Stopping Identities for Stochastic Integrals

Stopping the Integral or Stopping the Integrand?

Stopping a process at a random time changes its value after that time, but it does not erase what the process has accumulated beforehand. For an Ito integral, this idea has a precise form: stopping the integral at $\tau$ is equivalent to including the integrand only up to $\tau$. The endpoint convention is worth stating explicitly, even though including or excluding the single time $\tau$ does not change the Ito integral.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ adapted to the filtration. Fix $T<\infty$. Throughout this tutorial, an Ito integral on $[0,T]$ has a progressively measurable integrand $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, that is, $H\in\mathcal{L}^2$. We use the continuous version of $M_t=\int_0^t H_s\,dB_s$, whose existence is part of the standard construction of the $\mathcal{L}^2$ Ito integral.

Definition: For a stopping time $\tau$, the stopped process is $M_{t\wedge\tau}$, where $t\wedge\tau=\min(t,\tau)$. The integrand stopped at $\tau$ is $H_s\mathbf{1}_{\{s\leq\tau\}}$. The convention here includes the endpoint $s=\tau$.

The indicator in this definition is adapted and progressively measurable, so its product with $H$ is an admissible integrand. On any fixed finite interval, its square-integrability follows from $\mathbb{E}\int_0^T H_s^2\mathbf{1}_{\{s\leq\tau\}}\,ds\leq\mathbb{E}\int_0^T H_s^2\,ds$.

The Stopping Identity

Theorem: Let $H\in\mathcal{L}^2$ on $[0,T]$, and let $\tau$ be any stopping time, possibly taking the value $+\infty$. Then, up to indistinguishability on $[0,T]$, $$ \left(\int_0^\cdot H_s\,dB_s\right)_{t\wedge\tau} = \int_0^t H_s\mathbf{1}_{\{s\leq\tau\}}\,dB_s,\qquad 0\leq t\leq T. $$

Proof. First take a stopping time $\sigma\leq T$ that takes values in a finite deterministic grid $0=t_0<t_1<\cdots<t_m=T$. Fix $t$ in this grid. On an interval $(t_{j-1},t_j]$, the indicator $\mathbf{1}_{\{s\leq\sigma\}}$ agrees, except possibly at the interval's endpoint, with the random variable $\mathbf{1}_{\{\sigma\geq t_j\}}$. Because $\sigma$ takes grid values, $\{\sigma\geq t_j\}=\{\sigma>t_{j-1}\}$, which belongs to $\mathcal{F}_{t_{j-1}}$. For an event $A\in\mathcal{F}_{t_{j-1}}$, the defining construction of the Ito integral gives $\int_{t_{j-1}}^{t_j}\mathbf{1}_A H_s\,dB_s=\mathbf{1}_A\int_{t_{j-1}}^{t_j}H_s\,dB_s$; this follows first for adapted step integrands and then for $\mathcal{L}^2$ integrands by the Ito isometry. Summing over the grid intervals up to $t$, only the intervals ending no later than $\sigma$ contribute. Their sum is $M_{t\wedge\sigma}$, proving the identity for this $\sigma$ and grid time $t$.

