Quadratic Variation and the Doob–Meyer Decomposition
For a smooth path, sums of squared increments disappear as the partition is refined. Brownian motion has a nonzero quadratic variation, and this changes the rules of integration. We first describe quadratic variation as the increasing process that must be subtracted from the square of a martingale. The Doob–Meyer decomposition then places this construction in a more general setting: it separates a supermartingale into a martingale part and accumulated decrease.
Work on a filtered probability space satisfying the usual conditions. \(B\) denotes Brownian motion, and stochastic integrands are predictable with the stated integrability. A local martingale need not be a true martingale: the moment conditions in each claim are essential. See Brownian motion and martingale inequalities.
A filtration specifies the information available to a stochastic process at each time \(t\). To fix ideas, we say \(\{\mathcal F_t\}_{t \geq 0}\) on a probability space \((\Omega, \mathcal F, \mathop{\mathrm{\mathbf{P}\!\,}})\) is called a filtration if it is a collection of sub-\(\sigma\)-algebras of \(\mathcal F\) such that \(\mathcal F_s \subset \mathcal F_t\) for all \(s \leq t\). The space \((\Omega,\mathcal F,\{\mathcal F_t\}_{t \geq 0},\mathop{\mathrm{\mathbf{P}\!\,}})\) is called a filtered probability space. Given a filtration, the left- and right-limits at any time can be defined as
The usual conditions mean that the filtration is right-continuous and that \(\mathcal F_0\) contains every subset of every \(\mathbb P\)-null set in \(\mathcal F\). A process \(X\) is adapted when \(X_t\) is \(\mathcal F_t\)-measurable. Its natural filtration is \(\mathcal F_t^X=\sigma(X_s:s\leq t)\). A martingale is square-integrable if \(X_t\in L^2\) for each finite \(t\). We use the stronger term \(L^2\)-bounded when \(\sup_{t\geq0}\mathbb{E} X_t^2<\infty\); this is equivalent to being closed by an \(L^2\) terminal variable, as follows.
For a continuous martingale \(X\), the following are equivalent: \(a\) \(X_t=\mathbb{E}[X_\infty\mid\mathcal F_t]\) for some \(X_\infty\in L^2(\mathcal F_\infty)\); \(b\) \(\sup_{t\geq0}\mathbb{E} X_t^2<\infty\); \(c\) \(\mathbb{E}\sup_{t\geq0}|X_t|^2<\infty\).
Assume (a). For \(s\leq t\), the tower property gives \(\mathbb{E}[X_t\mid\mathcal F_s]=\mathbb{E}[\mathbb{E}[X_\infty\mid\mathcal F_t]\mid\mathcal F_s]=X_s\). Conditional Jensen then gives the uniform bound
This implies (b).
Assume (b) holds. Then, Doob’s maximal inequality implies for all \(N \in \mathbb N\),
Writing \(Y = \sup_{t \geq 0} X_t^2\) and \(Y_N = \sup_{0 \leq t \leq N} X_t^2\), which are clearly nonnegative, we have
This implies (c).
Finally, under \(c\), the discrete-time martingale convergence theorem gives \(X_n\to X_\infty\) almost surely. The square-integrable dominating variable \(\sup_t|X_t|\) then gives \(L^2\) convergence by dominated convergence. For any fixed \(t\), the martingale identity \(X_t=\mathbb{E}[X_n\mid\mathcal F_t]\), valid for integers \(n\geq t\), passes to the limit in \(L^2\). Hence \(X_t=\mathbb{E}[X_\infty\mid\mathcal F_t]\), proving \(a\). ◻
For an \(L^2\)-bounded continuous martingale, the identity \(\mathbb E X_\tau=\mathbb E X_0\) holds at every finite stopping time \(\tau\). Here are the approximation details. For a stopping time taking finitely many grid values, write the stopped variable as \(X_0\) plus the sum of increments multiplied by the events that stopping has not yet occurred; each multiplier is measurable before its increment, whose conditional expectation is zero. For a general finite \(\tau\), round \(\tau\wedge n\) upwards to the grid of mesh \(2^{-n}\). The resulting finite-valued stopping times converge almost surely to \(\tau\). Continuity gives convergence of the sampled values, and domination by \(\sup_{t\geq0}|X_t|\in L^2\) gives convergence in \(L^1\). This proves the expectation identity used next.
