Two Components of Continuous Stochastic Motion
Many continuous stochastic processes combine two different kinds of motion. One part is a local martingale, whose increments have no predictable drift in the local sense; the other has paths of finite variation and records accumulated drift. A continuous semimartingale is a process that can be represented as the sum of these parts. The representation is useful because it identifies which component carries the stochastic fluctuation and which accumulates ordinary time-integrated change.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). A process \(M\) is a local martingale if there are stopping times \(\tau_n\) increasing to infinity almost surely such that each stopped process \(M_{t\wedge\tau_n}\) is a martingale. A martingale, unlike a local martingale, must itself be integrable at each time and satisfy \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\). A local martingale is not automatically a martingale: localization does not, by itself, guarantee global integrability or preservation of expectations in the limit.
The finite-variation component need not be deterministic or increasing. It may depend on the path, and it may increase and decrease. For example, \(A_t=\int_0^t b_s\,ds\) has finite variation whenever \(\int_0^T|b_s|\,ds<\infty\) almost surely. The local-martingale component need not be a true martingale. When it is a true martingale, its integrability and conditional-expectation property can be used directly; when it is only local, those conclusions require additional hypotheses.
Why the Decomposition Is Unique
The key fact is that a continuous local martingale whose paths have finite variation cannot have nonconstant motion. Here is a proof that uses partition sums and localization.
Proof. Set \(N=M-M'\). Then \(N=A'-A\), so \(N\) is both a continuous local martingale and a continuous finite-variation process. Subtract its initial value and write \(L_t=N_t-N_0\). This is a continuous local martingale with \(L_0=0\), and it has finite-variation paths. Write \(V_t\) for the total variation of \(L\) on \([0,t]\). For each positive integer \(n\), stop \(L\) when \(|L_t|+V_t\) first reaches \(n\), and call that time \(\rho_n\). Continuity implies that the stopped process \(L^{\rho_n}\) and its total variation on each finite interval are bounded by \(n\). It is therefore a bounded local martingale, hence a true martingale. For a fixed \(T\), take partitions of \([0,T]\) with mesh tending to zero. Martingale increments on disjoint partition intervals are orthogonal in expectation, so $$ \mathbb{E}\left[(L_{T\wedge\rho_n})^2\right] =\mathbb{E}\left[\sum_i\big(L_{t_i\wedge\rho_n}-L_{t_{i-1}\wedge\rho_n}\big)^2\right]. $$ For each path, the sum on the right is bounded above by the largest absolute increment times the total variation of \(L^{\rho_n}\) on \([0,T]\). The largest increment tends to zero by continuity, and the total variation is bounded by \(n\). The sum therefore tends to zero pathwise. It is bounded by \(n^2\), so dominated convergence gives \(\mathbb{E}[(L_{T\wedge\rho_n})^2]=0\). Thus \(L_{T\wedge\rho_n}=0\) almost surely. For each path and fixed \(T\), \(\rho_n>T\) for all sufficiently large \(n\), because \(L\) is bounded and has finite variation on \([0,T]\). Hence \(L_T=0\) almost surely. Applying this at every fixed \(T\) shows that \(N_t=N_0\) at each time, and continuity gives the constant-process identity. Since \(N=A'-A\), the finite-variation parts differ by the opposite constant. If both start at zero, that constant is zero. \(\square\)
The theorem says that normalization matters only for the constants: once the finite-variation part is required to start at zero, both components are determined. This is a structural statement, not a claim that the martingale component is necessarily a true martingale.
Three Basic Classifications
Worked Example: Brownian Motion
Classify \(X_t=B_t\) as a continuous semimartingale and verify the martingale property of its martingale component.
Take \(M_t=B_t\) and \(A_t=0\). Brownian paths are continuous, and the zero process has finite variation. Brownian motion is integrable since \(B_t\sim N(0,t)\), so \(\mathbb{E}|B_t|<\infty\). For \(s\le t\), the increment \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has mean zero. Therefore $$ \mathbb{E}[B_t\mid\mathcal{F}_s] =B_s+\mathbb{E}[B_t-B_s\mid\mathcal{F}_s] =B_s. $$ Thus \(B\) is a true martingale, and \(B_t=B_t+0\) is its normalized semimartingale decomposition.
