discontinu-ities at both min[0,t] p(τ) and max[0,t] p(τ). The way that Brownian pathsavoid these problems is that they are changing directions so fast that they,as distinguished from a well-behaved path, never slow down. Of course, onemight think that their speed could cause other problems. Namely, if theyare moving so fast, how do they spend enough time anywhere to have theiroccupancy time recordable on a Lebesgue scale? The explanation is in thefuzz alluded to above. That is, although a Brownian path exits small inter-vals too fast to have their presence recorded on a Lebesgue scale, it does notdepart when it exits but, instead, dithers back and forth.

From a more analytic point of view, the existence of the Brownian localtime functional is a stochastic manifestation of the fact that points have pos-itive capacity in one dimension. This connection is an inherent ingredientin Tanaka’s derivation. In fact, his argument works only because the thefundamental solution, x∨y, is locally bounded, which is also the reason whypoints have positive capacity in one dimension. In two and more dimen-sions, the fundamental solution is too singularity for points to have positivecapacity.

6.1.2. Some Properties of Local Time. Having constructed `, we nowwant to derive a couple of its elementary properties. To begin with, supposethat f : [0,∞) × R × Ω −→ R is a bounded, measurable function with theproperties that (t, y) f(t, y, ω) is continuous for P-almost every ω. UsingRiemann sum approximations, one can easily see that, for P-almost every ω,

(6.1.5)∫ t

0

f(τ, β(τ, ω), ω

)dτ =

∫R

(∫ t

0

f(τ, y, ω)`(dτ, y, ω))dy, t ≥ 0,


where `(dτ, y, ω) denotes integration with respect to the measure on [0,∞)determined by the non-decreasing function `( · , y, ω). In particular, if η ∈C∞

(R; [0, 1]

)vanishes on [−1, 1] and equals 1 off (2, 2), ψ ∈ C∞c

(R; [0,∞)

)vanishes off [−1, 1] and has total integral one, and

fε,R(t, y, ω) = ε−1η(R(β(t, ω)− y0

))ψ(ε−1(y − y0)

),

then, by (6.1.5), P-almost surely,

ε−1

∫Rψ(ε−1(y − y0)

)(∫ t

0

η(R(β(τ)− y0

))`(dτ, y)

)dy = 0

whenever 0 < ε < R−1. Hence, after first letting ε 0 and then R ∞,we see that, for each y ∈ R,

(6.1.6) `(τ ≥ 0 : β(τ) 6= y, y

)= 0 P-almost surely.

An interesting consequence of (6.1.6) is that, for each y, `( · , y) is P-almostsurely a singular, continuous, non-decreasing function. Indeed, `( · , y) isP-almost surely supported on τ : β(τ) = y, whereas

EP[Leb

(τ : β(τ) = y

)]= EP

[∫ ∞0

1y(β(τ)

)dτ

]=∫ ∞

0

P(β(τ) = y

)dτ = 0.

In order to take the next step, notice that, from (6.1.3),

β(t) ∧ 0 = β(t)− β(t) ∨ 0 =∫ t

0

1(−∞,0)

(β(τ)

)dβ(τ)− `(t, 0)

2.

Hence, after subtracting this from (6.1.3), we find that

(6.1.7)∣∣β(t)

∣∣ = ∫ t

0

sgn(β(τ)

)dβ(τ) + `(t, 0),

where, for the sake of definiteness1 we will take sgn(0) = 1.Among the many interesting consequences of (6.1.7) are the two discussed

in the next statement, both of which were discovered by P. Levy. For muchmore information about local time, the reader should consult the books [15]and [27].

1 It really does not matter how one takes sgn at 0 since, P-almost surely, β spends no time

there.


6.1.8 Theorem. With P-probability 1, `(t, 0) > 0 for all t > 0. Further-

more, if, for t > 0 and ε > 0, Nε(t) denotes the number of times |β| [0, t]downcrosses2 the interval [0, ε], then

(6.1.9) limε0

∥∥εNε( · )− `( · , 0)∥∥

[0,t]= 0 P-almost surely for each t > 0.

Proof: To prove the first statement, set B(t) =∫ t

0sgn(β(τ)

)dβ(τ). In

view of (6.1.7), it suffices for us to prove that

(*) P(∀t > 0 ∃τ ∈ (0, t] B(τ) < 0

)= 1.

For this purpose, we will first show that(B(t),Ft,P

)is a Brownian motion.

To do this, apply Ito’s formula to see that(exp

(λB(t)− λ2t

2

),Ft,P

)is a martingale for every λ ∈ C. In particular, this means that, for all0 ≤ s < t,

EP[e√−1 ξ(B(t)−B(s))∣∣Fs

]= e−

ξ2(t−s)2 P-almost surely,

which, together with B(0) = 0, is enough to identify the distribution ofω B( · , ω) as that of a Brownian motion.

Returning to (*), there are several ways in which it can be approached.One way is to apply Blumenthal’s 0–1 Law (cf. Theorem 8.1.20 in [36])which says that, because B is a Brownian motion under P, any set from⋂

t>0 σ(B(τ) : τ ∈ [0, t]

)has P-probability 0 or 1. In particular, this

means thatP(∀ t > 0 ∃τ ∈ [0, t] B(τ) < 0

)∈ 0, 1

Hence, since

P(∀ t > 0 ∃τ ∈ [0, t] B(τ) < 0

)= lim

t0P(∃τ ∈ [0, t] B(τ) < 0

)≥ lim

t0P(B(t) < 0

)=

12,

(*) follows. Another, less abstract approach is to use

P(∃τ ∈ [0, t] B(τ) < 0

)≥ lim

δ0P(∃τ ∈ [δ, t] B(τ) < 0

)= lim

δ0

1√2πδ

∫Re−

x22δ P(∃ τ ∈ [0, t− δ] B(τ) < −x

)dx

=12

+ limδ0

1√2πδ

∫ ∞0

e−x22δ 2P

(B(t− δ) ≤ −x

)dx = 1,

2 That is, Nε(t) ≥ m if and only if there exist 0 ≤ τ1 < τ2 < · · · < τ2m ≤ t such that

|β(τ2`−1)| ≥ ε and β(τ2`) = 0 for 1 ≤ ` ≤ m.


where, in the passage to the last line, we used the reflection principle (cf.(4.3.5) in [36]) when we replaced P

(∃ τ ∈ [0, t− δ] B(τ) ≤ −x

)by 2P

(B(t−

δ) ≤ −x)

for x < 0.In order to prove (6.1.9), we introduce the stopping times ζ` : ` ≥ 0 so

that ζ0 ≡ 0, ζ2`+1 = inft ≥ ζ2` : |β(t)| ≥ ε for ` ≥ 0, and ζ2` = inft ≥ζ2`−1 : β(t) = 0 for ` ≥ 1. Next, set I` = [ζ2`−1, ζ2`), and observe that

∞∑`=1

(∣∣β(t ∧ ζ2`

)∣∣− ∣∣β(t ∧ ζ2`−1

)∣∣) = −εNε(t) +(∣∣β(t)

∣∣− ε) ∞∑

`=1

1I`(t).

At the same time, because β(τ) 6= 0 when τ ∈ I` and therefore `(t ∧ ζ2`

)=

`(t ∧ ζ2`−1

), (6.1.7) tells us that

∞∑`=1

(∣∣β(t ∧ ζ2`

)∣∣− ∣∣β(t ∧ ζ2`−1

)∣∣) =∞∑

`=1

∫ t

0

1I`(τ)sgn

(β(τ)

)dβ(τ)

=∣∣β(t)

∣∣− ∞∑`=0

∫ t

0

1J`(τ)sgn

(β(τ)

)dβ(τ)− `(t, 0),

where J` ≡ [ζ2`, ζ2`+1). Hence, we have now shown that

εNε(t)− `(t, 0) = −∣∣β(t)

∣∣ ∞∑`=0

1J`(t)− ε

∞∑`=0

1I`(t) +

∫ t

0

θε(τ) dβ(τ),

where θε(τ) ≡∑∞

`=0 1J`(τ)sgn

(β(τ)

). Finally, notice that t ∈ J` =⇒

|β(t)| ≤ ε, and so the absolute values of the first two terms on the rightare each dominated by ε. As for the stochastic integral term on the right,observe that |θε(τ)| ≤ 1[−ε,ε]

(β(τ)

), and therefore

EP

[∥∥∥∥∫ ·0

θε(τ) dβ(τ)∥∥∥∥2

[0,t]

]≤ 8ε

∫ t

0

1√2πτ

dτ.

Thus, by taking ε = 1m2 , we conclude that

limm→∞

∥∥m−2N 1m2− `( · , 0)

∥∥[0,t]

= 0 P-almost surely for each t > 0.

But

(m+ 1)−2 ≤ ε < m−2 =⇒ (m+ 1)−2N 1m2≤ εNε ≤ m−2N 1

(m+1)2,

and so the asserted result follows.

It should be clear that the results in Theorem 6.1.8 are further evidenceof the “fuzziness” alluded to in Remark 6.1.4.


6.1.3. Exercises.

Exercise 6.1.10. Let(β(t),Ft,P

)be a one-dimensional Brownian motion,

and let `( · , 0) be its local time at 0.

(i) Show that `(t, 0) is measurable with respect to the P-completion Gt ofσ(|β(τ)| : τ ∈ [0, t]

)and that σ

(sgn

(β(τ)

): τ ≥ 0

)is P-independent of

P-completion G of σ(|β(τ)| : τ ≥ 0

).

(ii) Set B(t) =∫ t

0sgn(β(τ)

)dβ(τ), and show that

(B(t),Gt,P

)is a one-

dimensional Brownian motion.