Clearly, if \(X = \{X_t\}_{t \geq 0}\) is a L²-bounded martingale, and \(S \leq T\) are finite stopping times, then \(\mathbb{E}[X_T \mid \mathcal F_S]=X_S\). To see this, we simply need to truncate the process \(X\) at the stopping times in an appropriate way. Define, for any \(A \in \mathcal F_S\),
Therefore,
This means \(U\) is itself a stopping time. The expectation identity just proved then implies
This clearly implies \(\mathbb{E}[X_S ; A] = \mathbb{E}[X_T ; A]\), as desired. \(\blacksquare\)
Remark 1. We can immediately obtain the following results from the discussion above:
Suppose \(X\) is a L²-bounded martingale and \(T\) is a stopping time. Then, \(X_{t \wedge T}\) is a martingale with respect to the filtration \(\{F_{t \wedge T}\}\).
If \(\mathbb{E}X_T = 0\) for every bounded stopping time \(T\), and \(X\) is a process adapted to \(\{\mathcal F_t\}_{t \geq 0}\) such that \(X_t\) is integrable for all \(t\), then \(X_t\) is a martingale.
Optional sampling lets us use the martingale identity at random times. The next calculation records the corresponding second-moment identity, which will later control stopped increments.
Relatedly, note that a direct calculation
So taking expectations again, we have \(\mathbb{E}[(X_T - X_S)^2] = \mathbb{E}X_T^2 - \mathbb{E}X_S^2\). \(\blacksquare\)
Remark 2. A zero-start continuous local martingale of locally finite variation is identically zero. Stop it when its total variation reaches \(n\), and at time \(n\). The stopped process is a bounded martingale. Orthogonality of its increments gives \(\mathbb{E}X_T^2=\mathbb{E}\sum_i(\Delta X_i)^2\leq\mathbb{E}[\max_i|\Delta X_i|\operatorname{Var}_{[0,T]}(X)]\to0\) along vanishing-mesh partitions, by continuity and the stopped bounds. Remove the stops, and then use rational times and continuity. This also proves uniqueness whenever two continuous compensators differ by a local martingale.
1. Prequel: Doob-Meyer Decomposition
A continuous local martingale \(X\) starting at zero has a unique continuous adapted increasing process \(\langle X\rangle\), starting at zero, such that \(X^2-\langle X\rangle\) is a local martingale. If \(X\) is \(L^2\)-bounded, the compensated square is a true martingale. For Brownian motion, \(\langle B\rangle_t=t\). Brownian motion is square-integrable at every finite time; it is not \(L^2\)-bounded over the entire half-line. Stopping at a deterministic horizon supplies the global bound used in the initial integral construction.
Here is the construction behind this assertion. For a bounded zero-start martingale and a partition \(\pi\) of \([0,T]\), let \(Y_t^\pi=\sum_{[u,v]\in\pi}X_u(X_{t\wedge v}-X_{t\wedge u})\). On a common refinement, differences of two such sums are predictable martingale transforms. Orthogonality and Cauchy–Schwarz give
where \(D_j\) is the difference between the two left-endpoint coefficients. If \(|X|\leq D\), expansion and conditional orthogonality give \(\mathbb{E} Q_\rho^2\leq12D^4\); continuity makes the first factor tend to zero. Doob’s inequality then gives a continuous martingale limit \(Y\), uniformly in \(L^2\). The identity \(X_{t_i}^2-2Y_{t_i}^\pi=\sum_{j\leq i}(\Delta X_j)^2\) makes \(A=X^2-2Y\) increasing, by approximation of times with partition points. Remark 2 proves uniqueness. Continuous hitting times of \(|X|\), with deterministic cutoffs, extend the construction to local martingales. The full common-refinement estimates and the treatment of nonzero initial values appear in Theorem 2 and Lemma 3.
