Which Information Is Available at Time \(t\)?
A Brownian motion does not determine a filtration by itself. Its paths generate a natural supply of information, but one may also add null events or other information. The choice matters: the predictable representation property is a statement about martingales in a particular filtration, not merely about the distribution of \(B_t\).
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Fix a finite horizon \(T>0\). Let \(\mathcal{N}\) denote all subsets in \(\mathcal{F}\) of \(\mathbb{P}\)-null sets. The filtration generated by the Brownian path up to time \(t\) is its natural filtration. Completing it adds the null events, while taking its right-continuous augmentation also imposes right-continuity.
The raw filtration records only events determined by the path up to the stated time. Completion changes events only on sets of probability zero. Right-continuous augmentation asks that information available immediately after \(t\), no matter how soon after, already be regarded as available at \(t\). For Brownian motion, the completed natural filtration is right-continuous; the usual augmentation is its standard formulation. The proof of this right-continuity fact is a standard result treated with null-set completion and right-continuous Brownian filtrations.
In the usual augmented filtration, \(B_t\) is still Brownian motion: its paths are continuous, and for \(s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s^u\) and has distribution \(N(0,t-s)\). Completion alone preserves independence because adding or removing null sets does not change probabilities of events. The right-continuity assertion for the usual augmentation is part of the standard Brownian filtration result just noted.
Worked Example: A Null Event Added by Completion
Take \(T=1\). Consider \(A=\{B_{1/2}>0\}\) and \(N=\{B_1=0\}\). Identify when these events are available in the raw and completed natural filtrations.
By definition, \(A\in\mathcal{F}_{1/2}^0\), since it is determined by \(B_{1/2}\). It is therefore also in \(\mathcal{F}_{1/2}^c\) and \(\mathcal{F}_{1/2}^u\). No claim that \(A\) is unavailable earlier is needed here. Since \(B_1\sim N(0,1)\) has a continuous density, \(\mathbb{P}(N)=0\). Thus \(N\in\mathcal{N}\), so \(N\in\mathcal{F}_t^c\) for every \(t\ge0\), including \(t=0\). It also belongs to every \(\mathcal{F}_t^u\).
Check: completion does not make the event \(\{B_1=0\}\) positive-probability information. Its probability remains \(0\); completion only ensures the event is measurable at every time.
The Predictable Representation Property
For stochastic integration, the relevant integrand class on \([0,T]\) is \(\mathcal{L}^2\): progressively measurable processes \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Progressive measurability means that, for every \(t\), the restriction of \((s,\omega)\mapsto H_s(\omega)\) to \([0,t]\times\Omega\) is measurable with respect to \(\mathcal{B}([0,t])\otimes\mathcal{F}_t\). An Ito integral below is taken with respect to \(B_t\) under \(\mathbb{P}\), with an integrand in this class. The Ito integral is not the Stratonovich integral: for a continuous semimartingale integrand \(H\), the conversion is \(\int_0^t H_s\circ dB_s=\int_0^t H_s\,dB_s+\tfrac12[H,B]_t\), when the indicated integrals are defined.
The initial value \(M_0\) is retained in the formula. In the usual augmented natural Brownian filtration, \(\mathcal{F}_0^u\) contains only events of probability \(0\) or \(1\), by the standard Brownian zero-one law; hence \(M_0\) is constant almost surely. The representation theorem does not say every martingale in every filtration is a Brownian integral. The filtration hypothesis is essential.
Worked Example: Representing the Brownian Terminal Value
On \([0,1]\), set \(X=B_1\) and define \(M_t=\mathbb{E}[X\mid\mathcal{F}_t^u]\). Find and verify its representation.
For \(t\le1\), write \(B_1=B_t+(B_1-B_t)\). The increment \(B_1-B_t\) is independent of \(\mathcal{F}_t^u\) and has mean \(0\). Therefore \(M_t=B_t\). Since \(B_0=0\), this is \(M_t=\int_0^t 1\,dB_s\). The integrand \(H_s=1\) is progressively measurable and satisfies \(\mathbb{E}\int_0^1 1^2\,ds=1<\infty\).
For \(s\le t\), independence and the zero mean of \(B_t-B_s\) give \(\mathbb{E}[M_t\mid\mathcal{F}_s^u]=\mathbb{E}[B_t\mid\mathcal{F}_s^u]=B_s=M_s\). Also, \(\mathbb{E}[M_t^2]=t<\infty\), so this is a square-integrable martingale. At \(t=1\), the integral equals \(B_1=X\), as required.
Worked Example: A Quadratic Terminal Variable
On \([0,1]\), let \(X=B_1^2\). Compute \(M_t=\mathbb{E}[X\mid\mathcal{F}_t^u]\) and express it as an Ito integral plus its initial value.
