Before Applying a Representation Theorem
Brownian martingale representation is powerful, but it is not a statement about every martingale on every filtered probability space. Its conclusion depends on the filtration, the integrability of the martingale, and the kind of stochastic integral being used. Even when a representation exists, its integrand is unique only up to equality almost everywhere with respect to time and probability. These qualifications are not technical footnotes: each one rules out a common, concrete error.
Throughout, let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) be a filtered probability space carrying a standard Brownian motion \(B_t\). Unless stated otherwise, a Brownian filtration means the completed, right-continuous natural filtration generated by \(B\). An Ito integral \(\int_0^t H_s\,dB_s\) is taken with a progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) on each stated finite time interval. For a local Ito integral, the corresponding requirement is local square-integrability: there are stopping times increasing to infinity such that the stopped integrands satisfy this finite-interval condition.
Notice the initial value \(M_0\). A stochastic integral starting at zero cannot represent a nonzero initial value. For terminal variables, the constant term is \(\mathbb{E}[X]\); for a filtration with nontrivial \(\mathcal{F}_0\), the relevant initial term is instead \(\mathbb{E}[X\mid\mathcal{F}_0]\), and the Brownian-filtration theorem above cannot simply be transferred to that setting.
Failure 1: The Filtration Contains Extra Information
Adding information to a filtration can create martingales that change in ways Brownian stochastic integrals cannot. To see this without any integrability complication, take \(T=1\), let \(Y\) be independent of \(B\), and suppose \(\mathbb{P}(Y=1)=\mathbb{P}(Y=-1)=\tfrac12\). Fix \(a=\tfrac12\). Use the usual augmented filtration that observes the Brownian path up to time \(t\), and also reveals \(Y\) at time \(a\): before \(a\), \(Y\) is not observed; from \(a\) onward, it is observed. This is an enlarged filtration, not the Brownian natural filtration.
Worked Example: A Martingale with a Newly Revealed Jump
Define \(M_t=0\) for \(0\le t<a\) and \(M_t=Y\) for \(a\le t\le1\). Verify that \(M\) is a square-integrable martingale, and decide whether it can be represented as a Brownian Ito integral plus its initial value.
Before \(a\), \(\mathbb{E}[Y\mid\mathcal{F}_t]=\mathbb{E}[Y]=0\), because \(Y\) is independent of the Brownian information observed so far. At and after \(a\), \(Y\) is known, so \(\mathbb{E}[Y\mid\mathcal{F}_t]=Y\). Thus \(M_t=\mathbb{E}[Y\mid\mathcal{F}_t]\) at every time, which verifies the martingale property by the tower property of conditional expectation. Also \(\mathbb{E}[M_t^2]\le\mathbb{E}[Y^2]=1\), so it is square-integrable. Its initial value is \(M_0=0\), and its jump at \(a\) is \(\Delta M_a=Y\), which is nonzero with probability one.
For every admissible integrand \(H\), the process \(I_t=\int_0^t H_s\,dB_s\) has continuous paths (up to indistinguishability). Therefore \(M_t=M_0+I_t\) would have continuous paths, contradicting \(\Delta M_a=Y\ne0\) almost surely. No such representation exists. Check: the failure is not a lack of square-integrability: \(\mathbb{E}[Y^2]=1\). It is the extra information revealed at \(a\).
Why the continuity argument works. Ito integrals with respect to Brownian motion have continuous versions. Adding the constant \(M_0\) preserves continuity. Consequently, if a martingale has a jump with positive probability, it cannot equal a Brownian integral plus its initial value. This necessary condition is useful even when the filtration is not explicitly described: a jump is enough to disprove a proposed Brownian-only representation.
Failure 2: Integrability and Localization Are Omitted
The L2 theorem requires \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The Ito isometry, \(\mathbb{E}\left[\left(\int_0^T H_s\,dB_s\right)^2\right]=\mathbb{E}\int_0^T H_s^2\,ds\), applies to integrands in this square-integrable class. It does not license treating an arbitrary progressively measurable process as an L2 integrand. When only local square-integrability is available, stochastic integration is defined after stopping; the resulting process is a continuous local martingale, but need not be a true martingale or have a square-integrable terminal value.
