The law of B φ viewed as a regular conditional probability

For the moment let us think of our original Brownian sheetB = (Bxt; (x, t)∈

[0,1]2_{), which is almost surely continuous on [0,1]}2_{, as a canonical process.}

We take (C([0,1]2_),

B(C([0,1]2_{))) to be our underlying measurable space, on}

which we define a probability measure µ by µ(A) = P(B _∈ A) for all A _∈ B(C([0,1]2_{)). We now define a measurable map}

L : C([0,1]2₎

→ C([0,1]) by Lf(r) =f(x(r), t(r)) forf _∈C([0,1]2_{) andr}

∈[0,1]. _Lcorresponds toLin the sense thatµ(_{L ∈}A) =P(L_∈A) forA_{∈ B}(C([0,1])).

Definition 2.1. Let(Ω,F_{,P)be a probability space,(E,E)a measurable space}

andX : Ω→E a measurable function. We say that ν : E×F _{→ [0,1] is a}

regular conditional probability with respect toX if

• ν(x,·) is a probability measure onF _{for allx∈E;} • x7→ν(x, A)isE measurable function for allA∈F_; • for allA_∈F _{andD∈ E,}P(A_∩X−1_{(D)) =}R

Dν(x, A)PX(dx),

wherePX is the law ofX on(E,E), that isPX(D) =P(X ∈D) =P(X−1(D)).

Our setting then is the probability space (C([0,1]2_),

B(C([0,1]2_{)), µ), whilst}

(E,_E) = (C([0,1]),_B(C([0,1]))) andX =_L. In this case we know there exists a regular conditional probability with respect to L (see for example [RW87]), which we denote byP(B∈ ·|L=φ) for allφ∈C([0,1]). Note thatµL−1_=P

L.

Suppose now thatφ∈C([0,1]) is in supp(P_L), the topological support of P_L. By this we mean that for any N ∈ B(C([0,1])) such that φ ∈ N, PL(N) >

0. One soon sees in examples that we require this condition on φ if we are to solve (2.2.3) for Bφ_{. We further suppose that for all such φ, the equa-}

tion (2.2.3) has a solution Bφ _{in C([0,1]}2_{), and we define a measure µ}

φ on

(C([0,1]2_),

B(C([0,1]2_{))) byµ}

like to show that (µφ;φ ∈ C([0,1])) is a regular conditional probability with

respect to _L. Of course, we have not defined µφ yet forφ /∈supp(PL). How-

ever, if D ∈ B(C([0,1])), then there exists an open set D1 such that D1 ⊃

D∩supp(P_L)c _{and µ(L}−1_(D

1)) = 0. It follows that for allA ∈ B(C([0,1]2)),

R

Dµφ(A)PL(dφ)) = R

D∩supp(PL)µφ(A)PL(dφ)). Thus we may chooseµφ how-

ever we wish forφ /_∈supp(PL) and not affect the third property of the regular

conditional probability. We shall set µφ = 0 for such φ. It is now clear that

the first property of the regular conditional probability holds. Furthermore, assuming that supp(PL)c is in B(C([0,1])), the second property follows if we

can show that the mapφ_7→µφ(A) restricted to supp(PL) is measurable for any

A_{∈ B}(C([0,1]2_{)), for example if we can show this restricted map is continuous.}

Note that since the underlying probability space is assumed to be complete, µ, µφ and PL have natural extensions to the completions of B(C([0,1]2)) and

B(C([0,1])). Therefore, since supp(P_L)c _{is contained in a set of P}

L measure

zero, we can drop the above assumption by working with the completions of B(C([0,1]2_{)) and}

B(C([0,1])) under the measuresµand PL respectively. This

adjustment does not affect the first and third properties at all.

