Why Stochastic Proofs Use Cutoffs
Many stochastic-calculus identities are first valid when every quantity is bounded, or when the stochastic integrand is square-integrable. A process may be continuous and finite at every fixed time while still having no deterministic bound over all times. Stopping at the first exit from a large bounded region creates a version on which calculations are controlled. The essential proof strategy is to establish an identity for each stopped process, then use the increasing cutoffs to recover information on any fixed finite time interval.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). An Ito integral \(\int_0^t H_s\,dB_s\) below is taken for an adapted, progressively measurable integrand \(H_s\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) on each stated finite horizon. A stopping time \(\tau\) is a random time for which \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\). The stopped process is \(X_{t\wedge\tau}\).
What the Exit Times Guarantee
Proof. Continuity and adaptedness imply that \(\sup_{0\le s\le t}|X_s|\) is \(\mathcal{F}_t\)-measurable: the supremum can be computed over a countable dense set of times, together with \(t\). Continuity also gives \(\{\tau_n\le t\}=\{\sup_{0\le s\le t}|X_s|\ge n\}\), so \(\tau_n\) is a stopping time. The exit set for level \(n+1\) is contained in the exit set for level \(n\), giving \(\tau_n\le\tau_{n+1}\). For any fixed \(T<\infty\), each continuous sample path is bounded on \([0,T]\). Thus, for all sufficiently large \(n\), that path does not exit \((-n,n)\) by time \(T\). This holds for every finite \(T\), so \(\tau_n\to\infty\) almost surely. Before \(\tau_n\), continuity gives \(|X_t|<n\); at a finite first exit, continuity gives \(|X_{\tau_n}|=n\). Since \(|X_0|<n\), the stopped process therefore satisfies the asserted bound. \(\square\)
The same idea controls coefficients. Suppose \(X_t\) satisfies the stochastic differential equation \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\), where \(\mu\) is the drift and \(\sigma\) is the diffusion coefficient. If both coefficients are bounded on every compact set \([0,T]\times[-n,n]\), then before \(\tau_n\) they are bounded by constants depending on \(T\) and \(n\). In particular, for some finite \(C_{n,T}\), \(|\sigma(s,X_s)|\le C_{n,T}\) for \(s\le T\wedge\tau_n\). The stopped stochastic integrand is progressively measurable and satisfies
Thus the stopped Ito integral is square-integrable. The Ito isometry gives its finite second moment; the martingale property follows from the construction of the Ito integral, first for adapted step integrands by conditional expectation of Brownian increments and then by the \(\mathcal{L}^2\) limit. This conclusion requires the displayed integrability bound. A local martingale, by itself, need not be a true martingale, and stopping does not remove the need to check integrability.
Worked Uses of Cutoffs
Worked Example: An Explicit First Exit
Let \(X_t=3t\) for \(t\ge0\), and stop when \(X\) first reaches level \(2\). Find the exit time and stopped path, and check the bound.
Since \(3t\) is increasing, the first time \(3t\ge2\) is the solution of \(3t=2\), namely \(\tau_2=\tfrac23\). For \(t\le\tfrac23\), \(t\wedge\tau_2=t\), so \(X_{t\wedge\tau_2}=3t\). For \(t\ge\tfrac23\), \(t\wedge\tau_2=\tfrac23\), so \(X_{t\wedge\tau_2}=3(\tfrac23)=2\). Therefore
Check: at \(t=\tfrac12\), the stopped value is \(\min(\tfrac32,2)=\tfrac32\); at \(t=1\), it is \(\min(3,2)=2\). The path reaches the boundary exactly at \(t=\tfrac23\) and never exceeds it after stopping.
Worked Example: Ito's Formula Before and After Exit
Let \(X_t=B_t\), start at \(0\), and set \(\tau_1=\inf\{t\ge0:|B_t|\ge1\}\). Apply Ito's formula to \(f(x)=x^3\) up to \(t\wedge\tau_1\), and verify that the stochastic integrand is square-integrable on a fixed horizon \(T\).
The derivatives are \(f'(x)=3x^2\) and \(f''(x)=6x\). Since \(dB_t\,dB_t=dt\), the Ito correction is \(\tfrac12 f''(B_t)\,dt=3B_t\,dt\). Consequently, the stopped formula is
On the stopped interval \(|B_s|\le1\). The Ito integrand \(H_s=\mathbf{1}_{\{s\le\tau_1\}}3B_s^2\) is progressively measurable and satisfies \(\mathbb{E}\int_0^T H_s^2\,ds\le9T\), so its Ito integral is a true square-integrable martingale. The ordinary integral is also bounded in absolute value by \(3T\). Check: before exit, the differential terms are \(3B_t^2\,dB_t+3B_t\,dt\), precisely \(f'(B_t)\,dB_t+\tfrac12f''(B_t)\,dt\); if \(t\ge\tau_1\), both integrals stop at \(\tau_1\), so the identity remains the formula for the stopped value.
