Localization and Stopping Identities for Semimartingale Integrals

Why Stopping an Integral Matters

Stopping a process at a random time changes which increments are included in an integral. The corresponding change to the integrand is a cutoff at that stopping time. This simple-looking rule is essential when an integral is first defined after localization: it lets us compare the localized constructions and recover a single integral on the original time interval.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Let $X$ be a continuous semimartingale, written $X_t=X_0+M_t+A_t$, where $M$ is a continuous local martingale and $A$ is a continuous finite-variation process. The integral against $X$ is the semimartingale integral, not necessarily an Ito integral against Brownian motion alone. In particular, for $X=B$, an adapted, progressively measurable integrand $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ belongs to the Brownian Ito class $\mathcal{L}^2$. For a general continuous semimartingale, the relevant requirement is local semimartingale integrability: locally, $\int_0^T H_s^2\,d[M]_s<\infty$ and $\int_0^T |H_s|\,d|A|_s<\infty$, almost surely. We take integrands to be predictable; a continuous adapted integrand is predictable.

Definition: A stopping time $\tau$ is a random time such that $\{\tau\le t\}\in\mathcal{F}_t$ for every $t$. The stopped process is $X^\tau_t=X_{t\wedge\tau}$. For a predictable integrand $H$, the stopped integrand is $H_s\mathbf{1}_{(0,\tau]}(s)$. The interval convention includes the value at $\tau$; for continuous integrators, a single endpoint does not create a jump contribution.

A local martingale is a process that becomes a martingale after stopping along a localizing sequence; it need not itself be a martingale. A martingale, in contrast, must be integrable at each time and satisfy $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ for $s\le t$. The stopping identities below are pathwise equalities up to indistinguishability; they do not by themselves assert that an integral is a true martingale.

The Stopping Identity

Theorem: Let $X$ be a continuous semimartingale, let $H$ be integrable with respect to $X$, and let $\tau$ be a stopping time. Then, for all $t\ge0$, up to indistinguishability, $$ (H\cdot X)_{t\wedge\tau} =\int_0^t H_s\mathbf{1}_{(0,\tau]}(s)\,dX_s =\int_0^t H_s\,dX^\tau_s. $$ Here the notation $H\cdot X$ means the semimartingale integral of $H$ against $X$.

Proof. First take a predictable elementary integrand $H_s=\sum_{j=0}^{m-1}\xi_j\mathbf{1}_{(t_j,t_{j+1}]}(s)$, where each $\xi_j$ is $\mathcal{F}_{t_j}$-measurable. Its integral is the sum of the corresponding coefficients times increments of $X$. Stopping that integral at $\tau$ retains exactly the parts of those increments occurring no later than $\tau$. Equivalently, each elementary interval is cut off by $\mathbf{1}_{(0,\tau]}$, which gives the first equality. The stopped semimartingale has increments $dX^\tau_s=\mathbf{1}_{(0,\tau]}(s)\,dX_s$, so the same calculation gives the second equality for elementary integrands.

For a general integrable $H$, the semimartingale integral is constructed by localization and approximation by elementary predictable integrands: locally, their integrals converge uniformly on compact time intervals in probability (ucp). If $Y^n\to Y$ ucp, then $Y^{n,\tau}\to Y^\tau$ ucp, because for each finite $T$, $\sup_{t\le T}|Y^n_{t\wedge\tau}-Y_{t\wedge\tau}|\le\sup_{t\le T}|Y^n_t-Y_t|$. Apply this to the elementary integrals. The stopped elementary integrands give the integral of $H\mathbf{1}_{(0,\tau]}$ by the same localized construction, and the stopped integrators give the integral against $X^\tau$. Uniqueness of the ucp limit proves both equalities. Removing the localization yields the asserted identity on every finite time interval, hence on $t\ge0$. $\square$

The indicator is a cutoff in time: before $\tau$, the original integrand is used; after $\tau$, it is zero. The result also gives an interval identity. If $\tau\le\sigma$ are stopping times, subtracting the stopped integrals at $\sigma$ and $\tau$ yields

\int_0^t H_s\mathbf{1}_{(\tau,\sigma]}(s)\,dX_s =(H\cdot X)_{t\wedge\sigma}-(H\cdot X)_{t\wedge\tau}.

