Constructing Integrals Against Continuous Semimartingales

From Two Integrals to One

Brownian integration and pathwise integration against finite-variation processes address different parts of stochastic calculus. A continuous semimartingale combines both: it can be written as a continuous local martingale plus a continuous finite-variation process. The integral against the semimartingale is constructed by integrating against each part and adding the results. The key issue is whether this prescription depends on the chosen decomposition; uniqueness of the decomposition ensures that it does not.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ with respect to $(\mathcal{F}_t)$. A continuous semimartingale $X_t$ has a decomposition $X_t=X_0+M_t+A_t$, where $M_0=A_0=0$, $M$ is a continuous local martingale, and $A$ is a continuous adapted process of finite variation on every bounded time interval. Here finite variation means that the total variation of each path on $[0,T]$ is finite, almost surely, for each finite $T$. The term $X_0$ is kept separate so that both components start at zero.

Definition: An integrand $H$ is admissible for $X=X_0+M+A$ on $[0,T]$ if it is adapted and progressively measurable, has a predictable version for the local-martingale integral, and satisfies, almost surely, $\int_0^T H_s^2\,d[M]_s<\infty$ and $\int_0^T |H_s|\,d\operatorname{Var}(A)_s<\infty$. Here $[M]$ is the quadratic variation of $M$, and $\operatorname{Var}(A)$ is the pathwise total-variation process. For local integrability on all finite horizons, these conditions are imposed locally: there are stopping times increasing to infinity for which the stopped integrals are finite. Brownian integrands in $\mathcal{L}^2$ are progressively measurable processes satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; since $[B]_t=t$, this gives the required Brownian square-integrability.

The predictable-version qualification is part of the usual Itô integration convention; it holds directly, for example, for continuous adapted $H$. For the finite-variation part, the integral is the pathwise Lebesgue–Stieltjes integral against the signed measure $dA_s$. The displayed absolute-integrability condition ensures it is defined. The local-martingale part is the standard Itô integral, constructed from predictable simple integrands and extended by the Itô isometry after localization. It is a local martingale; it need not be a true martingale without additional integrability.

Definition: For an admissible $H$, define the integral against $X$ by $$ (H\mathbin{\cdot}X)_t=\int_0^t H_s\,dM_s+\int_0^t H_s\,dA_s,\qquad t\ge0. $$ The first integral is an Itô integral against a continuous local martingale, and the second is a pathwise finite-variation integral. The dot is notation for the resulting process, not ordinary multiplication.

Thus the two components have distinct roles. The local-martingale component carries stochastic fluctuations; the finite-variation component contributes an ordinary signed-measure integral. For $X_t=B_t+\int_0^t b_s\,ds$, for instance, the latter is the drift contribution $\int_0^t H_sb_s\,ds$. It is not a diffusion coefficient. The construction here uses the Itô/semimartingale integral, not a Stratonovich integral; no Stratonovich correction is being applied.

Why the Decomposition Is Unique

Theorem: Suppose a continuous process has two decompositions $X=X_0+M+A=X_0+\widetilde M+\widetilde A$, where both local-martingale parts are continuous and start at zero, and both finite-variation parts are continuous, adapted, and start at zero. Then $M=\widetilde M$ and $A=\widetilde A$ indistinguishably. Consequently, the definition of $H\mathbin{\cdot}X$ is independent of the decomposition.

Proof. Subtract the two decompositions and set $N=M-\widetilde M=\widetilde A-A$. The first expression shows that $N$ is a continuous local martingale starting at zero. The second shows that its paths have finite variation on every bounded interval. A continuous finite-variation path has zero quadratic variation: on a partition, the sum of squared increments is bounded by the largest absolute increment times the total variation; the largest increment tends to zero as the mesh tends to zero by continuity. Hence $[N]_t=0$ for every $t$. To conclude that $N$ is identically zero, stop it at a localizing sequence and at its first exit from $[-k,k]$. Each resulting process is a bounded local martingale, hence a square-integrable martingale, and the Itô isometry gives $\mathbb{E}[N_{t\wedge\tau}^2]=\mathbb{E}[[N]_{t\wedge\tau}]=0$. Therefore each such stopped process is zero almost surely. Letting the localizing and exit levels increase, and using continuity, gives $N_t=0$ for all $t$, outside one null set. Thus $M=\widetilde M$, and the original decompositions then give $A=\widetilde A$. The two component integrals in the definition are consequently the same for either decomposition, proving independence. $\square$

This argument also explains why the normalization matters. If constants were allowed to move freely between $M$ and $A$, their initial values would not be separately determined. Requiring both to start at zero fixes that ambiguity. The theorem concerns decompositions with a local martingale and a finite-variation process; it does not say that an arbitrary split into two processes is unique.

Worked Constructions

Worked Example: A Simple Integrand Against Brownian Motion with Drift

Let $X_t=B_t+2t$ on $[0,2]$, and let $H_t=3$ for $0\le t\le1$ and $H_t=-2$ for $1<t\le2$. Compute $(H\mathbin{\cdot}X)_2$, its mean, and its variance.

The decomposition is $M_t=B_t$ and $A_t=2t$. The integrand is predictable up to its irrelevant value at the single time $1$, and $\mathbb{E}\int_0^2 H_s^2\,ds=9+4=13<\infty$, so its Brownian integral is in $\mathcal{L}^2$. The simple-integrand definition gives $\int_0^2H_s\,dB_s=3(B_1-B_0)-2(B_2-B_1)=3B_1-2(B_2-B_1)$. The finite-variation integral is $2\int_0^2H_s\,ds=2(3\cdot1-2\cdot1)=2$. Therefore

(H\mathbin{\cdot}X)_2=3B_1-2(B_2-B_1)+2.