Worked Example: Localizing an Unbounded Diffusion Coefficient
Consider \(X_t=\exp(B_t-\tfrac12t)\), which satisfies \(dX_t=X_t\,dB_t\), and let \(\tau_n=\inf\{t\ge0:X_t\ge n\}\), with \(n>1\). Apply Ito's formula to \(f(x)=x^2\) up to \(T\wedge\tau_n\).
First, for \(g(t,b)=\exp(b-\tfrac12t)\), the derivatives are \(g_t=-\tfrac12g\), \(g_b=g\), and \(g_{bb}=g\). Ito's formula gives \(dX_t=(-\tfrac12g+\tfrac12g)\,dt+g\,dB_t=X_t\,dB_t\), verifying the stated equation. For \(f(x)=x^2\), \(f'(x)=2x\) and \(f''(x)=2\), so the correction is \(\tfrac12f''(X_t)X_t^2\,dt=X_t^2\,dt\). The stopped identity is
Because \(X_0=1<n\), continuity gives \(0<X_{s\wedge\tau_n}\le n\). Thus the stochastic integrand \(H_s=\mathbf{1}_{\{s\le\tau_n\}}2X_s^2\) is progressively measurable and \(\mathbb{E}\int_0^T H_s^2\,ds\le4n^4T\). Its Ito integral is therefore a square-integrable martingale. The drift term is bounded above by \(n^2T\), so its expectation is finite. Taking expectations is legitimate and yields
Check: directly, \(X_t^2=\exp(2B_t-t)\). Applying Ito to this expression gives \(d(X_t^2)=2X_t^2\,dB_t+X_t^2\,dt\), agreeing with the stopped formula before exit. After exit, both integrals stop and the left side stays at \(X_{\tau_n}^2\).
Using the Cutoffs Correctly
For a fixed finite \(T\), the exit-time proposition implies that eventually \(T<\tau_n\) on almost every sample path. Hence \(X_{T\wedge\tau_n}=X_T\) eventually path by path. This observation lets one pass from identities for stopped processes toward identities for the original process. It does not, on its own, justify passing expectations or limits through an integral. Such a step requires an additional convergence or integrability argument, such as a suitable dominating bound or uniform integrability; those are separate conditions, not consequences of \(\tau_n\to\infty\).
For coefficients that depend on time and state, stopping at \(|X_t|\ge n\) bounds the coefficients only when they are bounded on the corresponding compact time-state region. If coefficients are merely locally bounded, this compact-region condition is exactly what is needed. If they can become singular inside that region, the state-exit time alone may not suffice; one can also stop when an accumulated coefficient quantity reaches a prescribed level, provided that quantity is continuous and adapted.
Common mistake. A bound on \(X_{t\wedge\tau}\) is a bound on the stopped process, not on \(X_t\) after the exit. For \(X_t=t\) and \(\tau=1\), the stopped process is \(X_{t\wedge1}=\min(t,1)\). At \(t=2\), it equals \(1\), while \(X_2=2\). The incorrect estimate \(\int_0^2X_s^2\,ds\le2\), based on the stopped bound \(X_{s\wedge1}\le1\), fails: the unstopped integral is \(\int_0^2s^2\,ds=\tfrac83\). The correctly stopped integral is \(\int_0^{2\wedge1}s^2\,ds=\tfrac13\), which is bounded by \(1^2\cdot1=1\). Always keep the stopping time in the integral as well as in the process being bounded.
Check Your Understanding
For each question, state the assumptions needed for the cutoff or expectation argument.
- For a continuous adapted process with bounded initial value, explain why its first exit times from \((-n,n)\) increase to infinity almost surely.
- If \(|\sigma(s,X_s)|\le C\) before a stopping time on \([0,T]\), give the bound that verifies the stopped Ito integrand belongs to \(\mathcal{L}^2\).
- For \(f(x)=x^4\) and \(X_t=B_t\), write the first- and second-derivative terms in Ito's formula stopped at \(\tau_n\).
- In the example \(X_t=t\), compare \(\int_0^2 X_s^2\,ds\) with \(\int_0^{2\wedge1} X_s^2\,ds\). Which bound comes from stopping at \(1\)?
- Why does almost-sure eventual equality \(X_{T\wedge\tau_n}=X_T\) not by itself permit taking a limit inside an expectation?