Stopping-Time Versions of the Brownian Ito Formula

Why Stopping the Formula Matters

For a Brownian motion $B_t$, Ito's formula gives an identity of processes: it holds at every deterministic time, and, on a single event of probability one, at every time in a bounded interval. A stopping time $\tau$ selects a random time using information available up to that time. Evaluating the process identity at $\tau$ therefore stops both integrals at $\tau$; it does not turn an integral over a deterministic interval into an integral whose integrand is evaluated only at the endpoint.

Throughout, let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ satisfy the usual conditions, and let $B_t$ be Brownian motion with respect to this filtration. In particular, for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. For an Ito integral, the integrand is progressively measurable and belongs to $\mathcal{L}^2$ on the stated horizon: $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. These conditions make the integral a square-integrable martingale, not merely a local martingale.

Stopped Ito formula: Suppose $f\in C^2(\mathbb{R})$, $\tau\le T$ is a stopping time, and $f'(B_s)\mathbf{1}_{\{s\le\tau\}}$ belongs to $\mathcal{L}^2$ on $[0,T]$. Then $$ f(B_\tau)=f(B_0)+\int_0^T\mathbf{1}_{\{s\le\tau\}}f'(B_s)\,dB_s +\frac12\int_0^T\mathbf{1}_{\{s\le\tau\}}f''(B_s)\,ds. $$ Equivalently, the two integrals are written with upper limit $\tau$. If the displayed integrability condition is not known, the identity is first used after localization; the resulting integrals are then local Ito integrals.

The value of the indicator at the single time $s=\tau$ does not affect either integral. The process $\mathbf{1}_{\{s\le\tau\}}$ is adapted and left-continuous as a function of $s$, hence predictable. The stopped stochastic integrand is therefore an admissible Ito integrand whenever it satisfies the stated square-integrability condition.

Why the Bounded-Time Formula Holds

Proposition: Let $f\in C^2(\mathbb{R})$, let $T<\infty$, and assume $f'(B_s)\in\mathcal{L}^2$ on $[0,T]$. For every stopping time $\tau\le T$, the stopped Ito identity in the box above holds almost surely.

Proof. Ito's formula on $[0,T]$ states that, simultaneously for $0\le t\le T$, $$ f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds. $$ The stochastic integral is a continuous martingale, and the ordinary time integral is continuous wherever its integrand is locally integrable; localization gives this property if needed. Thus this process identity can be evaluated at $t=\tau$. By the definition of the stopped Ito integral, $\int_0^\tau f'(B_s)\,dB_s=\int_0^T\mathbf{1}_{\{s\le\tau\}}f'(B_s)\,dB_s$. The same change of upper limit holds for the ordinary integral. Substitution of $t=\tau$ proves the displayed identity. $\square$

The assumption $f'(B_s)\in\mathcal{L}^2$ ensures the stochastic integral is a true martingale on $[0,T]$. In particular, its stopped value has expectation zero: the stopped integrand is also in $\mathcal{L}^2$, and the Ito integral of that integrand is a square-integrable martingale. Without such integrability, the integral obtained by localization is in general only a local martingale; a local martingale need not have the conditional-expectation property of a true martingale.

Worked Example: A Random Bounded Stopping Time

Set $t_0=\tfrac14$, $T=\tfrac12$, and define $\tau=t_0$ on $\{B_{t_0}\ge0\}$, and $\tau=T$ on $\{B_{t_0}<0\}$. This is a stopping time: before $t_0$, $\{\tau\le t\}$ is empty; for $t_0\le t<T$, it equals $\{B_{t_0}\ge0\}\in\mathcal{F}_t$; and for $t\ge T$, it is all of $\Omega$. The event boundary is assigned to the first case, so the two cases partition the sample space.

Apply Ito's formula to $f(x)=x^2$. Its derivatives are $f'(x)=2x$, $f''(x)=2$, and $f(0)=0$. The correction term is $\tfrac12 f''(B_s)\,ds=ds$, so $$ B_\tau^2=2\int_0^\tau B_s\,dB_s+\tau. $$ The integrand is square-integrable because $\mathbb{E}\int_0^T\mathbf{1}_{\{s\le\tau\}}(2B_s)^2\,ds\le4\int_0^{1/2}s\,ds=\tfrac12$. Hence the stopped stochastic integral has mean zero.