Check: the finite-variation part has initial value zero, as required, and the conditional expectation calculation returns exactly the present value \(B_s\). By uniqueness, no nonzero finite-variation term starting at zero can also appear in this decomposition.
Worked Example: Drifted Brownian Motion
Let \(X_t=2+3t-2B_t\). Find its normalized decomposition and compute its mean and conditional mean.
Write \(X_t=X_0+M_t+A_t\) with $$ X_0=2,\qquad M_t=-2B_t,\qquad A_t=3t. $$ The process \(M\) is a continuous martingale: it is integrable, and for \(s\le t\), $$ \mathbb{E}[M_t\mid\mathcal{F}_s] =-2\mathbb{E}[B_t\mid\mathcal{F}_s] =-2B_s=M_s. $$ The process \(A\) is continuous and has variation \(3T\) on \([0,T]\). Since \(\mathbb{E}[B_t]=0\), the mean is $$ \mathbb{E}[X_t]=2+3t. $$ Conditioning on \(\mathcal{F}_s\) gives $$ \mathbb{E}[X_t\mid\mathcal{F}_s] =2+3t-2B_s =X_s+3(t-s). $$ The additional term \(3(t-s)\) is the accumulated drift between \(s\) and \(t\).
Check: subtracting \(X_s=2+3s-2B_s\) from the conditional mean gives \(3t-3s=3(t-s)\), consistent with the finite-variation increment \(A_t-A_s\). If the coefficient of \(t\) were zero, the conditional mean would instead equal \(X_s\).
Itô Processes as Semimartingales
Let \(b_s\) be progressively measurable with \(\int_0^T|b_s|\,ds<\infty\) almost surely, and let \(H_s\) be progressively measurable with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The latter condition says \(H\in\mathcal{L}^2\) on \([0,T]\). The integral \(\int_0^t H_s\,dB_s\) is the Itô integral, not a Stratonovich integral. The standard Itô-integral martingale theorem states that it is a square-integrable martingale and gives \(\mathbb{E}[\int_0^t H_s\,dB_s\mid\mathcal{F}_u]=\int_0^u H_s\,dB_s\) for \(u\le t\). This result follows from the construction of the Itô integral and its isometry.
Proof. The time integral is continuous and its total variation on \([0,T]\) is at most \(\int_0^T|b_s|\,ds\), which is finite almost surely. The Itô integral has a continuous version and, by the stated square-integrable martingale theorem, is a continuous martingale. Thus \(X=X_0+M+A\) has the required form. Both \(M\) and \(A\) start at zero, so uniqueness follows from the theorem above. \(\square\)
Worked Example: A Deterministic Integrand
Consider \(X_t=\int_0^t(1+s)\,ds+\int_0^t s\,dB_s\). Identify its parts and calculate its mean and variance at \(t=2\).
Here \(b_s=1+s\) and \(H_s=s\). The drift is finite because \(\int_0^T(1+s)\,ds=T+\tfrac12T^2\). The integrand is deterministic and progressively measurable, and $$ \mathbb{E}\int_0^T s^2\,ds=\frac{T^3}{3}<\infty. $$ Thus the decomposition has \(A_t=t+\tfrac12t^2\) and \(M_t=\int_0^t s\,dB_s\). The Itô-integral martingale property gives \(\mathbb{E}[M_t]=0\), so \(\mathbb{E}[X_t]=t+\tfrac12t^2\). By the Itô isometry, \(\mathbb{E}[M_t^2]=\int_0^t s^2\,ds=t^3/3\). Since \(A_t\) is deterministic, \(\operatorname{Var}(X_t)=t^3/3\). At \(t=2\), substitution gives $$ \mathbb{E}[X_2]=2+\frac12(2^2)=4,\qquad \operatorname{Var}(X_2)=\frac{2^3}{3}=\frac83. $$
Check: the drift integral at \(t=2\) is \([s+\tfrac12s^2]_0^2=2+2=4\), while the stochastic integral has mean zero and variance \([s^3/3]_0^2=8/3\), matching both calculations.