(iii) Refer to part (ii), but think of t B(t) as being a Brownian motionrelative of the σ-algebras t Ft, instead of t Gt. Show that

β(t) =∫ t

0

sgn(β(τ)

)dB(τ), t ≥ 0,

but that β(t) is not measurable with respect to the P-completion of σ(B(τ)

: τ ≥ 0).

The conclusion drawn in (iii) demonstrates that it is important to makea distinction between the notion of strong and weak solutions to a stochasticdifferential equation. Without going into details, suffice it to say that strongsolutions are those which are measurable functions of the driving Brownianmotion, whereas weak solutions are those which simply solve the stochasticdifferential equation relative to some family of σ-algebras with respect towhich the driving process happens to be a Brownian motion. For moredetails, see [27].

6.2 An Extension of the Cameron–Martin Formula

In Exercise 5.1.20 we discussed the famous formula of Cameron and Mar-tin for the Radon–Nikodym derivative Rh of the (τh)∗P0 with respect to P0

when h(t) =∫ t

0η(τ) dτ for some η ∈ L2

([0,∞); Rn

). Because, on average,

Brownian motion has no preferred direction, it is reasonable to think of theaddition of h to a Brownian path p as the introduction of a drift. That is,because h does have a velocity, it forces p+h to have a net drift which p by it-self does not possess. Thus, the Cameron–Martin formula can be interpretedas saying that the introduction of a drift into a Brownian motion producesa path whose distribution is equivalent to that of the unperturbed Brown-ian motion as long as the perturbation has a square integrable derivative.Further, as I. Segal (cf. Exercise 5.2.38 in [36]) showed, only such perturba-tions leave the measure class unchanged. In fact, τhP0 is singular to P0 if hfails to have a square integrable derivative. The intuitive reason underlyingthese results has to do with the “fuzz” alluded to in Remark 6.1.4. Namely,

6.2 An Extension of the Cameron–Martin Formula. 161

because the “fuzz” is manifested in almost sure properties, the perturbationmust be smooth enough to not disturb the “fuzz.” For example, the driftmust not alter P0-almost sure properties like

limN→∞

2N∑m=1

∣∣∣p(m2−N ))− p((m− 1)2−N

∣∣∣2 = n.

In this section, we will give an extension of their formula, one which will al-low us to introduce Radon-Nikodym derivatives to produce a random “drift”.

6.2.1. Introduction of a Random Drift. The key result here is thefollowing application of Ito’s formula.

6.2.1 Theorem. Suppose that β : [0,∞) × C([0,∞); Rn)

)−→ Rn is

a Bt : t ≥ 0-progressively measurable function, and assume that P is a

probability measure on C([0,∞); Rn)

)for which

(β(t),Bt

P,P)3 is a Brow-

nian motion. If θ ∈ Θ2loc(P; Rn) and (cf. (5.1.10)) EP[Eθ(T )

]= 1 for all

T ≥ 0, then there exists a unique Q ∈M1

(C([0,∞); Rn)

))such that

Q(A) = EP[Eθ(T ), A]

for all T ≥ 0 and A ∈ BTP.

Furthermore, if η is a Bt : t ≥ 0-progressively measurable satisfying∫ T

0|η(τ)| dτ <∞ P-almost surely for all T ∈ [0,∞) and if X(t) ≡

∫ t

0η(τ) dτ ,

then, for each ϕ ∈ C1,2(Rm × Rn; C

),4

(ϕ(X(t), β(t)

)−∫ t

0

Hϕ(τ) dτ,BtP,Q)

with

Hϕ(t) ≡∫ t

0

((η(τ), gradxϕ

)Rm + 1

2∆yϕ+(θ(τ), gradyϕ

)Rn

)(X(τ), β(τ)

)dτ

is a local martingale.

Proof: In order to prove the first assertion, note that, by part (ii) ofExercise 5.3.4,

(Eθ(t),Bt

P,P)

is a martingale. Hence, if the Borel probability

measure QT is determined on BTP

by Q(A) = EP[Eθ(T ), β ∈ A]

for all

A ∈ BTP, then the family QT : T ≥ 0 is consistent in the sense that, for

each 0 ≤ T1 < T2, QT1 is the restriction of QT2 to BT1 . Thus, by a minor

3 Given a probability measure P defined on a σ-algebra F , FPis used to denote the

P-completion of F .4 In the following, the subscripts “x” and “y” are used to indicate to which variables theoperation is being applied. The “x” refers to the first m coordinates, and the “y” refers

to the last n coordinates. Also, ∆ denotes the standard Euclidean Laplacian for Rn.


variant (cf. Theorem 1.3.5 in [41]) of Kolmogorov’s Extension Theorem, thereis a unique Borel probability measure Q whose restriction to BT is QT foreach T ≥ 0.

To prove the second part, define

ζR(p) = inft ≥ 0 : |β(t)| ∨

∫ t

0

∣∣η(τ)∣∣ ∨ ∣∣θ(τ)∣∣2 dτ ≥ R

.

Because ∫ T

0

|η(τ)| dτ ∨∫ T

0

|θ(τ)|2 dτ <∞ P-almost surely

and Q BTP P BT

Pfor each T ≥ 0, we know that ζR ∞ Q-

almost surely as R ∞. Thus, it is enough for us to show that, for eachϕ ∈ C1,2

(Rm × Rn; C

), 0 ≤ t1 < t2, A ∈ Bt1

P, and R > 0:

(*) EQ[M(t2 ∧ ζR), A]

= EQ[M(t1 ∧ ζR), A],

where

M(t) ≡ ϕ(X(t), β(t)

)−∫ t

0

Hϕ(τ) dτ.

By the definition of Q, (*) is equivalent to

EP[Eθ(t2)M(t2 ∧ ζR), A]

= EP[Eθ(t2)M(t1 ∧ ζR), A],

which, because(Eθ(t),Bt

P,P)

is a martingale, is equivalent to

(**) EP[M(t2 ∧ ζR), A]

= EP[M(t1 ∧ ζR), A],

where

M(t) ≡ Eθ(t)ϕ(X(t), β(t)

)−∫ t

0

Eθ(τ)Hϕ(τ) dτ.

But, by Ito’s formula and (5.3.5),

Eθ(t)ϕ(X(t), β(t)

)− ϕ

(0, 0)−∫ t

0

Eθ(τ)Hϕ(τ) dτ

=∫ t

0

Eθ(τ)(ϕ(X(τ), β(τ)

)θ(τ) + gradyϕ

(X(τ), β(τ)

), dβ(τ)

)Rn.


Finally, from the definition of ζR and the estimate in (5.1.25), it is an easymatter to check that

EP

[∫ T∧ζR

0

Eθ(τ)2∣∣∣ϕ(X(τ), β(τ)

)θ(τ) + gradyϕ

(X(τ), β(τ)

)∣∣∣2 dτ]

≤ EP[Eθ(T ∧ ζR)2

∫ T∧ζR

0

∣∣∣ϕ(X(τ), β(τ))θ(τ)

+ gradyϕ(X(τ), β(τ)

)∣∣∣2 dτ] <∞,

where, in passing to the second line, we have used the fact that(Eθ(t)2,Bt

P,

P)

is a submartingale. But (cf. Exercise 5.1.20), this means that(M(t∧ζR),

Bt,P)

is a square-integrable martingale, and so (**) holds.

The preceding leaves open the problem of determining when EP[Eθ(T )]

=1 for all T ≥ 0. Of course, Novikov’s condition (cf. part (iv) of Exercise5.3.4) gives one such condition. In particular, it tells us that there is noproblem when θ is bounded. However, when θ is not bounded, Novikov’scondition is often too crude, and one needs to use the sort of considerationsgiven in our next result, which is a variant on the Cameron–Martin themeusually attributed to I.V. Girsanov.

6.2.2 Corollary. Let θ : [0,∞)×C([0,∞); Rn)

)−→ Rn be a Bt : t ≥

0-progressively measurable function with the property that∫ T

0

∣∣θ(τ, p)∣∣2 dτ<∞ for each (T, p) ∈ [0,∞)×C

([0,∞); Rn)

). Then, for each x ∈ Rn, there

is at most one Qx ∈M1

(C([0,∞); Rn)

))with the property that

p ∈ C([0,∞); Rn)

)7−→ β(t, p) ≡ p(t)− x−

∫ t

0

θ(τ, p) dτ ∈ C([0,∞); Rn)

)is a Qx-Brownian motion relative to Bt

Qx : t ≥ 0. Moreover, if P0x ≡

(τx)∗P0 and (t, p) Eθ(t, p) is defined relative to the Brownian motion(p(t)− x,Bt

P0x ,P0

x

), then Qx exists if and only if

EP0x[Eθ(T )

]= 1 for all T ≥ 0,

in which case Qx BT is equivalent to P0x BT and dQx = Eθ(T ) dP0

x on BT

for each T ≥ 0.

Proof: First suppose that EP0x

[Eθ(T )

]= 1 for all T ≥ 0, and apply The-

orem 6.2.1 to produce Qx so that dQx BTP

= Eθ(T ) dP0x BT

Pfor each

T ≥ 0. Next, apply the second part of that same theorem to see that(exp(√−1(ξ, β(t)

)Rn + t

2 |ξ|2),Bt,Qx

)


is a local martingale, which, because it is bounded on [0, T ]×C([0,∞); Rn)

)for each T ≥ 0, must therefore be a martingale. In particular, this provesthat β(s+ t)− β(s) is Qx-independent of Bs and its Qx-distribution is thatof a centered Gaussian with covariance tI. Equivalently,

(β(t),Bt

Qx,Qx

)is

a Brownian motion.Now suppose that Q has the property that

(β(t),Bt

Q,Q)

is a Brownianmotion. We must show that dQ BT = Eθ(T ) dP0

x BT for each T ≥ 0. Forthis purpose, set

S(t) = exp(−∫ t

0

(θ(τ), dβ(τ)

)Rn −

12

∫ t

0

∣∣θ(τ)∣∣2 dτ) .The key step in our argument is to show that EQ[S(T )

]= 1 for all T ≥ 0.