Again write \(B_1=B_t+(B_1-B_t)\). The increment is independent of \(\mathcal{F}_t^u\), has mean \(0\), and has variance \(1-t\). Squaring and taking conditional expectation gives \(\mathbb{E}[B_1^2\mid\mathcal{F}_t^u]=B_t^2+(1-t)\), since the conditional expectation of the cross term \(2B_t(B_1-B_t)\) is \(0\). Thus \(M_t=B_t^2+1-t\) and \(M_0=1\).
To identify the integral, apply Ito's formula to \(f(x,t)=x^2+1-t\). Its derivatives are \(f_t=-1\), \(f_x=2x\), and \(f_{xx}=2\). The correction term is \(\tfrac12 f_{xx}\,dt=dt\), so \(d f(B_t,t)=-dt+2B_t\,dB_t+dt=2B_t\,dB_t\). Consequently \(M_t=1+\int_0^t 2B_s\,dB_s\). The integrand \(H_s=2B_s\) is progressively measurable, and \(\mathbb{E}\int_0^1(2B_s)^2\,ds=4\int_0^1s\,ds=2<\infty\). The martingale property also follows directly from the conditional-expectation calculation: \(M_t=\mathbb{E}[X\mid\mathcal{F}_t^u]\), with \(\mathbb{E}[X^2]=\mathbb{E}[B_1^4]=3<\infty\).
Check: at \(t=0\), the formula gives \(M_0=0^2+1=1=\mathbb{E}[B_1^2]\); at \(t=1\), it gives \(M_1=B_1^2=X\). Both endpoint values agree with the conditional-expectation definition.
Why the Filtration Hypothesis Matters
Adding null events is different from revealing new random information. Here is a concrete way to see why the predictable representation property cannot simply be transferred to an arbitrary larger filtration. Let \(U\) be independent of the entire Brownian path, with \(\mathbb{P}(U=0)=\mathbb{P}(U=1)=1/2\). Use the Brownian filtration before time \(1\), and reveal \(U\) at time \(1\): before time \(1\), take \(\mathcal{G}_t=\mathcal{F}_t^u\vee\sigma(U\mathbf{1}_{\{t\ge1\}})\) interpreted as adjoining no information about \(U\) for \(t<1\); from time \(1\) onward, take \(\mathcal{G}_t=\mathcal{F}_t^u\vee\sigma(U)\). Equivalently, the information at times \(t<1\) is just \(\mathcal{F}_t^u\), and at \(t\ge1\) it also contains \(U\).
The process \(M_t=\mathbb{E}[U\mid\mathcal{G}_t]\) is \(1/2\) for \(t<1\) and \(U\) for \(t\ge1\). This is a martingale: before the revelation its conditional mean is \(1/2\), and after revelation it stays equal to the known value \(U\). It has a jump at time \(1\) with probability \(1\), because \(U\) is either \(0\) or \(1\). Every Ito integral with respect to Brownian motion has continuous paths. Therefore this martingale cannot be represented as its initial value plus a Brownian Ito integral. The additional information is not a null-set modification of the Brownian filtration.
In a Brownian filtration, the representation theorem says that the martingale’s changes are captured by its Brownian integral. The integral is a martingale when its integrand is in \(\mathcal{L}^2\): the Ito isometry gives finite second moments, and the conditional-mean-zero property of Brownian increments gives the martingale property. A merely local martingale need not be a true martingale; the square-integrability condition in the theorem rules out that distinction here.
Common Pitfall: Changing the Filtration Silently
A tempting but incorrect argument is: “\(B_t\) is Brownian, so every martingale is an integral against \(B_t\).” The delayed-revelation example disproves this. Brownian motion can remain Brownian in a larger filtration while that filtration carries additional martingales. The correct theorem specifies the usual augmented natural Brownian filtration and a square-integrable martingale.
Another error is to confuse completion with new information. If \(N\) is a null event, completion places \(N\) in every completed sigma-field, but it does not change its probability or create a nontrivial random variable up to almost-sure equality. By contrast, the event \(\{U=1\}\) in the example has probability \(1/2\); revealing it changes what is known at time \(1\).
| Filtration | What is added? | Consequence |
|---|---|---|
| Raw natural | Only events generated by the Brownian path up to \(t\) | Path information is recorded without completing null sets |
| Completed natural | All subsets of \(\mathbb{P}\)-null events | Events equal almost surely are handled measurably |
| Usual augmentation | Completion and right-continuity | The standard filtration for the Brownian representation theorem |
| Enlarged by later information | New random information revealed at a later time | Additional martingales may not be Brownian integrals |
Check Your Understanding
Use the definitions and hypotheses above to answer these questions.
- What is added when passing from the raw natural filtration to its completion?
- For \(X=B_1\), compute \(\mathbb{E}[X\mid\mathcal{F}_t^u]\) and identify its integrand.
- For \(X=B_1^2\), explain why the Ito correction term cancels the time derivative in \(B_t^2+1-t\).
- Why does a Brownian integral representation require an integrability condition on \(H\)?
- In the delayed-revelation example, what feature of the martingale rules out a representation by a Brownian Ito integral?