Indeed, the difference of the two cutoff indicators is $\mathbf{1}_{(\tau,\sigma]}$, and the integral is linear in its integrand.

Worked Example: A Deterministic Stopped Interval

Let $X_t=2B_t+3t$, let $H_t=1$, and stop at the deterministic time $\tau=1$. Compute the integral at $t=2$, both by stopping the integral and by stopping the integrator.

Since $dX_s=2\,dB_s+3\,ds$, the integral up to time $u$ is $(H\cdot X)_u=2B_u+3u$, using $B_0=0$. Stopping this process at $t=2$ gives $(H\cdot X)_{2\wedge1}=2B_1+3$.

For the cutoff calculation, $\mathbf{1}_{(0,1]}(s)$ retains only the interval from $0$ to $1$, so $\int_0^2\mathbf{1}_{(0,1]}(s)\,dX_s=2(B_1-B_0)+3(1-0)=2B_1+3$. For the stopped integrator, $X^\tau_2=X_1=2B_1+3$, while $X^\tau_0=0$; integrating $H=1$ against it gives the same increment.

Check: the three calculations all give $2B_1+3$. The drift contribution is $3\cdot1=3$, not $3\cdot2=6$, because the integral is stopped at time $1$.

Localization and Consistency

A sequence of stopping times $(\tau_n)$ is a localizing sequence if $\tau_n$ increases and $\tau_n\to\infty$ almost surely. On any fixed finite horizon, such a sequence eventually exceeds the horizon for almost every outcome. Localization is useful because a process or integrand may have the required bounds only before these stopping times.

Theorem: Suppose $H$ is integrable with respect to a continuous semimartingale $X$, and $(\tau_n)$ is a localizing sequence. For every $n$, $$ (H\cdot X)^{\tau_n} =H\cdot X^{\tau_n} =(H\mathbf{1}_{(0,\tau_n]})\cdot X. $$ Consequently, if $X^n=X^{\tau_n}$, then the integrals against $X^n$ and $X$ agree up to $\tau_n$.

Proof. Apply the stopping identity with $\tau=\tau_n$. Its first equality gives $(H\cdot X)^{\tau_n}=(H\mathbf{1}_{(0,\tau_n]})\cdot X$, and its second gives $(H\cdot X)^{\tau_n}=H\cdot X^{\tau_n}$. Thus all three processes are equal. In particular, the integral against $X^n=X^{\tau_n}$ is the stopped integral, so it agrees with $H\cdot X$ at every time $t\le\tau_n$. Since $\tau_n\to\infty$ almost surely, for each fixed finite $T$ this agreement holds throughout $[0,T]$ once $n$ is large enough, almost surely. This is the consistency needed to pass from localized integrals to the original integral. $\square$

This theorem concerns equality of integrals, not an automatic martingale property. For example, an integral against a local martingale is generally only a local martingale unless additional integrability conditions show it is a true martingale. When the integrator is Brownian motion and the integrand lies in $\mathcal{L}^2$, the Ito isometry gives square integrability, and the Ito integral is a martingale. Stopping identities alone do not supply that integrability.

Worked Example: Stopping a Brownian Integral at a Level

Let $\tau_n=\inf\{t\ge0:|B_t|\ge n\}$, for a positive integer $n$, and fix $T>0$. Compute the stopped integral $\int_0^T B_s\mathbf{1}_{(0,\tau_n]}(s)\,dB_s$.