To this end, define

ζR(p) = inft ≥ 0 :

∫ t

0

∣∣θ(τ, p)∣∣2 dτ ≥ R

.

Then

S(t ∧ ζR) = SR(t) ≡ exp(∫ t

0

(θR(τ), dβ(τ)

)Rn −

12

∫ t

0

∣∣θR(τ)∣∣2 dτ) ,

where θR(t) ≡ 1[0,t](ζR)θ(t), and so

EQ[S(T ), ζR > T]

= EQ[SR(T )]− EQ[SR(T ), ζR ≤ T

].

To take the next step, note that EQ[SR(T )]

= 1 for each (T,R) ∈ [0,∞)2,and, for a fixed R > 0, use Theorem 6.2.1 to determine PR so that dPR BT = SR(T ) dQ BT for all T ≥ 0. By repeating the argument given in thefirst part of the present proof, one can apply the second part of Theorem

6.2.1 to see that(pR(t)− x,Bt

PR

,PR)

is a Brownian motion when

pR(t) ≡ p(t)−∫ t

0

(θ(τ)− θR(τ)

)dτ.

At the same time, because (cf. Exercise 4.3.45 in [36]) ζR is a Bt : t ≥0-stopping time and p [0, ζR(p)) = pR [0, ζR(p)), one can check thatζR(p) = ζR(pR) for all p ∈ C

([0,∞); Rn)

). Hence,

EQ[SR(T ), ζR ≤ T]

= PR(ζR pR ≤ T

)= P0

x

(ζR ≤ T

)−→ 0

as R→∞. That is, EQ[S(T )]

= 1.


Knowing that EQ[S(T )]

= 1 for all T ≥ 0, we can again apply Theorem6.2.1 to see that dP0

x BT = S(T ) dQ BT for all T ≥ 0. Hence, all thatremains is to show that

S(T, p)−1 = exp

(∫ T

0

(θ(τ, p), dp(τ)

)Rn −

12

∫ T

0

∣∣θ(τ, p)∣∣2 dτ)

P0x-almost surely, which is almost obvious. Indeed, because p(t) = x +β(t, p) +

∫ t

0θ(τ) dτ , it is only reasonable that

∫ t

0

(θ(τ), dp(τ)

)Rn =

∫ t

0

(θ(τ), dβ(τ)

)Rn +

∫ t

0

∣∣θ(τ)∣∣2 dτP0

x-almost surely. Unfortunately, we have not yet (cf. Exercise 7.2.12) de-veloped the machinery with which to justify this line of reasoning directly,and so we will have to take the following circuitous route. Let R > 0 begiven, and choose (cf. Lemma 5.1.8) ηN : N ≥ 0 ⊆ SΘ2(P0

x; Rn) so thatηN (t) = ηN ([t]N ) and

EQ

[∫ ζR

0

∣∣θ(τ)− ηN (τ)∣∣2 dτ]+ EP0

x

[∫ ζR

0

∣∣θ(τ)− ηN (τ)∣∣2 dτ] −→ 0.

By taking

ζNR = inf

t ≥ 0 :

∫ t

0

∣∣ηN (τ)∣∣2 dτ ≥ R

and replacing ηN (t) by θN (t) ≡ 1[0,ζN

R)(t)ηN (t), we preserve the L2-conver-

gence, still have θN (t) = θN ([t]N ) for t ∈ [0, ζNR ), and achieve the bound

∫ ζNR

0

∣∣θN (τ)∣∣2 dτ ≤ R for all N ≥ 0.

Now let T > 0 be given, set V N (p) equal to

exp( ∞∑

m=0

(θN (m2−N , p), β

(T ∧ ζN

R (p) ∧ (m+ 1)2−N , p)

− β(T ∧ ζN

R (p) ∧m2−N , p))

Rn

+12

∫ T∧ζNR (p)

0

(θ(τ, p), θN (τ, p)

)Rn dτ

),


and observe that another expression for V N (p) is

exp( ∞∑

m=0

(θN (m2−N , p), p

(T ∧ ζN

R (p) ∧ (m+ 1)2−N)

− p(T ∧ ζN

R (p) ∧m2−N))

Rn

− 12

∫ T∧ζNR (p)

0

(θ(τ), θN (τ, p)

)Rn dτ

),

Moreover, for any bounded, BT -measurable F ,

EP0x[V NF, ζR > T

]= EQ[S(T )V NF, ζR > T

]= EQ[S(T ∧ ζR)V NF, ζR > T

].

Using the second expression for V N , we see that, as N → ∞, V N −→Eθ(T ∧ ζR) in P0

x-probability. In fact, by taking into account the estimatesin (5.1.25), one sees that this convergence is happening in L1(P0

x; R). At thesame time, by using the first expression for V N , the same line of reasoningshows that V NS(T ∧ ζR) −→ 1 in L1(Q; R). Thus, we have proved that

EQ[F, ζR > T]

= EP0x[Eθ(T )F, ζR > T

],

and, after letting R ∞, we have proved that EP0x [Eθ(T )] = 1 and that

dQ BT = Eθ(T ) dP0x BT .

6.2.3 Corollary. Let b : [0,∞) × Rn −→ Rn be a locally bounded,

measurable function. Then there exists a Qbx ∈ M1

(C([0,∞); Rn)

))with

the property that (p(t)− x−

∫ t

0

b(τ, p(τ)

)dτ,Bt

Qbx ,Qb

x

)is a Brownian motion if and only if (cf. the notation in Corollary 6.2.2)

EP0x[Rb(T )

]= 1 for all T ≥ 0,

where

Rb(T, p) ≡ exp

(∫ T

0

(b(τ, p(τ)

), dp(τ)

)Rn− 1

2

∫ T

0

∣∣b(τ, p(τ))∣∣2 dτ) ,in which case dQb

x BT = Rb(T ) dP0x BT for each T ≥ 0. In particular,

there is at most one such Qbx.


Remark 6.2.4. There is a subtlety about the result in Corollary 6.2.3.Namely, it might lead one to believe that, for any bounded, measurableb : Rn −→ Rn and P0 almost every p ∈ C

([0,∞); Rn)

), one can solve the

equation

(*) X(t, p) = p(t) +∫ t

0

b(X(τ, p)

)dτ, t ≥ 0,

and that the solution is unique. However, this is not what it says! Instead,it says that there is a unique Qb

0 ∈ M1

(C([0,∞); Rn)

))with the property

that P0 is the Qb0-distribution of

p β( · , p) ≡ p( · )−∫ ·

0

b(p(τ)

)dτ.

There is no implication that p can be recovered from β( · , p). In other words,we are, in general, dealing here with the kind of weak solutions alluded toat the end of Exercise 6.1.10. Of course, if b is locally Lipschitz continuousand therefore the solution X( · , p) to (*) can be constructed (e.g., by Picarditeration) “p by p,” then the P0-distribution of p X( · , p) is Qb

0.

6.2.2. An Application to Pinned Brownian Motion. A remarkableproperty of Wiener measure is that if T ∈ (0,∞), x, y ∈ Rn, and

(6.2.5) pT,y(t) ≡ p(t ∧ T ) +t ∧ TT

(y − p(T )

)for p ∈ C

([0,∞); Rn)

),

then the P0x-distribution of p pT,y [0, T ] is the P0

x-distribution of p p [0, T ] conditioned on the event p(T ) = y. To be more precise, if F :C([0,∞); Rn)

)−→ R is a bounded, BT -measurable function and Γ ∈ BRn ,

then (cf. Theorem 4.2.18 in [36] or Lemma 8.3.6 below)

EP0x[F, p(T ) ∈ Γ

]=

1(2πT )

n2

∫Rn

∫C([0,∞);Rn)

F(pT,y

)P0(dp)

e−|y−x|2

2T dy.

For this reason, p pT,y [0, T ] is sometimes called pinned Brownianmotion, or, in greater detail, Brownian motion pinned to y at time T .

The fact that (6.2.5) gives a representation of Brownian motion condi-tioned to be at y at time T is one of the many peculiarities (especially itsGaussian nature) of Brownian motion which sets it apart from all other dif-fusions. For this reason, it is interesting to find another representation, onewhich has a better chance of admitting a generalization (cf. Remark 6.2.8


below). With this in mind, let t ∈ [0, T ) be given, notice that, by definition,for any bounded, Bt-measurable F ,

EP0x[F∣∣ p(T ) = y

]= EP0

[FγT−t

(y − p(t)

)γT (y − x)

],

where γτ (ξ) ≡ (2πτ)−n2 e−

|ξ|22τ denotes the centered Gauss kernel on Rn with

covariance τIRn . Next, recall that

∂τγT−τ (y − · ) + 12∆γT−τ (y − · ) = 0 on [0, T )× Rn,

where ∆ denotes the Euclidean Laplacian for Rn; and apply Ito’s formulato conclude that

γT−t

(y − p(t)

)γT (y − x)

= 1−∫ t

0

γT−τ

(y − p(τ)

)γT (y − x)

(p(τ)− y

T − τ, dp(τ)

)Rn

for t ∈ [0, T ); and so, by Exercise 5.3.4,

(6.2.6)

γT−t

(y − p(t)

)γT (y − x)

= exp

(−∫ t

0

(p(τ)− y

T − τ, dp(τ)

)Rn

− 12

∫ t

0

∣∣∣∣p(τ)− y

T − τ

∣∣∣∣2 dτ)

for t ∈ [0, T ). Furthermore, by the Chapman–Kolmogorov equation,

EP0x

[γT−t

(y − p(t)

)γT (y − x)

]=∫

Rn

γt(ξ − x)γT−t(y − ξ)γT (y − x)

dξ = 1.