Now check the expectation directly. Since $B_{1/4}$ is a centered normal random variable, symmetry and its variance give $\mathbb{P}(B_{1/4}\ge0)=\mathbb{P}(B_{1/4}<0)=\tfrac12$ and $\mathbb{E}[B_{1/4}^2\mathbf{1}_{\{B_{1/4}\ge0\}}]=\mathbb{E}[B_{1/4}^2\mathbf{1}_{\{B_{1/4}<0\}}]=\tfrac18$. Write $B_{1/2}=B_{1/4}+(B_{1/2}-B_{1/4})$. The increment is independent of $\mathcal{F}_{1/4}$, has mean zero, and has variance $\tfrac14$. Therefore $$ \mathbb{E}[B_\tau^2] =\tfrac18+\mathbb{E}[B_{1/2}^2\mathbf{1}_{\{B_{1/4}<0\}}] =\tfrac18+\tfrac18+\tfrac14\cdot\tfrac12 =\tfrac38. $$ Also, $\mathbb{E}[\tau]=\tfrac14\cdot\tfrac12+\tfrac12\cdot\tfrac12=\tfrac38$. This verifies the expectation of the stopped identity.

Time-Dependent Functions and the Stopping Rule

For $g\in C^{1,2}([0,T]\times\mathbb{R})$, Ito's formula before stopping is $$ g(t,B_t)=g(0,B_0)+\int_0^t g_x(s,B_s)\,dB_s +\int_0^t\left(g_t(s,B_s)+\frac12g_{xx}(s,B_s)\right)\,ds. $$ The derivatives in the stochastic and ordinary integrals are evaluated at the running time $s$, not at the terminal time. When the relevant stopped stochastic integrand is in $\mathcal{L}^2$, the bounded stopping version is $$ g(\tau,B_\tau)=g(0,B_0)+\int_0^T\mathbf{1}_{\{s\le\tau\}}g_x(s,B_s)\,dB_s +\int_0^T\mathbf{1}_{\{s\le\tau\}}\left(g_t(s,B_s)+\frac12g_{xx}(s,B_s)\right)\,ds. $$ The term $\tfrac12g_{xx}\,ds$ is the Ito correction arising from $dB_s\,dB_s=ds$; the rules $dB_s\,ds=0$ and $ds\,ds=0$ do not remove it. This is the Ito formula, not the ordinary chain rule. For a Stratonovich integral the conversion rule is different: for a sufficiently regular integrand, $\int H_s\circ dB_s=\int H_s\,dB_s+\tfrac12[H,B]$, where $[H,B]$ denotes quadratic covariation.

Worked Example: A Stopped Exponential Martingale

Let $g(t,x)=e^{-t/2}\cos x$. Its derivatives are $g_t=-\tfrac12e^{-t/2}\cos x$, $g_x=-e^{-t/2}\sin x$, and $g_{xx}=-e^{-t/2}\cos x$. In particular, $g_t+\tfrac12g_{xx}=-\tfrac12e^{-t/2}\cos x-\tfrac12e^{-t/2}\cos x=-e^{-t/2}\cos x$. Correction: this does not cancel, so $g$ is not the desired martingale test function.

Instead take $g(t,x)=e^{t/2}\cos x$. Then $g_t=\tfrac12e^{t/2}\cos x$, $g_x=-e^{t/2}\sin x$, and $g_{xx}=-e^{t/2}\cos x$, giving $g_t+\tfrac12g_{xx}=0$. Ito's formula therefore yields $$ e^{t/2}\cos B_t=1-\int_0^t e^{s/2}\sin B_s\,dB_s, $$ since $B_0=0$. For a bounded stopping time $\tau\le T$, the stopped integrand has square expectation bounded by $\int_0^T e^s\,ds=e^T-1<\infty$. The integral is a true martingale, so $\mathbb{E}[e^{\tau/2}\cos B_\tau]=1$. This check depends on boundedness and integrability; cancellation of the drift alone would establish only a local martingale without those conditions.

Unbounded Stopping Times Require Localization

If $\tau$ is unbounded, one cannot simply invoke the bounded-time proposition with $T=\infty$. First stop at $\tau_n=\tau\wedge n$. If the function or its derivatives are unbounded, one can also stop on leaving a compact region, for example at $\rho_m=\inf\{t\ge0:|B_t|\ge m\}$, and apply the formula at $\tau\wedge n\wedge\rho_m$. On each such bounded, localized interval, the derivatives are bounded on the visited compact set, so the required stochastic integrand is square-integrable.