The two Brownian increments are independent, have mean zero, and have variances $1$ and $1$. The stochastic part has mean $0$ and variance $3^2\cdot1+(-2)^2\cdot1=13$. Thus the full integral has mean $2$ and variance $13$. More generally, the process $t\mapsto\int_0^tH_s\,dB_s$ is a true square-integrable martingale: its increments have conditional mean zero by independence of Brownian increments from the past, and its second moment at time $2$ is $13$. The full integral is not a martingale, because it also contains the nonconstant finite-variation term.

Check: setting the drift $2t$ to zero removes exactly the constant $2$, leaving a mean-zero stochastic integral with variance $13$, as the decomposition predicts.

Worked Example: Integrating Along a Finite-Variation Path

Take $X_t=t^2$ on $[0,1]$ and $H_t=1+t$. Here $M=0$, $A_t=t^2$, and $dA_t=2t\,dt$. The integrand is deterministic, adapted and progressively measurable, and the finite-variation integrability condition holds because

\int_0^1 |1+t|\,d\operatorname{Var}(A)_t =\int_0^1(1+t)2t\,dt =\left[t^2+\frac{2}{3}t^3\right]_0^1 =\frac{5}{3}.

There is no local-martingale contribution. At every $t\in[0,1]$, direct integration gives $(H\mathbin{\cdot}X)_t=\int_0^t(1+s)2s\,ds=t^2+\frac23t^3$, and in particular $(H\mathbin{\cdot}X)_1=\frac53$.

Check: differentiating the computed expression gives $2t+2t^2=2t(1+t)$, the integrand $H_t$ multiplied by the pathwise derivative $dA_t/dt=2t$; the value at $t=0$ is $0$.

Worked Example: A Time-Weighted Brownian Integral

Let $X_t=B_t+t^2$ and $H_t=t$ on $[0,1]$. The integrand belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^1t^2\,dt=\frac13$, and its finite-variation integral is $\int_0^t s\,d(s^2)=\frac23t^3$. Thus $(H\mathbin{\cdot}X)_t=\int_0^t s\,dB_s+\frac23t^3$.

To compute the stochastic term in another form, apply Itô's formula to $f(t,x)=tx$ and $x=B_t$. Its derivatives are $f_t=x$, $f_x=t$, and $f_{xx}=0$. Since $dB_t\,dB_t=dt$, the formula, including its $\frac12 f_{xx}\,dt$ correction, is

d(tB_t)=B_t\,dt+t\,dB_t+\frac12\cdot0\cdot dt =B_t\,dt+t\,dB_t.

Integrating and rearranging gives $\int_0^t s\,dB_s=tB_t-\int_0^tB_s\,ds$, so the full answer is $tB_t-\int_0^tB_s\,ds+\frac23t^3$. The first integral is a true square-integrable martingale: its mean is zero, and its second moment is $\int_0^1s^2\,ds=\frac13$. The full integral at time $1$ therefore has mean $\frac23$ and variance $\frac13$, since its finite-variation term is deterministic.

Check: the alternate expression for the Brownian integral also has mean zero: $\mathbb{E}[B_1]=0$ and $\mathbb{E}\int_0^1B_s\,ds=\int_0^1\mathbb{E}[B_s]\,ds=0$. Its variance is $\frac13$ by the Itô isometry, agreeing with the direct integral calculation.

What the Construction Guarantees—and a Common Error

Whenever the stated local integrability conditions hold, the construction gives a continuous semimartingale integral whose local-martingale component is $\int H\,dM$ and whose finite-variation component is $\int H\,dA$. The first component is generally only a local martingale. If, in addition, $\mathbb{E}\int_0^T H_s^2\,d[M]_s<\infty$, the local-martingale integral is square-integrable and therefore a true martingale. For Brownian motion, this condition is exactly the $\mathcal{L}^2$ condition $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The finite-variation component is not generally a martingale and should not be included in a martingale claim about the stochastic component.

A common mistake is to keep only the local-martingale part. For $X_t=B_t+2t$ and $H_t=1$, that incorrect calculation gives $(H\mathbin{\cdot}X)_t=B_t$, omitting the finite-variation contribution. The definition instead gives $B_t+2t$. Substituting $t=1$, the incorrect result has mean $0$, whereas the constructed integral has mean $2$. The omitted term is not an Itô correction: it is the pathwise integral of $H$ against $A$.

Conversely, one cannot freely move a finite-variation term into the local martingale part. For example, $B_t+t$ is not a local martingale: if it were, subtracting the Brownian martingale $B_t$ would make the deterministic process $t$ a local martingale, which is impossible because a continuous finite-variation local martingale starting at zero is identically zero, as shown in the uniqueness proof. In the semimartingale decomposition, the drift belongs in $A$, not in $M$.

Check Your Understanding

For each question, identify which part of the construction is being used and check that the integrability assumptions are satisfied.

For $X_t=B_t+3t$ and $H_t=2$ on $[0,1]$, write $(H\mathbin{\cdot}X)_1$ as a Brownian increment plus a deterministic term. Find its mean and variance.
State the two local integrability conditions for $H\mathbin{\cdot}X$ when $X=X_0+M+A$.
For $A_t=t^3$ and $H_t=t$ on $[0,1]$, compute $\int_0^1H_t\,dA_t$ and check it by differentiating the corresponding expression at a general time $t$.
Why does uniqueness require the difference of two local-martingale parts to have finite variation?
Give a sufficient condition under which $\int_0^tH_s\,dM_s$ is a true martingale, rather than merely a local martingale.

Constructing Integrals Against Continuous Semimartingales

What You'll Learn