Hence, by Corollary 6.2.3, we have already proved a good deal of the follow-ing statement.

6.2.7 Theorem. Given T > 0, there a continuous map (y, p) ∈ Rn ×C([0,∞); Rn)

)7−→ Xy( · , p) ∈ C

([0, T ); Rn

)such that

Xy(t, p) = p(t)−∫ t

0

Xy(τ, p)− y

T − τdτ for t ∈ [0, T ).

Furthermore, for each x ∈ Rn and P0x-almost every p, limtT Xy(t, p) = y,

and so there exists a unique P0T,x,y ∈ M1

(C([0, T ]; Rn)

)with property that

the P0T,x,y-distribution of p p [0, T ) is the same as the P0

x-distribution of

p Xy( · , p) [0, T ). In fact, P0T,x,y

(p(T ) = y

)= 1, P0

T,x,y = (τx)∗P0T,0,y−x,


(x, y) P0T,x,y is continuous, and, if p(t) = p(T − t) for p ∈ C

([0, T ]; Rn

)and t ∈ [0, T ], then P0

T,y,x is the P0T,x,y-distribution of p p. Finally, for

any BT -measurable F which is either bounded or non-negative, EP0T,x,y [F ] is

the P0x-conditional expectation value EP0

x

[F∣∣ p(T ) = y

]of F given p(T ) = y

in the sense that

EP0x[F∣∣σ(p(T )

)]= EPT,x,p(T ) [F ] P0

x-almost surely.

Proof: The existence and continuity of (y, p) Xy( · , p) [0, T ) followeasily from the standard Picard iteration procedure. Moreover, by unique-ness, it is easy to see that Xy( · , τxp) = x + Xy−x( · , p), and so the dis-tribution of p Xy( · , p) under Px coincides with the P0-distribution ofp τxXy−x( · , p). Thus, if they exist, then P0

T,x,y = (τx)∗P0T,0,y−x.

By Corollary 6.2.3, (6.2.6), and the preceding discussion, we know that,for any t ∈ [0, T ) and A ∈ Bt,

(*) EP0x

[γT−t

(p(t)− y

)γT (y − x)

, A

]= P0

x

(Xy( · ) ∈ A

).

Our first application of (*) will be to show that the P0x distribution of

p Xy( · , p) (0, T ) coincides with the P0y-distribution of p Xx(T −

· , p) (0, T ); and for this purpose, it suffices to observe that, by repeatedapplication of (*), for any m ≥ 1 and 0 < t1 < · · · < tm,

P0x

(Xy(t1) ∈ dξ1, . . . Xy(tm) ∈ dξm

)=γt1(ξ1 − x) · · · γtm−tm−1(ξm − ξm−1)γ(y − ξm)

γT (y − x)dξ1 · · · dξm

= P0y

(Xx(T − tm) ∈ dξm, . . . Xx(T − t1) ∈ dξ1

).

In particular, we have shown that

P0x

(limtT

Xy(t) = y

)= P0

y

(limt0

Xx(t) = y

)= 1,

and so we now know that P0T,x,y exists, P0

T,x,y

(p(T ) = y

)= 1, and that

P0T,y,x is the P0

T,x,y-distribution of p p.We next want to check the continuity of (x, y) P0

T,x,y. To this end,remember (cf. the end of the first paragraph of this proof) that we knowP0

T,x,y = (τx)∗P0T,0,y−x. Hence, it is enough for us to show that y P0

T,0,y

is continuous. Next, observe that, for any δ ∈ (0, 1),∥∥Xy′( · , p)−Xy( · , p)∥∥

[0,T )

≤(|y′ − y|+

∥∥Xy′( · , p)−Xy( · , p)∥∥

[0,(1−δ)T ]

)log

1δ

+∫ T

(1−δ)T

|Xy′(τ, p)− y′|T − τ

dτ +∫ T

(1−δ)T

|Xy(τ, p)− y|T − τ

dτ.


Since, for any δ ∈ (0, 1) and ε > 0, we know that

limy′→y

∥∥Xy′( · , p)−Xy( · , p)∥∥

[0,(1−δ)T ]= 0 for all p,

we will know that

(**) limy′→y

P0(∥∥Xy′( · )−Xy( · )

∥∥[0,T ]

≥ ε)

= 0,

once we show that, for each R > 0,

limδ0

sup|y|≤R

∫ T

(1−δ)T

EP0[|Xy(τ)− y|]

T − τdτ = 0.

But ∫ T

(1−δ)T

EP0[|Xy(τ)− y|]

T − τdτ =

∫ δT

0

EP0y[|X0(τ)− y|τ

dτ,

and it is an easy matter to check that

sup|y|≤R

supτ∈(0, T

2 ]

EP0y[|X0(τ)− y|

]|

√τ

<∞.

After combining these, we know that (**) holds and therefore that (x, y) P0

T,x,y is continuous.All that remains is to identify P0

T,x,y as the conditional distribution ofP0

x BT given p(T ) = y. However, from (*), we know that, for any Γ ∈ BRn ,

EP0x[A ∩ p(T ) ∈ Γ

]=∫

Γ

PT,x,y(A) γT (y − x) dy

for all A ∈ Bt when t ∈ [0, T ). Since the set of A ∈ BT for which this relationholds is a σ-algebra, it follows that it holds for all A ∈ BT .

Remark 6.2.8. Certainly the most intriguing aspect of the preceding resultis the conclusion that limtT Xy(t, p) = y for P0

x-almost every p. Intuitively,one knows that the drift term −X(t,p)−y

T−t is “punishing” X(t, p) for not beingclose to y, and that, as t T , the strength of this penalty becomes infi-nite. On the other hand, this intuition reveals far less than the whole story.Namely, it completely fails to explain why the penalization takes this partic-ular form instead of, for example, −2X(t,p)−y

T−t or − (X(t,p)−y)3

T−t . As a careful

examination of the preceding reveals, the reason why −X(t,p)−yT−t is precisely

the correct drift is that it is equal to gradX(t,p) log gT−t( · , y) where gτ ( · , y)is the fundamental solution (in this case γτ ( · − y)) to the heat equation.It is this observation which allows one to generalize the considerations todiffusions other than Brownian motion.


6.2.3. Exercises.

Exercise 6.2.9. The purpose of this exercise is to interpret the meaning of(cf. Corollary 6.2.3) EP0[

Rb(T )]

failing to be 1.Let b : Rn −→ Rn be a locally Lipschitz continuous function. Choose

ψ ∈ C∞(Rn; [0, 1]

)so that ψ ≡ 1 on BRn(0, 1) and ψ ≡ 0 off BRn(0, 2),

and set bR(x) = ψ(R−1x)b(x) for R > 0 and x ∈ Rn. Given R > 0 andp ∈ C

([0,∞); Rn)

), define ζR(p) = inf

t ≥ 0 : |p(t)| ≥ R

and determine

XR( · , p) ∈ C([0,∞); Rn)

)by the equation

XR(t, p) = p(t) +∫ t

0

bR(X(τ, p)

)dτ t ∈ [0,∞).

(i) Given 0 < R1 < R2, show that XR2(t, p) = XR1(t, p) for 0 ≤ t ≤ζR1 XR1( · , p), and conclude that

ζR1 XR1( · , p) = ζR1 XR2( · , p) ≤ ζR2 XR2( · , p).

Next, define the explosion time e(p) = limR∞ ζR XR( · , p), and determinet ∈[0, e(p)

)7−→ X(t, p) ∈ Rn so that X(t, p) = XR(t, p) for 0 ≤ t ≤ ζR(p).

(ii) Define Rb as in Corollary 6.2.3, show that

EP0[Rb(T ), ζR > T

]= P0

(ζR XR( · ) > T

),

and conclude that EP0[Rb(T )

]= P0(e > T ). That is, EP0[

Rb(T )]

is equal tothe probability that X( · , p) has not exploded by time T .

Exercise 6.2.10. When the drift b in Corollary 6.2.3 is a gradient, theRadon–Nikodym Rb(T ) becomes much more tractable. Namely, apply Ito’sformula to show that if b(t, x) = gradxU(t, · ), then

(6.2.11)Rb(T, p) = exp

(U(T, p(T )

)− U

(0, p(0)

)− 1

2

∫ T

0

V(τ, p(τ)

)dτ

)where V (t, x) ≡

∣∣gradxU(t, · )∣∣2 +

(∂t + 1

2∆)U(t, x).

(i) Recall the Ornstein–Uhlenbeck process p′ X( · , x, p′) given by (3.3.1),and let Qx denote the distribution of p X( · , x, , p) under P0. Using thepreceding, show that

EQx[ϕ(p(T )

)]= EP0

x

[ϕ(p(T )

)exp

(βnT − |p(T )|2 + |x|2

2− β2

2

∫ T

0

|p(τ)|2 dτ

)]for any measurable ϕ : Rn −→ [0,∞).


(ii) After combining (i) with the considerations at the end of § 3.3.1, cometo the conclusion that

EP0x

[ϕ(p(T )

)exp

(−1

2

∫ T

0

|p(τ)|2 dτ

)]

= e−nT+|x|2

2

∫Rn

ϕ(y)e|y|22 γ 1

2 (1−e−2T )

(y − e−Tx

)dy.

In particular, conclude from this that

EP0x

[exp

(−1

2

∫ T

0

|p(τ)|2 dτ

)]=(√

coshT)−n exp

(−|x|

2

2tanhT

).

