From a Fixed Time to a Bounded Stopping Time
Dynkin’s formula relates the expected change of a function of a diffusion to the expected time integral of its generator. At a fixed time, Ito’s formula gives the relation directly, and the mean-zero property of the Ito integral removes its stochastic term. The same idea applies when the time is random, provided that the stopping time is bounded and the stochastic integral is a true martingale. Boundedness alone is not enough: the integrand and the generator term must also satisfy suitable integrability conditions.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Suppose that \(X_t\) is a continuous diffusion satisfying \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\), with initial value \(X_0=x\). Here \(\mu\) is the drift coefficient and \(\sigma\) is the diffusion coefficient. For a twice continuously differentiable function \(f:\mathbb{R}\to\mathbb{R}\), its generator is \(\mathcal{L}_t f(y)=\mu(t,y)f'(y)+\tfrac12\sigma(t,y)^2f''(y)\).
In this setting, Ito’s formula has a second-order correction because Brownian motion has quadratic variation \([B]_t=t\). The differential bookkeeping is \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\); this is not the ordinary chain rule for differentiable paths. For \(f(X_t)\), the correction is \(\tfrac12 f''(X_t)\sigma(t,X_t)^2\,dt\), which is the second term in the generator.
Dynkin’s Formula at Bounded Stopping Times
Proof. Ito’s formula, with \(f'(X_s)\) and \(f''(X_s)\) as the first and second derivatives, gives \(f(X_t)=f(X_0)+\int_0^t\mathcal{L}_s f(X_s)\,ds+M_t\), where \(M_t=\int_0^t H_s\,dB_s\) and \(H_s=f'(X_s)\sigma(s,X_s)\). The assumed square-integrability makes \(H\) an admissible Ito integrand in \(\mathcal{L}^2\). By the Ito isometry, \(M_t\) is square-integrable. Its martingale property follows from the construction of the Ito integral: for \(0\le u\le t\le T\), \(\mathbb{E}[M_t-M_u\mid\mathcal{F}_u]=0\), first for adapted step integrands by independence and zero mean of Brownian increments, then for \(\mathcal{L}^2\) integrands by approximation in \(L^2\).
For this continuous Ito martingale, bounded optional stopping gives \(\mathbb{E}[M_\tau]=\mathbb{E}[M_0]=0\). To see the zero-mean assertion directly, the stopped-integral identity gives \(M_\tau=\int_0^T\mathbf{1}_{\{s\le\tau\}}H_s\,dB_s\). The stopped integrand is progressively measurable and satisfies \(\mathbb{E}\int_0^T(\mathbf{1}_{\{s\le\tau\}}H_s)^2\,ds\le\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Thus this is an Ito integral in \(\mathcal{L}^2\), with expectation zero. Applying Ito’s identity at \(\tau\) and taking expectations is justified by the assumed absolute integrability of the generator term and the square-integrability of the martingale. The result is precisely the formula stated. \(\square\)
The theorem concerns the deterministic interval \([0,T]\) in both integrability assumptions. It does not suffice to bound a stopped process and then infer integrability of a function of the unstopped process \(X_s\). This distinction matters in exit problems, where the process can leave the interval after its first exit even though its stopped version remains inside it.
Worked Applications
Worked Example: Expected Exit Time from an Interval
Let \(X_t=x+B_t\), where \(0<x<2\), and let \(\tau=\inf\{t\ge0:X_t\notin(0,2)\}\). This exit time need not be bounded, so first use the bounded stopping time \(\tau_T=\tau\wedge T\). Find \(\mathbb{E}[\tau]\).
Use \(f(y)=y(2-y)=2y-y^2\). Its derivatives are \(f'(y)=2-2y\) and \(f''(y)=-2\). For this Brownian diffusion, \(\mu=0\) and \(\sigma=1\), so the generator is \(\mathcal{L}f(y)=0\cdot f'(y)+\tfrac12 f''(y)=-1\). Ito’s formula explicitly reads \(df(X_t)=(2-2X_t)\,dB_t-1\,dt\); the \(\tfrac12 f''(X_t)\,dt=-dt\) term is essential.