Worked Example: A Locally Defined Integral That Is Not L2 on the Closed Interval
On \(0\le t<1\), take the deterministic integrand \(H_t=(1-t)^{-1/2}\). Check whether it is an L2 integrand on \([0,1]\), and identify what can still be said before time \(1\).
For every \(u<1\), \(H\) is square-integrable on \([0,u]\), since \(\int_0^u H_s^2\,ds=\int_0^u(1-s)^{-1}\,ds=-\log(1-u)<\infty\). For the full interval, however, \(\int_0^1(1-s)^{-1}\,ds=\infty\). Thus \(H\) does not belong to the L2 integrand class on \([0,1]\); the L2 representation theorem and the Ito isometry on that closed interval cannot be invoked for it.
For \(t<1\), \(I_t=\int_0^t(1-s)^{-1/2}\,dB_s\) is well-defined and has \(\mathbb{E}[I_t]=0\) and \(\mathbb{E}[I_t^2]=-\log(1-t)\). These values follow from the Ito isometry and the zero-mean property for square-integrable Ito integrals on \([0,t]\). As \(t\) increases to \(1\), this second moment tends to infinity. Check: at \(t=\tfrac12\), the integrated squared integrand is \(-\log(\tfrac12)=\log 2\), finite; at \(t=1\), the integral diverges. This example licenses a local integral before \(1\), not an L2 representation through time \(1\).
For a continuous local martingale, localization means choosing stopping times \(\tau_n\) increasing to the terminal horizon (or to infinity) so that each stopped process is a true martingale with the required integrability. A local representation theorem asserts a representation under appropriate Brownian-filtration and local-integrability hypotheses; it does not say that the unstopped martingale is automatically square-integrable or that its terminal value lies in L2. A local martingale may fail to be a true martingale because its expectations need not be preserved without uniform integrability or another suitable condition. Do not replace a stated L2 hypothesis with the word “local” and then use L2 conclusions.
Failure 3: Treating Integrands as Pointwise Unique
Uniqueness of the representing integrand has a specific meaning. Two progressively measurable integrands represent the same Ito integral on \([0,T]\) if they agree almost everywhere with respect to the product measure \(dt\,d\mathbb{P}\). Pointwise equality for every time and outcome is not required. Conversely, a difference on a set of positive product measure cannot be dismissed as a mere choice of version.
Proof. Put \(G=H-K\). It is progressively measurable and L2, since \((H-K)^2\le2H^2+2K^2\). Linearity of the Ito integral gives \(\int_0^tG_s\,dB_s=0\) for every \(t\), up to indistinguishability. At \(t=T\), the Ito isometry yields \(\mathbb{E}\int_0^T G_s^2\,ds=\mathbb{E}[(\int_0^T G_s\,dB_s)^2]=0\). A nonnegative random variable with expectation zero is zero almost surely, so \(\int_0^T G_s^2\,ds=0\) almost surely. This is precisely \(G=0\) almost everywhere with respect to \(dt\,d\mathbb{P}\). Conversely, if that equality holds, then \(\mathbb{E}\int_0^T G_s^2\,ds=0\); the Ito isometry gives \(\mathbb{E}[(\int_0^T G_s\,dB_s)^2]=0\). Applying the same argument at each time (or using the continuous versions and a countable dense set of times) shows the integral process is identically zero up to indistinguishability. Therefore the two integrals agree. \(\square\)
Worked Example: Different Pointwise Values, the Same Integral
On \([0,1]\), define \(H_t=1\) when \(t\) is rational and \(H_t=0\) when \(t\) is irrational. Compare its integral with that of \(K_t=0\).