(iii) By applying the Brownian scale invariance property, note that theP0-distributions of

p exp

(−1

2

∫ λT

0

|p(τ)|2 dτ

)and p exp

(−1

2λ2

∫ T

0

|p(τ)|2 dτ

)

are the same for every choice of λ > 0. Hence,

EP0

[exp

(−λ

2

∫ T

0

|p(τ)|2 dτ

)]=(√

cosh√λT

)−n

.

(iv) Suppose that, for some A > 0,

EP0

[exp

(A2

2

∫ T

0

|p(τ)|2 dτ

)]<∞,

and show that

z ∈ BC(0, A2) 7−→ EP0

[exp

(−z

2

∫ T

0

|p(τ)|2 dτ

)]∈ C

must be a halomorphic function. After combining this with (iii), concludethat A ≤ π

2T . That is

A >π

2T=⇒ EP0

[exp

(A2

2

∫ T

0

|p(τ)|2 dτ

)]= ∞.

(v) The preceding can be used to bring out the inherent weakness incriteria like (cf. Exercise 5.3.4) Novikov’s . Namely, show that when θ(t) =


p(t) and Eθ(t) is defined as in (5.1.10) relative to the Brownian motion(p(t),Bt,P0

),(Eθ(t),Bt,P0

)is a martingale but that

T >π

2=⇒ EP0

[exp

(12

∫ T

0

|p(τ)|2 dτ

)]= ∞.

Exercise 6.2.12. The purpose of this exercise is to give an example of a θfor which the corresponding Eθ(t) is definitely not a martingale. Through-out, n = 1.

(i) Given p ∈ C([0,∞); R

), suppose that X( · , p) ∈ C

([0, T (p)]; R

)satis-

fies,

X(t, p) = p(t) +∫ t

0

(2 +X(τ, p)

)2dτ for t ∈

[0, T (p)

].

Assuming that p(t) ≥ −1 for t ∈ [0, T (p)], show that X(t, p) ≥ 11−t − 2, and

conclude that T (p) < 1.

Hint: First note that X( · , p) must be obtainable from p via Picard’s iter-ation method starting with X0( · , p) = p. Proceeding by induction, showthat XN ( · , p) ≥ YN ( · ), where YN : N ≥ 0 corresponds to X( · , p) whenp ≡ −1.

(ii) Take θ(t) =(2 + p(t)

)2, and define Eθ(t) relative to(p(t),Bt,P0

).

Show that

EP0[Eθ(1)

]≤ P0

(inf

t∈[0,1]p(t) ≤ −1

)< 1.

In particular,(Eθ(t),Bt

P0

,P0)

is not a martingale, and so, by Novikov’scriterion,

EP0[exp

(12

∫ 1

0

(2 + p(t)

)4dt

)]= ∞.

(iii) Show that if θ(t) = −sgn(p(t)

)(2 + p(t)

)2, then(Eθ(t),Bt,P0

)is a

martingale. Hence, in view of the last part of (ii), we see once again thatNovikov’s criterion is too weak.

Exercise 6.2.13. Set Ω = Rn × C([0,∞); Rn)

), and consider the Borel

probability measure Q = ΓT × P0 on Ω, where ΓT is the centered Gaussmeasure on Rn with covariance TI.

(i) Referring to Theorem 6.2.7, define

(y, p) ∈ Ω 7−→ Y(· , (y, p)

)≡ Xy( · , p) ∈ C

([0, T ); Rn

),

and show that, Q-almost surely, limtT Y(t, (y, p)

)= y. Thus, Q-almost

surely, Y ( · , ω) extends as a continuous path on [0, T ].


(ii) Show that the Q-distribution of ω ∈ Ω 7−→ Y ( · , ω) ∈ C([0, T ]; Rn

)is

the same as the P0-distribution of p p [0, T ]. Equivalently. if

β(t, ω) = Y (t∧T, ω)+(p(t)− p(t∧T )

)for (t, ω) =

(t, (y, p)

)∈ [0,∞)×Ω,

then (β(t), σ

(β(τ) : τ ∈ [0, t]

),Q)

is a Brownian motion.(iii) Remember (cf. part (ii) of Exercise 2.4.16) that almost every Brow-

nian path is Holder continuous of every order strictly less than 12 . Thus, if(

β(t),Ft,P)

is an Rn-valued Brownian motion, then

t ∈ [0,∞) 7−→ β(t) ≡ β(t)−∫ t∧T

0

β(T )− β(τ)T − τ

dτ

is P-almost surely well-defined. Show that(β(t), σ

(β(τ) : τ ∈ [0, t]

),P)

is again a Brownian motion. Hence, if C14([0,∞); Rn

)is the space of paths

which are Holder continuous of order 14 and we use P0 again to denote

P0 C14([0,∞); Rn

), then the map on C

14([0,∞); Rn

)into itself given by

[Φ(p)](t) = p(t)−∫ t∧T

0

p(T )− p(τ)T − τ

dτ

is P0-measure preserving.

6.3 Homogeneous Chaos

Anyone who has read the second part, I am a Mathematician, of NorbertWiener’s remarkable autobiography knows that Wiener attributed much ofhis own mathematical achievement to his deep insight into spectral theory.Wiener’s insight into the subject was not only deep, it was highly imagi-native. Indeed, he understood that the basic principles underlying spectraldecomposition apply to situations where the rest of us would probably nothave seen that they might apply. One of his most imaginative applicationsof spectral decomposition was to randomness. More precisely, he had themarvelous idea that random noise should be decomposable into componentsof what he, with his flair for words, called spaces of homogeneous chaos.

Wiener’s own treatment of this subject is fraught with difficulties,5 all ofwhich were resolved by Ito. Thus, we, once again, will be guided by Ito.

5 One reason why Wiener is difficult to read is that he insisted on doing all integrationtheory with respect to Lebesgue measure on the interval [0, 1]. He seems to have thought

that this decision would make engineers and other non-mathematicians happier.

6.3 Homogeneous Chaos. 175

6.3.1. Multiple Stochastic Integrals. Throughout, we will be workingwith the canonical Rn-valued Brownian motion

(p(t),Bt,P0

).

Our first order of business is to define multiple stochastic integrals. Thatis, if for m ≥ 1 and t ∈ [0,∞], (m)(t) ≡ [0, t)m and (m) ≡ (m)(∞), wewant to assign a meaning to expressions like

I(m)F (t) =

∫(m)(t)

(F (~τ), d~p(~τ)

)(Rn)m

when F ∈ L2((m); (Rn)m

).

With this goal in mind, when m = 1 and F = f ∈ L2([0,∞); Rn

), we

take I(1)F (t) = If (t), where If (t) is the Paley–Wiener integral of f . When

m ≥ 2 and F = f1 ⊗ · · · ⊗ fm for some f1, . . . , fm ∈ L2([0,∞); Rn

),6 we use

induction to define I(m)F (t) so that

(6.3.1) I(m)f1⊗···⊗fm

(t) =∫ t

0

I(m−1)f1⊗···⊗fm−1

(τ)(fm(τ), dp(τ)

)Rn ,

where now we need Ito’s integral. Of course, in order to do so, we are obligedof check that τ fm(τ)I(m−1)

f1⊗···⊗fm−1(t) is square integrable. But, assuming

that I(m−1)f1⊗···⊗fm−1

is well defined, we have that

EP0

[∫ T

0

∣∣fm(τ)I(m−1)f1⊗···⊗fm−1

(τ)∣∣2 dτ]

=∫ T

0

∣∣fm(τ)∣∣2EP0

[∣∣I(m−1)f1⊗···⊗fm−1

(τ)∣∣2] dτ.

Hence, at each step in our induction procedure, we can check that

EP0[∣∣I(m)

f1⊗···⊗fm(T )∣∣2] =

∫M(m)(T )

∣∣f1(τ1)∣∣2 · · · ∣∣fm(τm)∣∣2 dτ1 · · · dτm,

where M(m) (t) ≡(t1, . . . , tm) ∈ (m) :≤ t1 < · · · < tm < t

; and so, after

polarization, we arrive at

EP0[I(m)f1⊗···⊗fm

(T )I(m)f ′1⊗···⊗f ′m

(T )]

=∫

M(m)(T )

(f1(τ1), f ′(τ1)

)Rn · · ·

(fm(τm), f ′m(τm)

)Rn dτ1 · · · dτm.

6 Here we are identifying f1⊗· · ·⊗fm with the (Rn)m-valued function F on [0,∞)m such

that (Ξ, F (t1, . . . , tm))(Rn)m = (ξ1, f1(t1))RN · · · (ξm, fm(tm))Rn for Ξ = (ξ1, . . . , ξm).


We next introduce

(6.3.2) I(m)f1⊗···⊗fm

(t) ≡∑

π∈Πm

I(m)fπ(1)⊗···⊗fπ(m)

(t),

where Πm is the symmetric group (i.e., the group of permutions) on 1, . . . ,m.By the preceding, we see that


(T )I(m)f ′1⊗···⊗f ′m

(T )]

=∑

π,π′∈Πm

∫M(m)(T )

m∏`=1

(fπ(`)(τ`), f ′π′(`)(τ`)

)Rn dτ1 · · · dτm

=∑

π∈Πm

∫(m)(T )

m∏`=1

(f`(τ`), f ′π(`)(τ`)

)Rn dτ1 · · · dτm

=∑

π∈Πm

m∏`=1

(f`, f

′π(`)

)L2([0,T );Rn)

.