For the third condition we need to demonstrate for all D _{∈ B}(C([0,1]))_∩ supp(PL) andA∈ B(C([0,1]2)) thatµ(A∩ L−1(D)) =

R

Dµφ(A)PL(dφ)). Not-

ing thatµ(A_{∩ L}−1_{(D)) =P(}

{B_∈A_{} ∩ {}L_∈D_}), the third condition becomes

P(_{B_∈A_{} ∩ {}L_∈D_}) =

Z D

µφ(A)PL(dφ).

Suppose that the third property holds for some countable set {An;n ∈N} ⊂

B(C([0,1]2_{)), and we may as well assume that}

{An;n∈N} is increasing. It is

clear that the third property holds for Ac

1, and it also holds for

S

n∈NAn by

holds if we can show it holds for a collection of setsAwhich is large enough to generate_B(C([0,1]2_)).

If we can show that the third property holds for someA∈ B(C([0,1])), it follows that

Z

1D(φ) (P({B ∈A}|L=φ)−µφ(A))PL(dφ) = 0

for all D _{∈ B}(C[0,1]). This means that µφ(A) = P({B ∈ A}|L = φ) for

everyφexcept those belong to some _{N ∈ B}(C([0,1])) with PL(N) = 0. If we

now wish to pick some φ and condition on L = φ, we have a problem since we do not know if φ _{∈ N}. To overcome this, we instead look to show that φ _7→ µφ(A) is continuous (where convergence inC([0,1]) is in the supremum

norm). Recall that this would also confirm the second property, and thus that (µφ(A) : A ∈ B(C([0,1]2)), φ ∈ C([0,1])) is a regular conditional probability.

Our aim will be to demonstrate thatNc_{is dense in supp(P}

L). In this case we can

deduce thatµφ(A) =P(B∈A|L=φ) for allφ∈supp(PL). If we can show this

for enoughAto generateB(C([0,1]2)), we may deduce thatµφ=P(B∈ ·|L=φ)

for all suchφ.

2.2.4 Returning to our introductory example.

Let us return to the example of section (2.1.4). We can read from our previous calculations that the Brownian sheetB is given by

Bxt=xL(t) + (1−x) Z x 0 Z t 0 1 1−ydMys. In particular, if we takeA_{∈ B}(C([0,1]2_{)) of the form}

then we may expressP(B_∈A_|L) as some function

ψ(x1, . . . , xn, t1, . . . , tn, L(t1), . . . , L(tn)).

At the same time,µφ, the law onC([0,1]2) ofBφ, is given by

µφ(A) =ψ(x1, . . . , xn, t1, . . . , tn, φ(t1), . . . , φ(tn)).

by clear analogy for anyφ_∈supp(PL). We may now deduce that

P(_{B_∈A_{} ∩ {}L_∈D_}) =E[1 {L∈D}ψ(x1, . . . , xn, t1, . . . , tn, L(t1), . . . , L(tn))] = Z D µφ(A)PL(dφ). (2.2.5)

This shows that forPL almost everyφ∈supp(PL),P(B ∈A|L=φ) =µφ. (It

is not hard to see thatφ_∈supp(PL) if and only ifφ∈C([0,1]) andφ(0) = 0.)

It is also straightforward to see that φ _7→ µφ(A) is continuous in φ. Thus

(µφ;φ ∈ C([0,1]), φ(0) = 0) defines a regular conditional probability with

respect toL. We would also like to say thatP(B ∈A|L =φ) =µφ(A) for all

φ∈supp(P_L). This follows if we can show that any set ofP_Lmeasure 1 is dense under the supremum norm in supp(P_L). This is indeed the case, and the proof is very similar to that of the example in the next section.

It is not obvious whether we can find a general method for solving equation (2.2.3). We first need to calculate_C−1

y κy(1[0

,t]−σyt), if indeed we can actually

do this in non-trivial cases. Our best bet is to solve equation (2.1.3), but it is not difficult to find examples for which finding an explicit solution is difficult. One would hope to find explicit solutions in order to write an explicit form of (2.2.3), but even when we can do this, it is another matter altogether to solve

which is tractable and does not appear to be discussed in the literature.