Pathwise Integration Against Finite-Variation Processes

Why Integrate Pathwise Against Finite Variation?

A continuous finite-variation process can be an integrator without using the Ito construction. On each sample path, its increments define a finite signed measure, and an integrand that is measurable and integrable for that measure can be integrated by the ordinary Lebesgue integral. This covers drift terms such as \(\int_0^t H_s\mu(s,X_s)\,ds\), as well as integrals against random increasing clocks that need not have a deterministic rate of increase.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Fix a finite horizon \(T\). Let \(A_t\) be an adapted process whose paths are continuous and have finite variation on \([0,T]\), almost surely. Adaptedness means that \(A_t\) is \(\mathcal{F}_t\)-measurable for each \(t\). Its total variation up to time \(t\) is \(V_t=\operatorname{Var}_{[0,t]}(A)\), where variation is the supremum, over all finite partitions of \([0,t]\), of the sum of the absolute increments of \(A\).

Definition: Let \(H\) be progressively measurable, so that its restriction to \([0,t]\times\Omega\) is measurable with respect to \(\mathcal{B}([0,t])\otimes\mathcal{F}_t\) for every \(t\). The pathwise integrability condition on \([0,T]\) is \(\int_{(0,T]} |H_s|\,dV_s<\infty\) almost surely. Under this condition, \(\int_0^t H_s\,dA_s\) denotes the Lebesgue integral against the signed Lebesgue–Stieltjes measure of the path \(A\), for \(0\le t\le T\).

This condition is different from the \(L^2\) condition \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) used to define an Ito integral. The integral here is pathwise, with respect to \(dA\); it is not an Ito integral against \(dB\). In particular, pathwise integrability against \(dV\) does not require that \(H\) belong to that \(L^2\) class.

Construction from a Signed Measure

For each path on which \(A\) has finite variation, the Lebesgue–Stieltjes signed measure \(\mu_A\) is specified by \(\mu_A((s,t])=A_t-A_s\) for \(0\le s\le t\le T\). Its total variation measure is \(\mu_V\), specified by \(\mu_V((s,t])=V_t-V_s\). Since \(A\) is continuous, \(\mu_A\) has no atoms: the measure of a singleton is the jump of \(A\) there, and every such jump is zero. The measure \(\mu_V\) is also atomless because \(V\) is continuous.

For each such path, the section \(s\mapsto H_s(\omega)\) is measurable, and the assumed condition says it is integrable with respect to \(\mu_V\). The standard Lebesgue construction approximates it in \(L^1(\mu_V)\) by simple measurable functions. For a simple function \(\sum_j c_j\mathbf{1}_{E_j}\), its integral against \(\mu_A\) is \(\sum_j c_j\mu_A(E_j)\). The inequality \(|\mu_A(E)|\le\mu_V(E)\) makes these simple-function integrals converge as the approximations converge in \(L^1(\mu_V)\). Their limit defines the integral of \(H\) against \(\mu_A\), independently of the chosen approximating sequence. Restricting the measure to \((0,t]\) gives the integral up to time \(t\).

Theorem: If \(A\) is adapted, continuous, and of finite variation on \([0,T]\), and \(H\) is progressively measurable with \(\int_{(0,T]}|H_s|\,dV_s<\infty\) almost surely, then the pathwise integrals \(I_t=\int_0^t H_s\,dA_s\) are well-defined. Almost surely, for \(0\le s\le t\le T\), \(|I_t-I_s|\le\int_{(s,t]}|H_u|\,dV_u\), and \(\operatorname{Var}_{[0,T]}(I)\le\int_{(0,T]}|H_u|\,dV_u\). The process \(I_t\) is adapted and has continuous paths.

Proof. The pathwise construction above defines \(I_t\) for every \(t\). By additivity of the signed measure, \(I_t-I_s=\int_{(s,t]}H_u\,d\mu_A(u)\). The defining total-variation inequality for a signed measure gives \(|I_t-I_s|\le\int_{(s,t]}|H_u|\,d\mu_V(u)\), proving the increment bound. For any partition \(0=t_0<\cdots<t_n=T\), apply that bound on every interval and add: \(\sum_{i=1}^n|I_{t_i}-I_{t_{i-1}}|\le\sum_{i=1}^n\int_{(t_{i-1},t_i]}|H_u|\,dV_u=\int_{(0,T]}|H_u|\,dV_u\). Taking the supremum over partitions proves the variation bound. The pathwise integral up to \(t\) is measurable with respect to \(\mathcal{F}_t\): \(A\), \(H\), and their variation measure restricted to \([0,t]\) are determined by the processes up to \(t\), and the measurable-integral construction preserves \(\mathcal{F}_t\)-measurability. Thus \(I\) is adapted. Finally, for each fixed path with \(|H|\) integrable against the finite measure \(dV\), the measure \(|H|\,dV\) has no atoms because \(dV\) has no atoms. Its cumulative mass on \((s,t]\) therefore tends to zero as \(t\) decreases to \(s\) or \(s\) increases to \(t\). The increment bound then gives continuity of \(I\). \(\square\)

In particular, if \(|H_s|\le K\) for all \(s\) outside a pathwise null set, then \(|I_t-I_s|\le K(V_t-V_s)\), and \(\operatorname{Var}_{[0,T]}(I)\le K V_T\). These are pathwise bounds; they do not assert that \(I\) is a martingale. No martingale property is needed for this construction.

Worked Integrals

Worked Example: A Drift Integral with a Time-Dependent Integrand

On \([0,T]\), let \(A_t=t^2+t\) and \(H_t=3t\). Both are deterministic and adapted. The process \(A\) is increasing, with \(dA_t=(2t+1)\,dt\), so its total variation on \([0,T]\) is \(V_T=T^2+T\). The pathwise integral is

I_t=\int_0^t 3s\,dA_s =\int_0^t 3s(2s+1)\,ds =\int_0^t(6s^2+3s)\,ds =2t^3+\frac{3}{2}t^2.