In preparation for the next step, let gj : j ≥ 1 be an orthonormal basisin L2

([0,∞); Rn

), and note that

gj1⊗· · ·⊗gjm

: (j1, . . . , jm) ∈ (Z+)m

is anorthonormal basis in L2

((m)(∞); (Rn)m

). Next, for µ ∈ NZ+

, set ‖µ‖1 =∑∞1 µj , and take A to be the set of NZ+

with ‖µ‖1 < ∞. Finally, givenµ ∈ A with ‖µ‖1 = m, set Gµ = g⊗µ1 ⊗ · · · ⊗ g⊗µm , where the meaning hereis determined by the convention that g⊗µ1⊗· · ·⊗g⊗µ` = g⊗µ1⊗· · ·⊗g⊗µ`−1

if µ` = 0. In particular, Gµ ∈ L2((m); (Rn)m

), and, for µ, ν ∈ A,

(6.3.3)(I(‖µ‖1)Gµ

, I(‖ν‖1)Gν

)L2(P0;R)

= δµ,νµ!.

Now, given F ∈ L2((m); (Rn)m

), set

(6.3.4) F (t1, . . . , tm) ≡∑

π∈Πm

F(tπ(1), . . . , tπ(m)

),

and observe that

‖F‖2L2((m);(Rn)m) =∑

J∈(Z+)m

(F , gj1 ⊗ · · · ⊗ gjm

)2L2((m);(Rn)m

)=

∑‖µ‖1=m

(m

µ

)(F , Gµ

)2L2((m);(Rn)m

),where

(mµ

)is the multinomial coefficient m!

µ1!···µm! . Hence, after combiningthis with calculation in (6.3.3), we have that

EP0

∣∣∣∣∣∣∑‖µ‖1=m

(F , Gµ

)L2((m);(Rn)m)

µ!I(m)Gµ

(∞)

∣∣∣∣∣∣2 =

1m!‖F‖2L2((m);(Rn)m).

With these considerations, we have proved the following.


6.3.5 Theorem. There is a unique linear map

F ∈ L2((m); (Rn)m

)7−→ I

(m)F ∈M2(P0; R)

such that I(m)f1⊗···⊗fm

is given as in (6.3.2) and

EP0[I(m)F (∞)I(m)

F ′ (∞)]

=1m!(F , F ′

)L2((m);(Rn)m)

In fact, if gj : j ≥ 1 is an orthonormal basis in L2([0,∞); Rn

)and Gµ :

µ ∈ A is defined as above, then

I(m)F =

∑‖µ‖1=m

(F , Gµ

)L2((m);(Rn)m)

µ!IGµ

,

where µ! ≡∏

1∞‖µ‖1µj ! and the convergence is in L2(P0; R).

Although it is somewhat questionable to do so, we will, as indicated atthe beginning of this subsection, adopt the suggestive notation∫

(m)(t)

(F (~τ), d~p(~τ)

)(Rn)m

for I(m)F (t). The reason why this notation is questionable is that, although

it is suggestive, it may suggest the wrong thing. Specifically, in order toavoid stochastic integrals with non-progressively measurable integrands, ourdefinition of I(m)

F carefully avoids integration across diagonals, whereas thepreceding notation gives no hint of that fact.6.3.2. The Spaces of Homogeneous Chaos. Take Z(0) be the subspaceof L2(P0; R) consisting of the constant functions, and, for m ≥ 1, take

Z(m) =I(m)F (∞) : F ∈ L2

((m); (Rn)m

).

Clearly, each Z(m) is a linear subspace of L2(P0; R). Furthermore, if Fk :k ≥ 1 ⊆ L2

((m); (Rn)m

)and

I(m)Fk

(∞) : k ≥ 1

converges in L2(P0; R),then Fk : k ≥ 1

converges in L2

((m); (Rn)m

)to some symmetric function

G. Hence, since G = m!G, we see that I(m)Fk

(∞) −→ I(m)F (∞) in L2(P0; R)

where F = 1m!G. That is, each Z(m) is a closed linear subspace of L2(P0; R).

Finally, Z(m) ⊥ Z(m′) when m′ 6= m. This is completely obvious if either mor m′ is 0. Thus, suppose that 1 ≤ m < m′. Then


(∞)I(m′)f ′1⊗···⊗f ′

m′(∞)

]=∫

M(m′−m)

m′−m−1∏`=0

(fm−`(τ`), f ′m′−`(τ`)

)Rn

× EP0[I(m′−m)f ′1⊗···⊗f ′

m′−m

(τm′−m)]dτ1 · · · dτm′−m = 0,


which completes the proof.The space Z(m) is the space of mth order homogeneous chaos. The reason

why elements of Z(0) are said to be of 0th order chaos is clear: constantsare non-random. To understand why I

(m)F (∞) is of mth order chaos when

m ≥ 1, it is helpful to replace dp(τ) by the much more ambiguous p(τ) dτand write

I(m)F (∞) =

∫(m)

(F (τ1, . . . , τm),

(p(τ1), · · · , p(τm)

))(Rn)m

dτ1 · · · dτm.

In the world of engineering and physics, τ p(τ) is white noise.7 Thus,Z(m) is the space built out of homogeneous mth order polynomials in whitenoises evaluated at different times.8 In other words, the order of chaos isthe order of the white noise polynomial.

The result of Wiener, alluded at the beginning of this section, now becomesthe assertion that

(6.3.6) L2(P0; R) =∞⊕

m=0

Z(m).

The key to Ito’s proof of (6.3.6) is found in the following. See Exercise 6.3.18for an alternative approach.

6.3.7 Lemma. If f ∈ L2([0,∞); Rn

), then, for each λ ∈ C,

eλIf (∞)−λ2

2 ‖f‖2L2([0,∞);Rn) =

∞∑m=0

λm

m!I(m)f⊗m(∞) ≡ lim

M→∞

M∑m=0

λm

m!I(m)f⊗m(∞)

P0-almost surely and in L2(P0; R). In fact, if

RMf (∞, λ) ≡ e

λIf (∞)−λ22 ‖f‖

2L2([0,∞);Rn) −

M−1∑m=0

λm

m!I(m)f⊗m(∞),

then

EP0[∣∣RM

f (∞, λ)∣∣2] = e

|λ|2‖f‖2L2([0,∞);Rn) −

M−1∑m=0

(|λ|‖f‖L2([0,∞);Rn)

)2m

m!.

7 The terminology comes from the observation that, no matter how one interprets t p(t),it is a stationary, centered Gaussian process whose covariance is the Dirac delta functiontimes the identity. In particular, p(t1) is independent of p(t2) when t1 6= t2.8 Remember that our integrals stay away from the diagonal.


Proof: Set E(t, λ) = eλIf (t)−λ2

2 ‖1[0,t)f‖2L2([0,∞);Rn) . Using Ito’s formula, asin part (i) of Exercise 5.3.4, one sees that, for each λ ∈ C,

E(t, λ) = 1 + λ

∫ t

0

E(τ, λ)(f(τ), dp(τ)

)Rn .

Thus, if (cf. (6.3.1)) I(m)(t) ≡ I(m)f⊗m(t) = 1

m! I(m)f⊗m and

R0(t, λ) ≡ E(t, λ) and RM+1(t, λ) ≡ λ

∫ t

0

RM (τ, λ)(f(τ), dp(τ)

)Rn

for M ≥ 0, then, by induction, one sees that

E(t, λ) = 1 +M∑

m=1

λmI(m)(t) +RM+1(t, λ)

for all M ≥ 0. Finally, if α = Re(λ) and F (t) =∫ t

0|f(τ)|2 dτ , then

EP0[∣∣R0(t, λ)∣∣2] = e|λ|

2F (t)EP0[E(t, 2α)

]= e|λ|

2F (t),

and

EP0[∣∣RM+1(t, λ)∣∣2] = |λ|2

∫ t

0

EP0[∣∣RM (τ, λ)∣∣2]F (τ) dτ.

Hence, the asserted estimate follows by induction on M ≥ 0.

6.3.8 Theorem. The span of

R⊕I(m)f⊗m(∞) : m ≥ 1 & f ∈ L2

([0,∞); Rn

)is dense in L2(P0; R). In particular, (6.3.6) holds.

Proof: Let H denote smallest closed subspace of L2(P0; R) containing allconstants and all the functions I(m)

f⊗m(∞). By the preceding, we know thatcos If (∞) and sin If (∞) are in H for all f ∈ L2

([0,∞); Rn

).

Next, observe that the space of functions Φ : C([0,∞); Rn)

)−→ R which

have the formΦ = F

(If1(∞), . . . , IfL

(∞))

for some L ≥ 1, Schwartz class test function F : RL −→ R, and f1, . . . , fL ∈L2([0,∞); Rn

)is dense in L2(P; Rn). Indeed, this follows immediately from

the density in L2(P0; R) of the space of functions of the form

p F(p(t1), . . . , p(tL)

),


where L ≥ 1, F : (Rn)L −→ R is in the Schwartz test function class, and0 ≤ t0 < · · · < tL.

Now suppose that F : RL −→ R is given, and let F denote its Fouriertransform. Then, by elementary Fourier analysis,

FN (x) ≡(2(4N + 1)π

)−L ∑‖m‖∞≤4N

e−√−1 (2−Nm,x)RL F (m2−N ) −→ F (x),

both uniformly and boundedly, where m = (m1, . . . ,mL) ∈ ZL and ‖m‖∞= max1≤`≤L |m`|. Finally, since F (Ξ) = F (−Ξ) for all Ξ ∈ RL, we can write(

2(4N + 1)π)LFN

(If1(∞), . . . , IfL

(∞))

=∑

m∈NL:‖m‖∞≤4N

2(Re(F (m2−N )

)cos Im,N (∞)

+ Im(F (m2−N )

)sin Im,N (∞)

)∈ H,

where Im,N ≡ 2−NIfm and fm =∑L

`=1m`f`.