For each finite \(T\), verify the theorem’s hypotheses on the deterministic interval \([0,T]\). Since \(X_s=x+B_s\), \(\mathbb{E}[(1-X_s)^2]=(1-x)^2+s\). Therefore \(\mathbb{E}\int_0^T(2-2X_s)^2\,ds =4\int_0^T((1-x)^2+s)\,ds =4(1-x)^2T+2T^2<\infty\). Also \(\mathbb{E}\int_0^T|\mathcal{L}f(X_s)|\,ds=T<\infty\). The formula gives \(\mathbb{E}[f(X_{\tau_T})]=f(x)-\mathbb{E}[\tau_T]\), or equivalently \(\mathbb{E}[\tau_T]=f(x)-\mathbb{E}[f(X_{\tau_T})]\).
Before exit, \(X_t\in(0,2)\); at exit, continuity gives \(X_\tau\in\{0,2\}\). Thus \(X_{\tau_T}\in[0,2]\), so \(f(X_{\tau_T})\ge0\). It follows that \(0\le\mathbb{E}[\tau_T]\le f(x)\). As \(T\) increases, \(\tau_T\) increases to \(\tau\), so monotone convergence gives \(\mathbb{E}[\tau]\le f(x)<\infty\), and hence \(\tau<\infty\) almost surely. By continuity, \(X_{\tau_T}\to X_\tau\in\{0,2\}\), so \(f(X_{\tau_T})\to0\). Since \(0\le f(X_{\tau_T})\le\max_{0\le y\le2}f(y)=1\), dominated convergence gives \(\mathbb{E}[f(X_{\tau_T})]\to0\). Consequently \(\mathbb{E}[\tau]=f(x)=x(2-x)\).
Check: for \(x=1\), the answer is \(1(2-1)=1\). For \(x=\tfrac12\), it is \(\tfrac12(2-\tfrac12)=\tfrac34\). The answer tends to zero as \(x\) approaches either boundary, consistent with starting closer to an endpoint.
Worked Example: A Brownian Motion Stopped at a Barrier
Let \(X_t=B_t\), \(X_0=0\), \(T=1\), and \(\tau=1\wedge\inf\{t\ge0:B_t=1\}\). Show that \(\mathbb{E}[B_\tau]=0\).
Take \(f(y)=y\), so \(f'(y)=1\), \(f''(y)=0\), and \(\mathcal{L}f=0\) for standard Brownian motion. The Ito formula is \(df(B_t)=1\,dB_t+0\cdot dt\); in particular, the second-derivative correction is \(\tfrac12 f''(B_t)\,dt=0\). The deterministic-horizon integrability checks are \(\mathbb{E}\int_0^1(f'(B_s))^2\,ds=\int_0^1 1\,ds=1\) and \(\mathbb{E}\int_0^1|\mathcal{L}f(B_s)|\,ds=0\). Dynkin’s formula therefore gives \(\mathbb{E}[B_\tau]=f(0)+0=0\).
Check: the stopped-process representation is \(B_\tau=\int_0^1\mathbf{1}_{\{s\le\tau\}}\,dB_s\). Its integrand has squared integral at most \(1\), so the Ito isometry gives \(\mathbb{E}[B_\tau^2]=\mathbb{E}\int_0^1\mathbf{1}_{\{s\le\tau\}}\,ds=\mathbb{E}[\tau]\le1\). Thus \(B_\tau\) is integrable, and the mean-zero conclusion is consistent with a square-integrable stopped Ito integral.
Worked Example: A Stopping Rule Chosen at an Intermediate Time
For standard Brownian motion, define \(A=\{B_{1/2}>0\}\) and set \(\tau=\tfrac12\) on \(A\), \(\tau=1\) on \(A^c\). This is a stopping time: before \(1/2\) it has not occurred, between \(1/2\) and \(1\) the event \(\{\tau\le t\}\) is \(A\in\mathcal{F}_{1/2}\subseteq\mathcal{F}_t\), and at \(t=1\) it has certainly occurred. Compute \(\mathbb{E}[B_\tau^2]\) using Dynkin’s formula, then check it directly.