Check: Differentiating the answer gives \(6t^2+3t=3t(2t+1)\), the integrand \(H_t\) multiplied by the rate of increase of \(A_t\). The variation bound also holds: \(H_s\ge0\), so \(\operatorname{Var}_{[0,T]}(I)=I_T=2T^3+\tfrac32T^2\), while \(\int_0^T|H_s|\,dV_s=2T^3+\tfrac32T^2\). Here the bound is an equality.

Worked Example: A Drift Stopped at a Random Time

Let \(\tau\) be a stopping time bounded by \(T\), and take \(\mu\ge0\). Set \(A_t=\mu(t\wedge\tau)\) and \(H_s=s\). This \(A\) is adapted, continuous, and increasing. Pathwise, its associated measure is \(\mu\,ds\) before \(\tau\) and zero after \(\tau\); the value at the single time \(\tau\) does not affect the integral because the measure has no atoms. Thus

\int_0^t s\,dA_s =\mu\int_0^{t\wedge\tau}s\,ds =\frac{\mu}{2}(t\wedge\tau)^2.

Check: For each path, if \(t\le\tau\), the answer is \(\mu t^2/2\), whose derivative in \(t\) is \(\mu t\), the product of \(H_t=t\) and the rate \(\mu\) of \(A\). If \(t>\tau\), the answer stays at \(\mu\tau^2/2\), because \(A\) has no further increments. The variation bound gives \(\int_0^T s\,dA_s=\mu\tau^2/2\), equal to the total variation of this nondecreasing integral.

Worked Example: Integration Against a Random Clock

Define the random clock \(C_t=\int_0^t(1+B_s^2)\,ds\), where \(B\) is the Brownian motion on the stated filtered probability space, and let \(H_s=e^{-s}/(1+B_s^2)\). The clock is adapted, continuous, and increasing, with \(dC_s=(1+B_s^2)\,ds\). The integrand is continuous, adapted, and bounded by \(1\), so it is pathwise integrable on every finite interval. Then

\int_0^t H_s\,dC_s =\int_0^t \frac{e^{-s}}{1+B_s^2}(1+B_s^2)\,ds =\int_0^t e^{-s}\,ds =1-e^{-t}.

Check: The result is deterministic even though \(C\) is random. Substituting the clock’s rate and the integrand into the pathwise definition cancels the factor \(1+B_s^2\), leaving exactly \(e^{-s}\). The variation bound is exact: \(H_s\ge0\), so the integral is increasing and its total variation on \([0,T]\) is \(1-e^{-T}=\int_0^T|H_s|\,dC_s\).

What the Integral Does—and Does Not—Require

When \(A_t=\int_0^t a_s\,ds\) for a pathwise integrable rate \(a_s\), the change-of-measure construction reduces to \(\int_0^t H_s\,dA_s=\int_0^t H_sa_s\,ds\). But a continuous finite-variation path need not have such a density with respect to time. The definition through its Lebesgue–Stieltjes measure still applies. This is why a random clock can be a valid integrator even when replacing it by a time-dependent rate is not justified.

This pathwise construction is also distinct from integrating against Brownian motion. Brownian motion is not a finite-variation path, so its increments do not define a finite signed Lebesgue–Stieltjes measure of this kind. The Ito integral uses a different construction and different integrability conditions. In shorthand, the Ito multiplication rules include \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\); none of these rules is needed to define \(\int H\,dA\) here. The integral is an ordinary Lebesgue integral for each eligible path, not an Ito or Stratonovich integral.

One useful consequence of finite variation is that its squared increments vanish along fine partitions. Indeed, on any partition of \([0,T]\), \(\sum_i(\Delta A_i)^2\le\max_i|\Delta A_i|\sum_i|\Delta A_i|\le\max_i|\Delta A_i|V_T\). Continuity of the path makes \(\max_i|\Delta A_i|\) tend to zero as the mesh tends to zero, so the sum tends to zero. This contrasts with Brownian quadratic variation and helps explain why finite-variation terms are treated as ordinary measure integrals in stochastic calculus.

Common Pitfall: Assuming Every Clock Has a Time Density

A continuous finite-variation function can increase on a set of time-Lebesgue measure zero. For example, let \(A_t\) be the Cantor function on \([0,1]\), with \(A_0=0\) and \(A_1=1\). It is continuous and increasing, so \(dA\) is a finite measure and \(\int_0^1 1\,dA=A_1-A_0=1\). The Cantor function has derivative zero almost everywhere with respect to Lebesgue measure. The incorrect calculation \(\int_0^1 1\,dA=\int_0^1 A'_s\,ds=0\) assumes without justification that \(dA_s=A'_s\,ds\). That equality holds for absolutely continuous \(A\), not for every continuous finite-variation \(A\). The correction is to integrate against the Lebesgue–Stieltjes measure \(dA\), which gives \(1\) in this example.

Check Your Understanding

For each question, distinguish pathwise integration against \(dA\) from integration against time.

If \(A_t=2t\) and \(H_t=t^2\), compute \(\int_0^T H_s\,dA_s\) and its total variation.
State the pathwise integrability condition for a progressively measurable \(H\) against a continuous finite-variation process \(A\).
Why does continuity of \(A\) imply that its Lebesgue–Stieltjes measure has no atoms?
For \(A_t=t\wedge\tau\), where \(\tau\) is a bounded stopping time, compute \(\int_0^t 1\,dA_s\).
Explain why \(\int H\,dA=\int HA'_s\,ds\) cannot be assumed for every continuous finite-variation \(A\).