Remark 3. The exact increment identity
explains quadratic covariation. For smooth paths on a partition of \([0,t]\) into \(n\) pieces of length \(t/n\), the sum of increment products tends to zero. For \(X=Y=B\), its expectation is \(t\), and its variance is \(2t^2/n\), so it converges in \(L^2\) to \(t\). For independent Brownian motions the covariance sum instead converges to zero: its mean is zero and its variance is \(t^2/n\). Thus nonzero individual variations do not force nonzero mutual covariation. Polarisation defines \(\langle X,Y\rangle=(\langle X+Y\rangle-\langle X-Y\rangle)/4\).
Remark 4. For continuous semimartingales, squared-increment sums along stopping-time partitions with mesh tending to zero in probability converge uniformly on compact intervals in probability to quadratic variation. To see the martingale part, write the error as \(2\int_0^t(M_s-M_{\pi(s)})\,dM_s\), using the product formula. Stop so that \(M\) and its bracket are bounded. Continuity and the mesh assumption make the supremum of the integrand tend to zero in probability; boundedness gives convergence of its second moment, and the isometry and Doob’s inequality give uniform convergence of the error. Finite-variation terms vanish by the maximum-increment times variation bound. Polarisation treats two processes. This statement concerns continuous semimartingales; no regularity-free claim about arbitrary submartingales, or stronger semimartingale-topology convergence, is needed here.
Remark 5. It is not hard to see that if \(X\) is a cadlag martingale, then, for each \(t \geq 0\), \(\{X_\tau : \tau \leq t \text{ is a stopping time}\}\) is uniformly integrable. To see this, note the following general result: if \(X \in L^1(\mathop{\mathrm{\mathbf{P}\!\,}})\), then \(G \triangleq \{Y: Y = \mathbb{E}[X\mid\mathcal{G}] ~,~ \mathcal{G} \subset \mathcal{F} \}\) is uniformly integrable. Furthermore, optional sampling implies for any cadlag martingale \(X\), \(X_\tau = \mathbb{E}[X_t \mid \mathcal F_\tau]\) for stopping times \(\tau \leq t\). Combining both gives the desired statement.
A process \(Z\) is of class D when the family \(\{Z_T:T<\infty\text{ a stopping time}\}\) is uniformly integrable. The continuous Doob–Meyer theorem states that a continuous adapted supermartingale of class D has a unique decomposition
where \(A\) is continuous, adapted and increasing, and \(M\) is a continuous local martingale. The same assertion holds locally for continuous supermartingales. We give the continuous approximation argument. The theorem and this approach are discussed in Bass (2011, Theorem 9.12); the estimates below use nonoptimal constants and keep the terminal value in the averaging identity.
We first prove a comparison estimate. Suppose \(A^1,A^2\) are continuous adapted increasing processes, start at zero, and satisfy \(\mathbb{E}[A^j_\infty-A^j_t\mid\mathcal F_t]\leq K\) for every \(t\). Let \(C=A^1+A^2\), \(B=A^1-A^2\), and suppose \(V\geq0\), \(V\in L^2\), with
The conditional bounds extend to stopping times by approximation from above and right-continuous versions of the closed martingales. For a single \(A^j\), apply this at its first passage above \(x\). Continuity and Tonelli give
In particular \(\|C_\infty\|_2\leq2\sqrt2K\). The pathwise integration identity \(B_\infty^2=2\int_0^\infty(B_\infty-B_t)\,dB_t\) can be approximated by sums using right endpoints. Their increments are measurable at those endpoints, so conditioning there, rather than at the start of a future increment, yields
To justify the limiting sums, first restrict to a finite horizon and refine the partition. Their absolute values are dominated by a constant times \(C_\infty^2\), which is integrable. For the conditional bound use \(|\Delta B|\leq\Delta C\); the corresponding expected sum equals at most \(\mathbb{E}[V C_\infty]\). Then let the horizon increase. Finally \(B_t=\mathbb{E}[B_\infty\mid\mathcal F_t]-\mathbb{E}[B_\infty-B_t\mid\mathcal F_t]\), so Doob’s inequality and \((x+y)^2\leq2x^2+2y^2\) give
For comparison with continuous time, the discrete decomposition has a simple form. For a supermartingale \(Z_n\), set \(a_n=Z_n-\mathbb{E}[Z_{n+1}\mid\mathcal F_n]\geq0\), \(A_0=0\), \(A_n=\sum_{j<n}a_j\). Then \(A_{n+1}\) is \(\mathcal F_n\)-measurable and \(M_n=Z_n+A_n\) satisfies \(\mathbb{E}[M_{n+1}\mid\mathcal F_n]=M_n\).