Unbounded-time extension: Suppose $\tau<\infty$ almost surely and $f\in C^2(\mathbb{R})$. If $$ \mathbb{E}\int_0^\tau f'(B_s)^2\,ds<\infty \quad\text{and}\quad \int_0^\tau |f''(B_s)|\,ds<\infty\ \text{almost surely}, $$ then the stopped Ito identity holds at $\tau$. The stochastic integral converges in $L^2$ as the upper limits $\tau\wedge n$ increase to $\tau$; the ordinary integral converges almost surely.

Proof. Apply the bounded-time identity at $\tau\wedge n$, using compact localization as necessary and then removing it. Since $\tau<\infty$ almost surely, continuity of Brownian paths gives $B_{\tau\wedge n}\to B_\tau$ almost surely. For the stochastic terms, Ito's isometry gives $$ \mathbb{E}\left[\left(\int_{\tau\wedge n}^{\tau} f'(B_s)\,dB_s\right)^2\right] =\mathbb{E}\int_{\tau\wedge n}^{\tau} f'(B_s)^2\,ds\longrightarrow0. $$ The convergence follows by dominated convergence, since the tail integrals decrease to zero almost surely and are bounded by the integrable random integral in the hypothesis. The ordinary integrals converge by absolute integrability. Taking limits in the identity at $\tau\wedge n$ proves the result. $\square$

The conclusion is an identity of random variables; it does not automatically say that the stochastic integral has mean zero if the square-integrability hypothesis fails. Without $\mathbb{E}\int_0^\tau f'(B_s)^2ds<\infty$, localization may produce only a local martingale, and expectation cannot be passed through the limit without additional justification.

Worked Example: An Unbounded but Integrable Stopping Time

Enlarge the initial sigma-field by an independent random variable $U$ with exponential distribution of rate $1$, so $\mathbb{P}(U>u)=e^{-u}$ for $u\ge0$, and assume $B$ remains Brownian with respect to the enlarged filtration. Then $\tau=U$ is a stopping time, is finite almost surely, and is unbounded. For $f(x)=x^2$, the derivatives are $2x$ and $2$, giving $$ B_{U\wedge n}^2=2\int_0^{U\wedge n}B_s\,dB_s+U\wedge n. $$ Moreover, $\mathbb{E}\int_0^U(2B_s)^2ds=4\int_0^\infty s\,\mathbb{P}(U>s)\,ds=4\int_0^\infty se^{-s}ds=4$. Thus the stochastic integral converges in $L^2$. Also, $\mathbb{E}[U]=1$ and $\mathbb{E}[B_U^2]=\mathbb{E}[U]=1$, by conditioning on $U$ and using the variance of $B_u$. The limiting identity is $B_U^2=2\int_0^U B_s\,dB_s+U$, and its expectation checks as $1=0+1$.

Common Pitfalls

Stopping only the endpoint is not enough. The incorrect expression $f(B_\tau)=f(B_0)+\int_0^T f'(B_s)\,dB_s+\frac12\int_0^T f''(B_s)\,ds$ is the formula at $T$, not at $\tau$. The correction is to stop both integrals, using $\int_0^T\mathbf{1}_{\{s\le\tau\}}(\cdots)\,ds$ and the corresponding stopped Ito integral.

Bounded time is not a substitute for integrability. Even when $\tau\le T$, a rapidly growing $f'$ need not satisfy $\mathbb{E}\int_0^T f'(B_s)^2ds<\infty$. In that case state and use a localization argument rather than claiming the stochastic integral is a true martingale. For an unbounded $\tau$, passing from $\tau\wedge n$ to $\tau$ requires convergence of the stochastic and ordinary integral terms, not just the almost-sure convergence $B_{\tau\wedge n}\to B_\tau$.

Check Your Understanding

For each question, distinguish an identity of stopped processes from a claim about expectations.

For $f(x)=x^3$, write the Ito identity at a bounded stopping time $\tau$, including the explicit derivatives and correction term.
Why does $f'(B_s)\mathbf{1}_{\{s\le\tau\}}\in\mathcal{L}^2$ justify taking the expectation of the stopped stochastic integral?
State the two convergence requirements used to pass from $\tau\wedge n$ to an almost-surely finite unbounded $\tau$.
Explain why a local martingale with zero drift need not, without more conditions, have constant expectation.
For $g(t,x)=e^{t/2}\cos x$, calculate $g_t+\tfrac12g_{xx}$ and identify the stopped stochastic integrand.