Now fix an arbitrary deterministic $t\leq T$, set $\rho=\tau\wedge t$, and choose nested grids of $[0,t]$ whose meshes tend to zero. Round $\rho$ upward to the next grid point, calling the resulting grid-valued stopping time $\rho_n$. Then $\rho_n\downarrow\rho$. The finite-grid argument gives $M_{\rho_n}=\int_0^t H_s\mathbf{1}_{\{s\leq\rho_n\}}\,dB_s$. By continuity of $M$, $M_{\rho_n}\to M_\rho$ almost surely. Also, $\mathbf{1}_{\{s\leq\rho_n\}}\to\mathbf{1}_{\{s\leq\rho\}}$ for every $s$ and outcome. The Ito isometry and dominated convergence therefore give $$ \mathbb{E}\left[\left(\int_0^t H_s\big(\mathbf{1}_{\{s\leq\rho_n\}}-\mathbf{1}_{\{s\leq\rho\}}\big)\,dB_s\right)^2\right] = \mathbb{E}\int_0^t H_s^2\big(\mathbf{1}_{\{s\leq\rho_n\}}-\mathbf{1}_{\{s\leq\rho\}}\big)^2\,ds \longrightarrow 0. $$ Since $s\leq\rho$ is equivalent to $s\leq t$ and $s\leq\tau$, the limiting integral is the right-hand side of the theorem at time $t$, and $M_\rho=M_{t\wedge\tau}$. Thus the identity holds almost surely for each fixed $t$. Both sides have continuous versions; applying the fixed-time equality on a countable dense set of times and then using continuity gives equality simultaneously for all $t\in[0,T]$, up to indistinguishability. $\square$

Why the Endpoint Convention Is Harmless

Corollary: Under the theorem's assumptions, replacing $\mathbf{1}_{\{s\leq\tau\}}$ by $\mathbf{1}_{\{s<\tau\}}$ leaves the Ito integral unchanged on $[0,T]$.

Proof. The two integrands differ only when $s=\tau(\omega)$. For each outcome, the set of such times in $[0,T]$ contains at most one point and hence has Lebesgue measure zero. Consequently, $$ \mathbb{E}\int_0^T H_s^2\left(\mathbf{1}_{\{s\leq\tau\}}-\mathbf{1}_{\{s<\tau\}}\right)^2\,ds=0. $$ The Ito isometry says that the squared mean of the difference between the two integrals is this quantity, so that difference is zero almost surely at each fixed time. Continuity gives equality of the integral processes up to indistinguishability. $\square$

Including the endpoint is a convenient convention: it records that the process has been integrated through the stopping time. It does not assert that a Brownian increment occurs at a single instant. An Ito integral accumulates over time, and a singleton has zero time measure.

Worked Example: A Deterministic Stopping Time

Let $H_s=2$ on $[0,2]$, let $\tau=1$, and evaluate both sides at $t=\tfrac32$. The integrand belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^2 2^2\,ds=8$. The stopped integral is $$ M_{t\wedge\tau}=\int_0^1 2\,dB_s=2B_1. $$ The stopped-integrand side is $\int_0^{3/2}2\mathbf{1}_{\{s\leq1\}}\,dB_s=\int_0^1 2\,dB_s=2B_1$, as required.

Check: At $t=1$, both sides are still $2B_1$. Using the strict endpoint convention changes the integrand only at $s=1$, so the Ito isometry gives zero mean-square difference. At $t=\tfrac32$, the stopped time is $t\wedge\tau=1$, not $\tfrac32$.

Worked Example: A Stopping Time Chosen at Time One

Let $A=\{B_1\geq0\}$, and define $\tau=1$ on $A$ and $\tau=2$ on $A^c$. This is a stopping time: before time $1$ it has not occurred; for $1\leq t<2$, the event $\{\tau\leq t\}$ is $A\in\mathcal{F}_1\subseteq\mathcal{F}_t$; and for $t\geq2$ it is the whole space. Take $H_s=2$ on $[0,1]$ and $H_s=3$ on $(1,2]$, which is square-integrable.

At $t=2$, the stopped integral is $M_\tau$. On $A$, $\tau=1$, so its value is $2B_1$. On $A^c$, $\tau=2$, so its value is $2B_1+3(B_2-B_1)$. In one expression this is $$ M_\tau=2B_1+3\mathbf{1}_{A^c}(B_2-B_1). $$ The stopped-integrand calculation gives $\int_0^2 H_s\mathbf{1}_{\{s\leq\tau\}}\,dB_s$. Its first-interval contribution is $2B_1$, since $\tau\geq1$ on both events. Its second-interval contribution is $3\mathbf{1}_{A^c}(B_2-B_1)$, since that interval is included exactly when $\tau=2$. The two expressions agree.