Before $\tau_n$, continuity gives $|B_s|\le n$. Therefore the stopped integrand is progressively measurable and $\mathbb{E}\int_0^T B_s^2\mathbf{1}_{(0,\tau_n]}(s)\,ds\le n^2T<\infty$; it belongs to $\mathcal{L}^2$ on $[0,T]$. The stopping identity identifies the integral with $\int_0^{T\wedge\tau_n}B_s\,dB_s$.

Apply Ito's formula to $f(x)=x^2$. Its derivatives are $f'(x)=2x$ and $f''(x)=2$, so the correction term is $\tfrac12 f''(B_s)\,ds=ds$, and $d(B_s^2)=2B_s\,dB_s+ds$. Integrating up to $T\wedge\tau_n$ and rearranging gives $\int_0^T B_s\mathbf{1}_{(0,\tau_n]}(s)\,dB_s =\tfrac12(B_{T\wedge\tau_n}^2-(T\wedge\tau_n))$.

Check: substituting the right-hand side into the integrated Ito identity gives $2\cdot\tfrac12(B_{T\wedge\tau_n}^2-(T\wedge\tau_n))+(T\wedge\tau_n)=B_{T\wedge\tau_n}^2$, as required. Also, the integral is a true square-integrable martingale on $[0,T]$: its integrand is in $\mathcal{L}^2$, so the Ito integral has conditional expectation equal to its current value at every earlier time.

Worked Example: The Integral Between Two Stopping Times

Let $X_t=B_t+t$, take deterministic stopping times $\tau=1$ and $\sigma=2$, and let $H_t=2$. Find $\int_0^3 H_s\mathbf{1}_{(1,2]}(s)\,dX_s$ and verify it by subtracting stopped integrals.

On $(1,2]$, $dX_s=dB_s+ds$. Hence the cutoff integral is $2(B_2-B_1)+2(2-1)=2(B_2-B_1)+2$. Alternatively, $(H\cdot X)_u=2B_u+2u$. The value stopped at $2$ is $2B_2+4$, and the value stopped at $1$ is $2B_1+2$. Their difference is $(2B_2+4)-(2B_1+2)=2(B_2-B_1)+2$.

Check: the Brownian increment is multiplied by $2$, and the drift contribution is $2\cdot(2-1)=2$. The endpoints match the interval convention: the increment is from time $1$ to time $2$, not from time $0$ or time $3$.

A Common Stopping Error

A frequent mistake is to stop the integrator but continue integrating over the full original time interval without changing the integrand. For $X_t=t$, $H_t=1$, and $\tau=1$, the correct value at $t=2$ is $\int_0^2 1\,dX^\tau_s=X^\tau_2-X^\tau_0=1$. The incorrect uncut calculation $\int_0^2 1\,dX_s=X_2-X_0=2$ uses increments after the stopping time. The correction is to use the stopped process, or equivalently to insert $\mathbf{1}_{(0,\tau]}$ in the integrand; both give $1$.

The cutoff is also not $\mathbf{1}_{(\tau,\infty)}$: that indicator selects increments after the stopping time, which is the opposite of stopping the accumulated integral. Keeping track of whether an indicator retains times before or after $\tau$ prevents many sign and endpoint errors.

Check Your Understanding

For each question, state the integrability or stopping assumptions needed before applying an identity.

For a continuous semimartingale $X$ and an $X$-integrable predictable $H$, express $(H\cdot X)_{t\wedge\tau}$ as an integral over $[0,t]$.
Let $X_t=t$, $H_t=3$, and $\tau=2$. Compute $\int_0^4 H_s\mathbf{1}_{(0,\tau]}(s)\,dX_s$.
If $\tau\le\sigma$, write the integral over $(\tau,\sigma]$ as a difference of stopped integrals.
For $X=B$, what condition on $H$ ensures that $H$ belongs to $\mathcal{L}^2$ on $[0,T]$?
Why does stopping an integral against a local martingale not, by itself, prove that the unstopped integral is a true martingale?

Localization and Stopping Identities for Semimartingale Integrals

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