The following corollary is an observation made by Ito after he cleaned upWiener’s treatment of (6.3.6). It is often called Ito’s Representation Theoremand turns out to play an important role in applications of stochastic analysisto, of all things, models of financial markets.9

6.3.9 Corollary. The map

(α, θ) ∈ R×Θ2(P0; Rn) 7−→ α+ Iθ(∞) ∈ L2(P0; R)

is a linear, isometric surjection. Hence, for each Φ ∈ L2(P0; R) there is a

P0-unique θ ∈ Θ2(P0; Rn) such that

Φ = EP0[Φ] +

∫ ∞0

(θ(τ), dp(τ)

)Rn P0-almost surely.

In particular,

EP0[Φ∣∣Bt

]= EP0

[Φ] +∫ t

0

(θ(τ), dp(τ)

)Rn P0-almost surely for each t ≥ 0,

and therefore (t, p) EP0[Φ∣∣Bt

](p) can be chosen so that t EP0[

Φ∣∣Bt

](p)

is P0-almost surely continuous.

9 In fact, it shares with Ito’s formula responsibility for the wide spread misconception in

the financial community that Ito is an economist.


Proof: Since it is clear that the map is linear and isometric, the first as-sertion will be proved once we check that the map is onto. But, because itis a linear isometry, we know that its image is a closed subspace, and so weneed only show that its image contains a set whose span is dense. However,for each f ∈ L2

([0,∞); Rn

)and m ≥ 1,

I(m)f⊗m(∞) = m

∫ ∞0

(I(m−1)

f⊗(m−1)(τ)f(τ), dp(τ))

Rn ,

and so, by the first part of Theorem 6.3.8, we are done.Given the first assertion, the other assertions become essentially triv-

ial.

6.3.3. Exercises.

Exercise 6.3.10. Although our treatment of Wiener’s homogeneous chaosdecomposition makes no use of it, Wiener’s own treatment rested on theconnection between the decomposition in (6.3.6) and the Hermite polynomi-als. In order to understand this connection, recall the Hermite polynomialsHm : m ≥ 0 described in part (i) of Exercise 5.1.27. Next, given L ≥ 1and µ ∈ NL, define Hµ : RL −→ R so that

Hµ

(x1, . . . , xL

)=

L∏`=1

Hµ`(x`).

(i) Let ΓL denote the centered Gauss measure on RL with covariance IRL .Show that

‖Hµ‖2L2(ΓL;R) = µ!

and that Hµ : µ ∈ NL is an orthogonal basis in L2(ΓL; R).

(ii) Given ξ ∈ RL, show that

e(ξ,x)RL− 12 |ξ|

2=∑

µ∈NL

ξµ

µ!Hµ(x),

where ξµ ≡∏L

`=1 ξµ`

` .

(iii) Suppose that f1, . . . , fL is an orthonormal subset of L2([0,∞); Rn

),

and, given µ ∈ NL with m = ‖µ‖1 ≡∑L

`=1 µ`, show that

I(m)

f⊗µ11 ⊗···⊗f

⊗µLL

(∞) = Hµ

(If1(∞), . . . , IfL

(∞)),

where f⊗µ11 ⊗ · · · ⊗ f⊗µL

L is the element of L2((m); (Rn)m

)which is deter-

mined by the convention introduced in §6.3.1 just before (6.3.3).


Exercise 6.3.11. In this exercise, we will use the preceding to show that(6.3.6) can be interpreted as the spectral resolution for a nice self-adjointoperator. For this purpose, let gj : j ≥ 1 be an orthonormal basis forL2([0,∞); Rn

), and assume each gj ∈ C∞c

((0,∞); Rn

). Then (t, p)

Igj (t, p) is well defined as a continuous function on the whole of [0,∞] ×C([0,∞); Rn)

). Next, for

µ ∈ A ≡

µ ∈ NZ+

: ‖µ‖1 ≡∞∑1

µj <∞

,

set

Hµ(x) =‖µ‖1∏j=1

Hµj(xj) when x = (x1, . . . , x‖µ‖1) ∈ R‖µ‖1

andGµ ≡ g⊗µ1

1 ⊗ · · · ⊗ g⊗µ‖µ‖1‖µ‖1 .

Finally, define

Hµ(p) = Hµ

(Ig1(∞, p), . . . , Ig‖µ‖1

(∞, p))

and Iµ(t) = I‖µ‖1Gµ

(t).

(i) Use parts (ii) and (iii) of the preceding exercise to see that, for eachµ ∈ A,

Iµ(∞) = Hµ P0-almost surely,

and conclude that, for each m ∈ Z+, Hµ : ‖µ‖1 = m is an orthogonalbasis for Z(m).

(ii) Let D denote the space of continuous functions Φ : C([0,∞); Rn)

)−→

R with the property that

(*) Φ(p) = F(Ig1(∞, p), . . . , IgL

(∞, p))

for some L ≥ 1 and smooth function F : RL −→ R which, together withall its derivatives, has tempered (i.e., at most polynomial) growth. Givenf ∈ L2

([0,∞); Rn

), take hf ∈ C

([0,∞); Rn)

)so that hf (t) =

∫ t

0f(τ) dτ and

define DfΦ ∈ C(C([0,∞); Rn)

); R)

for Φ ∈ D so that

DfΦ(p) =d

dsΦ(p+ shf

) ∣∣∣s=0

for p ∈ C([0,∞); Rn)

).

Observe that when Φ is given by (*),

DfΦ(p) =L∑

j=1

(f, gj

)L2([0,∞);Rn)

∂jF(Ig1(∞, p), . . . , IgL

(∞, p)),


which shows that Df maps D into itself. Use these observations to checkthat number operator10

NΦ ≡∞∑

j=1

(Igj

(∞)DgjΦ−D2

gjΦ)

is a well-defined operator which takes D into itself.

(iii) After verifying that ∂jHµ = µjHµ(−j), where

µ(−j)i ≡µi if i 6= j or µj = 0µj − 1 if i = j and µj ≥ 1,

and that (xj − ∂j)Hµ = Hµ(+j), where

µ(+j)i ≡µi if i 6= j

µj + 1 if i = j,

show thatNHµ = ‖µ‖1Hµ, µ ∈ A,

and conclude that N is an essentially self-adjoint operator and for whichZ(m) is the eigenspace with eigenvalue m of the unique self-adjoint extensionN of N . See part (iv) in Exercise 6.3.17 below for more information.

Exercise 6.3.12. Continue with the notation in the preceding exercise,take Dj = Dgj

, and define D>j on D so that D>j Φ = Igj(∞)Φ−DjΦ.

(i) Show that D>j is the restriction to D of the adjoint of Dj . That is,show that

(6.3.13)(D>j Ψ,Φ

)L2(P0;R)

=(Ψ, DjΦ

)L2(P0;R)

for Φ,Ψ ∈ D.

Hint: Using the Cameron–Martin formula, show that

(ΦDj ,Ψ

)L2(P0;R)

=d

ds

∫Φ(p)eIsgj

(∞,p)− s22 Ψ(p− shj

)P0(dp)

∣∣∣∣s=0

,

where hj(t) =∫ t

0gj(τ) dτ .

(ii) Define Dµ : D −→ D inductively so that Dµ(+j) = Dj Dµ. WhenΦ ∈ D, use part (i) in Exercise 6.3.11 and (iv) above to show that

EP0[ΦIµ(∞)

]= EP0[

DµΦ],

10 The terminology here comes from Euclidean quantum field theory.


and conclude that

Φ = EP0[Φ] +

∞∑m=1

1m! I

(m)Fm

(∞),

where (cf. (*) in (ii))

Fm ≡∑‖µ‖1=m

EP0[DµΦ

]Gµ.

Thus, the resolution of Φ into its components of homogeneous chaos gives asort of Taylor’s expansion of Φ.

(iii) Set Λ = Leb[0,∞) × P0. Given θ ∈ Θ2(P0; Rn), show that theIto integral Iθ(∞) =

∫∞0

(θ(τ), dp(τ)

)Rn is the unique element of L2(P0; R)

with the property that11

(Ψ, Iθ(∞)

)L2(P0;R)

=∞∑

j=1

(gj ⊗DjΨ, θ

)L2(Λ;Rn)

for all Ψ ∈ D.

Hint: As an application of Corollary 6.2.2, show that the right hand sideof the preceding is equal to the derivative at s = 0 of (cf. (5.1.10)) s ∫

Φ(p)Esθ(∞, p) P0(dp).

(iv) Define the gradient operator D from D into L2(Λ; Rn) so that DΦ =∑∞j=1 gj ⊗ DjΦ. Since D is dense in L2(P0; R), D admits a well-defined

adjoint D> on L2(Λ; Rn). To the extent that D deserves to be called agradient, D> has to be called a divergence operator. Show that the result in(iii) can be summarized by the statement of Θ2(P0; Rn) ⊆ Dom(D>) andthat Iθ = D>θ for θ ∈ Θ2(P0; Rn).

Exercise 6.3.14. The considerations in (iv) of the preceding give a naturalway of extending to non-progressively measurable integrands Ito’s theoryof stochastic integration with respect to Brownian motion. Namely, givenθ ∈ Dom(D>), the extension we have in mind is provided by D>θ. Thepurpose of the present exercise is to examine the domain of D> and showthat D> determines the same extension of Ito’s theory as the one suggestedby A.V. Skorohod in [32].

(i) Given j ∈ Z+, m ∈ N, and F ∈ L2((m); (Rn)m

), show that gj⊗I(m)

F ∈Dom(D>), and that D>(gj ⊗ I

(m)F ) = I

(m+1)gj⊗F .

11 Here we are using f ⊗ Φ to denote the function on [0,∞) × C([0,∞); Rn)) which is

equal to f(t)Φ(p) at (t, p).