Both integrands are deterministic and Borel measurable, hence progressively measurable. The rational numbers in \([0,1]\) have Lebesgue measure zero, so \(\mathbb{E}\int_0^1(H_s-K_s)^2\,ds=\int_0^1 H_s^2\,ds=0\). Consequently, the Ito isometry gives \(\mathbb{E}[(\int_0^1H_s\,dB_s-\int_0^1K_s\,dB_s)^2]=0\). The integrals agree almost surely at time \(1\), and their continuous versions are indistinguishable on the whole interval.
Yet \(H_t\ne K_t\) at every rational \(t\in[0,1]\). Check: the difference occurs at many individual times, but on a set of zero time measure, so it does not change the stochastic integral. The correct uniqueness statement is equality \(dt\,d\mathbb{P}\)-almost everywhere, not equality at every time.
Failure 4: Representing a Discontinuous Martingale with Brownian Noise
Brownian stochastic integrals are continuous, so they cannot represent a martingale with jumps. This remains true even if the martingale is square-integrable and the Brownian motion is present in the filtration. An additional source of randomness, such as a counting process, may require additional martingale integrators; it cannot be silently absorbed into an integral against \(B\).
Worked Example: A Compensated Poisson Martingale
Let \(N_t\) be a Poisson process of rate \(\lambda=2\), independent of \(B\), and enlarge the filtration to observe both processes. On \([0,1]\), set \(M_t=N_t-2t\). Verify the martingale and integrability claims, then test Brownian representability.
For \(s\le t\), the increment \(N_t-N_s\) is independent of the information at time \(s\) and has mean \(2(t-s)\). Hence \(\mathbb{E}[M_t\mid\mathcal{F}_s]=N_s+2(t-s)-2t=N_s-2s=M_s\). Also \(M_t\) is square-integrable: \(N_t\) has mean \(2t\) and variance \(2t\), so \(\mathbb{E}[M_t^2]=\operatorname{Var}(N_t)=2t<\infty\). At each jump of \(N\), \(M\) jumps by \(1\); such jumps occur with positive probability on \([0,1]\).
If \(M_t=M_0+\int_0^tH_s\,dB_s\) for an admissible progressively measurable \(H\), the right-hand side would be continuous, while \(M\) has jumps with positive probability. This contradiction rules out the representation. Check: \(\mathbb{E}[M_1^2]=2\), so the obstruction is not failure of L2 integrability; it is the jump component of the martingale.
A Practical Diagnostic
Before asserting a Brownian representation, check each question in order:
- Which filtration? Is it the completed, right-continuous natural Brownian filtration required by the L2 theorem, or has it been enlarged with additional information?
- Which integrability regime? Is the terminal variable square-integrable, or are only local hypotheses available? Use the corresponding theorem and do not infer global L2 conclusions from localization alone.
- What is the initial value? A representation of a martingale has the form \(M_t=M_0+\int_0^tH_s\,dB_s\); the integral alone starts at zero.
- What does uniqueness mean? Compare integrands \(dt\,d\mathbb{P}\)-almost everywhere, not pointwise.
- Does the path have jumps? A Brownian Ito integral is continuous; a martingale with jumps cannot be represented by that integral alone.
The same discipline applies when checking a proposed integrand. Verify progressive measurability and the appropriate square-integrability or local square-integrability condition; then verify that the claimed integral has the required initial value and martingale behavior. A formal differential identity is not a substitute for these checks.
Check Your Understanding
For each question, identify the hypothesis or conclusion that needs checking before you apply a representation theorem.
- A square-integrable martingale is observed in a filtration that reveals an independent random variable at a fixed positive time. Which part of the Brownian representation argument can fail?
- For \(H_t=(1-t)^{-1/2}\) on \(0\le t<1\), compute \(\int_0^u H_s^2\,ds\) for \(u<1\), and explain why the L2 theorem on \([0,1]\) does not apply.
- Two L2 integrands differ on all rational times but agree at all irrational times. What is the correct criterion for their Ito integrals to agree?
- Why can a compensated Poisson process not equal a Brownian Ito integral plus a constant?
- State the form of the representation for a martingale whose initial value is not zero.