Choose \(f(y)=y^2\), with \(f'(y)=2y\), \(f''(y)=2\), and \(\mathcal{L}f(y)=\tfrac12(2)=1\). Ito’s formula is \(d(B_t^2)=2B_t\,dB_t+1\,dt\); its correction term is \(\tfrac12 f''(B_t)\,dt=dt\). On the deterministic horizon \([0,1]\), \(\mathbb{E}\int_0^1(2B_s)^2\,ds=4\int_0^1s\,ds=2<\infty\), and \(\mathbb{E}\int_0^1|\mathcal{L}f(B_s)|\,ds=1\). Since \(B_0=0\), Dynkin’s formula yields \(\mathbb{E}[B_\tau^2]=0+\mathbb{E}\int_0^\tau1\,ds=\mathbb{E}[\tau]\). The event \(A\) has probability \(1/2\), by symmetry of the centered normal variable \(B_{1/2}\). Hence \(\mathbb{E}[\tau]=\tfrac12\cdot\tfrac12+\tfrac12\cdot1=\tfrac34\), and \(\mathbb{E}[B_\tau^2]=\tfrac34\).
For a separate check, symmetry gives \(\mathbb{E}[\mathbf{1}_A B_{1/2}^2]=\tfrac12\mathbb{E}[B_{1/2}^2]=\tfrac12\cdot\tfrac12=\tfrac14\). Write \(B_1=B_{1/2}+(B_1-B_{1/2})\). The increment \(B_1-B_{1/2}\) is independent of \(\mathcal{F}_{1/2}\), has mean zero, and has second moment \(1/2\). Therefore \(\mathbb{E}[\mathbf{1}_{A^c}B_1^2]=\mathbb{E}[\mathbf{1}_{A^c}B_{1/2}^2]+\mathbb{P}(A^c)\cdot\tfrac12 =\tfrac14+\tfrac12\cdot\tfrac12=\tfrac12\). Adding the two contributions gives \(\mathbb{E}[B_\tau^2]=\tfrac14+\tfrac12=\tfrac34\), as before.
Why Boundedness and Integrability Both Matter
The bounded-stopping-time formula is useful because it turns a random terminal value into an expected accumulated generator. In the exit-time example, the formula was applied only to \(\tau\wedge T\). The passage to \(\tau\) required a separate argument: monotone convergence for \(\tau\wedge T\), almost-sure convergence of the stopped position, and dominated convergence for the bounded function \(f\) on the interval. Writing the formula at an unbounded \(\tau\) without such a limiting argument would leave its hypotheses unverified.
Common mistake. On the exit interval \((0,2)\), it is wrong to argue, “\(X_{s\wedge\tau}\in[0,2]\), so \(\mathbb{E}\int_0^T(2-2X_s)^2\,ds<\infty\).” The theorem’s integrand is evaluated at the unstopped process \(X_s\), including times after exit. The stopped-process bound alone does not establish this deterministic-horizon condition. For the Brownian example, the correct verification uses \(X_s=x+B_s\) and computes \(\mathbb{E}\int_0^T(2-2X_s)^2\,ds=4(1-x)^2T+2T^2<\infty\). The generator condition is checked separately: \(\mathcal{L}f=-1\), so \(\mathbb{E}\int_0^T|\mathcal{L}f(X_s)|\,ds=T\).
Also distinguish a true martingale from a local martingale. Ito’s formula initially supplies a local-martingale term in general. Here the explicit \(\mathcal{L}^2\) condition makes its stochastic integral square-integrable on \([0,T]\), hence a true martingale with the conditional-expectation property used above. Without such integrability, optional stopping cannot be invoked in this form merely because the stopping time is bounded.
Check Your Understanding
Use the bounded-stopping-time formula only after checking its hypotheses.
- For Brownian motion started at \(x\in(0,a)\), which function has generator \(-1\) and vanishes at both endpoints of \((0,a)\)?
- Why does \(\tau\le T\) not by itself guarantee that \(\mathbb{E}\int_0^T(f'(X_s)\sigma(s,X_s))^2\,ds\) is finite?
- For \(f(y)=y^2\) and standard Brownian motion, identify the first derivative, second derivative, generator, and Ito correction term.
- In the intermediate-time stopping example, explain why the event \(A=\{B_{1/2}>0\}\) is measurable at the time when the rule chooses whether to stop.
- In the interval-exit example, which convergence theorem justifies passing from \(\mathbb{E}[\tau\wedge T]\) to \(\mathbb{E}[\tau]\), and which justifies the limit of the terminal \(f\)-term?