Hint: Show that(I(m+1)gj⊗F ,Hν

)L2(P0;R)

=(gj ⊗ F ,Gν

)L2((m+1);(Rn)m+1)

= νj

(F , Gν(−j)

)L2((m);(Rn)m)

=(gj ⊗ IF , DF

)L2(Λ;Rn)

when ‖ν‖1 = m+ 1.

(ii) Given θ ∈ Θ2(P0; Rn) and m ∈ N, show that, for almost every t ∈[0,∞),

θ(m)(t) ≡∞∑

j=1

∑‖µ‖1=m

(θ, gj ⊗Hµ

)L2(Λ;Rn)

µ!gj ⊗Hµ

exists (i.e., the series converges) in L2(P0; R). In fact, if ΠZ(m) denotesorthogonal projection onto Z(m), show that Π(Z(m))nθ(t) = θ(m) for almostevery t in the sense that ΠZ(m)

(ξ, θ(t)

)Rn =

(ξ, θ(m)(t)

)Rn for all ξ ∈ Rn and

almost every t ≥ 0.

(iii) Again let θ ∈ L2(Λ; Rn) be given. Set

F θm+1 =

∑j=1

∑‖µ‖1=1

(θ, gj ⊗Hµ

)L2(Λ;Rn)

µ!gj ⊗Gµ,

show that F θm+1 ∈ L2

((m+1); (Rn)m+1

), and both that θ(m) ∈ Dom(D>)

and D>θ(m) = I(m+1)

F θm+1

.

(iv) Given θ ∈ L2(P0; Rn), show that θ ∈ Dom(DT ) if and only if (cf.part (iii))

(6.3.15)∞∑

m=0

1(m+ 1)!

∥∥F θm+1

∥∥2

L2((m+1);(Rn)m+1)<∞,

in which case D>θ =∑∞

m=0 I(m+1)

F θm+1

. This expression is the one which Sko-

rohod gave for what is often called the Skorohod integral of θ.

(v) Notice that∥∥F θm+1

∥∥L2((m+1);(Rn)m+1)

≤ (m+ 1)

∥∥∥∥∥∥∑j=1

∑‖µ‖1=m

(θ, gj ⊗Hµ

)L2(Λ;Rn)

µ!gj ⊗ Gµ

∥∥∥∥∥∥L2((m+1);(Rn)m+1)

= (m+ 1)(m!)12∥∥θ(m)

∥∥L2(Λ;Rn)

,


and conclude that θ ∈ Dom(D>) if

(6.3.16)∞∑

m=0

(m+ 1)∫ ∞

0

∥∥Π(Z(m))nθ(t)∥∥2

L2(P0;Rn)dt <∞.

Exercise 6.3.17. The condition in (6.3.15) is nearly impossible to checkdirectly in general, even though there are special cases in which the onedo so. In addition, (6.3.16) is reasonably tractable. The present exerciseexamines some aspects of these issues. Because we know that its adjointD> is densely defined, we know that D itself is closable. Let D denote theclosure of D in L2(P 0).

(i) In the case when θ ∈ Θ2(P0; Rn), we know thatD>θ = Iθ and therefore

∞∑m=0

∥∥F θm+1

∥∥2

L2((m+1);(Rn)m+1)

(m+ 1)!= ‖D>θ‖2L2(P0;R) = ‖θ‖2L2(Λ;\Rn).

Give a proof of this based on the observation that F θm+1(t, τ1, . . . , τm) =

0 for almost every (t, τ1, . . . , τ − m) with t < τ1 ∨ · · · ∨ τm. Thus, evenwithout knowing how D> acts on Θ2(P0; Rn), one can see that Θ2(P0; Rn) ⊆Dom(D>).

(ii) In connection with idea in (i), set Bt = σ(p(τ)− p(t) : τ ∈ [t,∞)

),

and introduce the space Θ←

2(P0; Rn) of θ ∈ L2(Λ; Rn) which are reverse

progressively measurable in the sense that θ [t,∞) × C([0,∞); Rn)

)is

B[t,∞) × Bt-measurable for each t ≥ 0. Given θ ∈ Θ←

2(P0; Rn), show that

F θm+1(t, τ1, . . . , τm) = 0 for almost everywhere (t, τ1, . . . , τm) with t > τ1 ∧· · · ∧ τm, and conclude that Θ

←2(P0; Rn) ⊆ Dom(D>) and ‖D>θ‖L2(P0;R) =

‖θ‖L2(Λ;Rn) for θ ∈ Θ←

2(P0; Rn). Given these observations, it is not hard tointroduce the appropriate notion of backward Ito stochastic integration whichgives D>θ for θ ∈ Θ

←2(P0; Rn).

(iii) In the rest of this exercise we will be developing another way ofthinking about the condition in (6.3.16). To this end, note that, because itsadjoint D> is densely defined, D admits a closure D. Show that Z(m) ⊆Dom(D) for every m ∈ N and that

DΦ =∞∑

j=1

∑µ:‖µ‖1=m & µj≥1

(Φ,Hµ

)L2(P0;R)

µ(−j)!gj ⊗Hµ(−j)

for m ∈ Z+ and Φ ∈ Z(m). In particular, conclude that, for all m ∈ N,

Φ ∈ Z(m) =⇒∥∥DΦ

∥∥2

L2(P0;R)= m‖Φ‖2L2(P0;R).


(iv) Given Φ ∈ L2(P0; R), show that Φ ∈ Dom(D) if and only if∞∑

m=1

m‖ΠZ(m)Φ‖2L2(P0;R) <∞,

in which case the preceding sum is equal to the ‖DΦ‖2L2(Λ;Rn). Also, showthat (cf. part (iii) of Exercise 6.3.11) Φ ∈ Dom(N ) if and only if Φ ∈Dom(D) and DΦ ∈ S(P0; Rn), in which case NΦ = D> DΦ.

(v) If θ ∈ L2(Λ; Rn) and θ(t) ∈ Dom(D)n for almost every t ∈ [0,∞),show that t Dθ(t) is Lebesgue measurable and that (6.3.16) holds if∫

[0,∞)

∥∥Dθ(t)∥∥2

L2(P0;Rn)dt <∞.

For other applications of the same condition to Skorohod integration, seethe article [24].

Exercise 6.3.18. The route which we have taken to Ito’s RepresentationTheorem, Corollary 6.3.9, is not very efficient and does not give much insightinto how one might go about constructing the θ corresponding to a given Φ.In this exercise we will develop another, more straight-forward, approach.

(i) Let g` : 1 ≤ ` ≤ L ⊆ C1([0,∞); Rn

)be an orthonormal subset of

L2([0,∞); Rn

), and define t ∈ [0,∞) 7−→ A(t) ∈ Hom(RL; RL) so that

A`,`′(t) =∫ ∞

t

(g`(τ), g`′(τ)

)Rn dτ.

Next, given F ∈ Cb(RL; R), set

UF (t, x) =∫

RL

F (x+ y) ΓA(t)(dy) for (t, x) ∈ [0,∞)× RL,

where ΓA denotes the centered Gauss measure on RL with covariance A, andshow that

∂tUF (t, x) +12

L∑`,`′=1

(g`(t), g`′(t)

)Rn∂`∂`′UF (t, x) = 0

with limt∞

UF (t, x) = F (x) and UF (0, x) =∫

RL

F (y) ΓI(dy).

(ii) Referring to part (i), define t ∈ [0,∞) 7−→ σ(t) ∈ Hom(Rn; RL) sothat σk`(t) is the kth coordinate of g`(t), note that σ(t)σ(t)> = −A(t), andshow that

EΓI[F 2]− EΓI

[F]2 =

∫ ∞0

(∫RL

∣∣σ(t)>gradyUF (t, · )∣∣2 ΓI−A(t)(dy)

)dt.


Hint: A proof of the preceding can be based on the observation that

d

dtEΓI−A(t)

[UF (t, · )2

]= EΓI−A(t)

[∣∣σ(t)>gradUF (t, · )∣∣2].

(iii) Referring to parts (i) and (ii), suppose that Φ = F(Ig1(∞), . . . ,

IgL(∞)

), and show that

Φ = EP0[Φ] +

∫ ∞0

(θΦ(t), dp(t)

)Rn ,

where

θΦ(t, p) ≡L∑

j=1

∂jF(Ig1(t, p), . . . , IgL

(t))gj(t).

In particular, notice that this result can substitute for Theorem 6.3.8 in theargument used to prove Corollary 6.3.9.

Exercise 6.3.19. Let X ∈ L1(P0; R) be given, and set M(t) = EP0[X | Bt

]for each t ≥ 0.

(i) Show that (t, p) ∈ [0,∞) × C([0,∞); Rn)

)7−→ M(t, p) ∈ R can be

chosen so that(M(t),Bt,P0

)is a continuous martingale. (See part (iii) of

Exercise 7.3.9 below for more information.)

Hint: Choose Xk∞1 ⊆ L2(P0; R) so that Xk −→ X in L1(P0; R), lett Mk(t) to be a P0-almost surely continuous version of t EP0[

Xk | Bt

],

and use Doob’s Inequality to see that

limk→∞

sup`≥k

P0(‖M` −Mk‖[0,∞) ≥ ε

)= 0 for all ε > 0.

(ii) As an application of (i) above and the Martingale Convergence Theorem,show that if X : [0,∞)×C

([0,∞); Rn)

)−→ R is a Bt : t ≥ 0-progressively

measurable function with the properties that X(t) : t ≥ 0 is uniformlyP-integrable and X(s) = EP0[

X(t)∣∣Bs

]for all 0 ≤ s ≤ t, then there exists

a Bt : t ≥ 0-progressively measurable X : [0,∞) × C([0,∞); Rn)

)−→ R

such that X( · , p) is continuous for each p ∈ C([0,∞); Rn)

)and X(t) = X(t)

P-almost surely for each t ∈ [0,∞).

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