Worked Example: An Itô Integral with Brownian Integrand
Let \(X_t=\int_0^t B_s\,dB_s\). Verify that this is a semimartingale and compute its mean and variance at \(t=1\).
The integrand \(H_s=B_s\) is progressively measurable, and $$ \mathbb{E}\int_0^1 B_s^2\,ds=\int_0^1 s\,ds=\frac12<\infty. $$ Thus \(H\in\mathcal{L}^2\), and the Itô integral is a square-integrable martingale. Its decomposition has \(M=X\) and \(A=0\). To check its moments independently, apply Itô's formula to \(f(x)=x^2\), for which \(f'(x)=2x\) and \(f''(x)=2\). The correction term is \(\tfrac12f''(B_s)\,ds=ds\), giving $$ d(B_t^2)=2B_t\,dB_t+dt,\qquad X_t=\frac12(B_t^2-t). $$ Since \(\mathbb{E}[B_1^2]=1\), this identity gives \(\mathbb{E}[X_1]=\tfrac12(1-1)=0\). Since \(B_1\sim N(0,1)\), \(\mathbb{E}[B_1^4]=3\), and hence $$ \operatorname{Var}(X_1) =\mathbb{E}[X_1^2] =\frac14\mathbb{E}[(B_1^2-1)^2] =\frac14(3-2+1) =\frac12. $$ The independent Itô-isometry calculation gives \(\mathbb{E}[X_1^2]=\mathbb{E}\int_0^1B_s^2\,ds=\tfrac12\), confirming the variance.
Common Mistake: Treating Every Local Martingale as a Martingale
A decomposition \(X=M+A\) does not say that \(M\) has a finite expectation at every time. It says that \(M\) becomes a martingale after suitable stopping. The distinction matters when using conditional expectations or computing means: those operations require the relevant integrability and martingale hypotheses, not merely a local-martingale label.
Wrong calculation: for an arbitrary local martingale \(M\), one might write \(\mathbb{E}[M_t]=\mathbb{E}[M_0]\) simply because “martingale increments have mean zero.” That argument uses a property that has not been established for \(M\). A local martingale is a martingale only if the additional conditions needed for integrability and the conditional-expectation identity hold.
Correction: first verify that \(M_t\) is integrable and that \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\), or work with a stopped process already known to be a true martingale. In the Itô-process examples above, the condition \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) supplies a square-integrable martingale, so taking expectations is justified.
The three classifications can be summarized by the role of each component:
| Process | Martingale part | Finite-variation part | Additional condition |
|---|---|---|---|
| Brownian motion \(B_t\) | \(B_t\) | \(0\) | Brownian motion is a true martingale |
| Drifted Brownian motion \(x+\mu t+\sigma B_t\) | \(\sigma B_t\) | \(\mu t\) | \(\mu,\sigma,x\) are constants |
| Itô process | \(\int_0^tH_s\,dB_s\) | \(\int_0^t b_s\,ds\) | \(H\in\mathcal{L}^2\) and the drift is pathwise integrable |
Check Your Understanding
For each question, state any integrability assumption needed before taking an expectation.
- Why does the uniqueness theorem require the difference of the martingale parts to have finite-variation paths?
- Write the normalized semimartingale decomposition of \(x+\mu t+\sigma B_t\), including the cases \(\mu=0\) or \(\sigma=0\).
- For an Itô process, which condition makes the drift part finite variation, and which condition places the stochastic integrand in \(\mathcal{L}^2\)?
- Explain why the conditional mean of a drifted Brownian motion over \([s,t]\) differs from its current value by \(\mu(t-s)\).
- For \(X_t=\int_0^tB_s\,dB_s\), verify the mean at time \(t\) using \(X_t=\tfrac12(B_t^2-t)\).