Check: On $A$, substitution of $\mathbf{1}_{A^c}=0$ gives $2B_1$. On $A^c$, substitution of $\mathbf{1}_{A^c}=1$ gives $2B_1+3(B_2-B_1)$. The event $B_1=0$ is assigned to $A$ by the definition, so the cases cover the whole sample space without an omitted boundary case.

A Stopped Integral Computed by Ito's Formula

Worked Example: Stopping the Integral of Brownian Motion

On $[0,1]$, let $H_s=B_s$, which is in $\mathcal{L}^2$ because $\mathbb{E}\int_0^1 B_s^2\,ds=\int_0^1 s\,ds=\tfrac12$. Take the deterministic stopping time $\tau=\tfrac12$ and evaluate at $t=1$. The stopping identity gives $$ \int_0^{1/2}B_s\,dB_s = \int_0^1 B_s\mathbf{1}_{\{s\leq1/2\}}\,dB_s. $$ To compute the left-hand side, apply Ito's formula to $f(x)=x^2$. Its derivatives are $f'(x)=2x$ and $f''(x)=2$, so the correction term is $\tfrac12 f''(B_s)\,ds=ds$. Therefore $$ d(B_s^2)=2B_s\,dB_s+ds,\qquad \int_0^{1/2}B_s\,dB_s=\tfrac12\left(B_{1/2}^2-\tfrac12\right). $$ The answer from either side is $\tfrac12 B_{1/2}^2-\tfrac14$.

Check: Since $\mathbb{E}[B_{1/2}^2]=\tfrac12$, the computed expression has expectation $\tfrac12(\tfrac12-\tfrac12)=0$. The stopped integrand has finite energy $\mathbb{E}\int_0^{1/2}B_s^2\,ds=\int_0^{1/2}s\,ds=\tfrac18$, consistent with the square-integrable Ito integral and its zero mean.

A Common Mistake

A tempting but incorrect calculation is to stop the upper limit and also multiply by an endpoint indicator as though it were an ordinary random scalar: $\int_0^{t\wedge\tau}H_s\mathbf{1}_{\{\tau\geq t\}}\,dB_s$. This wrongly removes the accumulated integral whenever $\tau<t$. For example, with $H_s=2$, $t=\tfrac32$, and $\tau=1$, this expression is $0$, because $\mathbf{1}_{\{1\geq3/2\}}=0$, whereas the stopped integral is $2B_1$. The correct operation is to stop the upper limit, or equivalently to use the time-dependent cutoff $H_s\mathbf{1}_{\{s\leq\tau\}}$ with upper limit $t$. The cutoff must depend on the integration time $s$, not on whether the stopping time exceeds the final evaluation time.

The identity is useful whenever a stopped integral must be expressed as an Ito integral with a specified integrand. It also clarifies the role of stopping in arguments: the process is frozen after $\tau$, while the integral continues to be written up to the deterministic time $t$ with its integrand switched off after $\tau$. The result applies directly to square-integrable integrands on a finite horizon; integrals defined only through localization require additional construction beyond this identity.

Check Your Understanding

For each question, state the integrand and endpoint convention before calculating.

For $H_s=1$, $\tau=2$, and $t=3$, write the stopped-integral identity explicitly.
Why is $\mathbb{E}\int_0^T H_s^2\mathbf{1}_{\{s\leq\tau\}}\,ds\leq\mathbb{E}\int_0^T H_s^2\,ds$?
In the random-time example, what is the stopped integral on $A=\{B_1\geq0\}$, and what is it on $A^c$?
Explain why the Ito integrals with cutoffs $\mathbf{1}_{\{s\leq\tau\}}$ and $\mathbf{1}_{\{s<\tau\}}$ agree.
For $H_s=B_s$ and $\tau=\tfrac12$, identify the second-derivative correction term when applying Ito's formula to $B_{t\wedge\tau}^2$.