Now assume first that \(|Z_t|\leq N\) and \(Z_t=Z_N\) for \(t\geq N\). Define its measurable modulus \(V_\delta=\sup_{|s-t|\leq\delta}|Z_s-Z_t|\), with the supremum taken over nonnegative rational times. Continuity and eventual constancy give \(V_\delta\to0\) almost surely and in \(L^2\). For \(h>0\), take a jointly measurable adapted version \(Q_s^h\) of \(\mathbb{E}[Z_{s+h}\mid\mathcal F_s]\), and set
Such a version is constructed below. Nonnegativity holds for almost every \((\omega,s)\), by the supermartingale inequality and Fubini. Thus \(A^h\) has continuous increasing adapted paths, starts at zero, and is constant after \(N\). Conditional Fubini and telescoping the finite time integrals give
The term \(Z_N\) is essential. The conditional tail is bounded by \(2N\). For \(0<k<h\), the difference of two such tails is bounded in absolute value by \(\mathbb{E}[V_h\mid\mathcal F_t]\). The comparison estimate proves
Completeness in the expected squared uniform norm gives a limit \(A\). A subsequence converges uniformly almost surely, so \(A\) is continuous, increasing, adapted and starts at zero. Passing to the limit in the conditional-tail identity yields
Hence \(M=Z+A\) is a true martingale in the bounded case.
Here are the promised measurability details. Put \(Y_t=Z_{t+h}\), and let \(Y_t^n=Y_{k2^{-n}}\) on \([k2^{-n},(k+1)2^{-n})\). On each interval take a right-continuous version of the bounded martingale \(\mathbb{E}[Y_{k2^{-n}}\mid\mathcal F_t]\), and join these versions to obtain \(Q^n\). It is right-continuous, adapted and jointly measurable. For \(n>m\), conditional domination, first at rational times and then by right-continuity, gives
Doob’s inequality makes the right side tend to zero in \(L^2\). A subsequence therefore converges uniformly almost surely to a right-continuous adapted process \(Q^h\). For each fixed \(t\), the \(L^2\) contraction of conditional expectation identifies its limit as \(\mathbb{E}[Y_t\mid\mathcal F_t]\). This supplies the required version. It need not be continuous: the time integral defining \(A^h\) is continuous regardless.
For general \(Z\), first restrict to the \(\mathcal F_0\)-measurable events \(E_m=\{|Z_0|\leq m\}\). On \(E_m\), stop at \(\tau_{m,n}=\inf\{t\geq0:|Z_t|\geq n\}\wedge n\), for \(n>m\). The process \(\mathbb1_{E_m}Z^{\tau_{m,n}}\) is a bounded continuous supermartingale, so it has the decomposition just constructed. Optional sampling is valid for the class-D process; for a merely local assertion one works with bounded finite-horizon stops. Two decompositions agree wherever their stopped original processes agree: their difference is a zero-start continuous finite-variation local martingale and vanishes by Remark 2. Thus the decompositions on \(E_m\) agree as \(m,n\) increase. Patching defines continuous \(A,M\) globally, with \(A_0=0\), and \(M\) a local martingale. For example, on \(E_n^c\) use localising time zero, and on \(E_n\) use the level \(n+1\) stop with time cutoff \(n\); the unmodified initial value on the complement is integrable. These times increase to infinity and give integrable stopped martingales. The same finite-variation argument proves uniqueness. ◻
Reading: Le Gall (2016), Brownian Motion, Martingales, and Stochastic Calculus.