Why Stop a Girsanov Transform?
Girsanov’s theorem changes the drift of Brownian motion by using a stochastic exponential as a density. Its direct use requires that the exponential be a true martingale, not merely a local martingale. For an unbounded adapted drift integrand, a global integrability test such as Novikov’s condition may fail or be difficult to verify. Stopping can control the accumulated squared integrand, give a valid density on each stopped problem, and permit arguments to proceed locally.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Fix a finite horizon \(T\). Assume that \(\theta_t\) is progressively measurable and locally square-integrable: \(\int_0^T\theta_s^2\,ds<\infty\) almost surely. This pathwise condition allows the stochastic integral to be defined locally; it does not assert that \(\mathbb{E}\int_0^T\theta_s^2\,ds<\infty\).
Here \(\int H_s\,dB_s\) denotes the Ito integral. For a fixed horizon, its square-integrable integrands are the progressively measurable \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); more general locally square-integrable integrands are defined by localization. A martingale is an integrable adapted process satisfying the conditional-expectation identity at every pair of times. A local martingale satisfies that identity after stopping along a sequence of stopping times increasing to the horizon, but need not itself be a martingale. The stochastic exponential above is initially only a local martingale.
Stopping at a Fixed Energy Level
Let \(A_t=\int_0^t\theta_s^2\,ds\), a continuous, increasing adapted process. For \(n>0\), set \(\tau_n=\inf\{t\ge0:A_t\ge n\}\), with the infimum of the empty set equal to infinity. The stopped kernel is \(\theta_s\mathbf{1}_{\{s\le\tau_n\}}\); changing its value at the single endpoint does not change either integral. Its accumulated squared size by time \(T\) is \(A_{T\wedge\tau_n}\le n\).
Proof. The stopped kernel has expected squared integral at most \(n\), so it belongs to the square-integrable class on \([0,T]\). Its stochastic exponential is a positive local martingale. Novikov’s condition for a progressively measurable kernel \(H\), namely \(\mathbb{E}\exp(\frac12\int_0^T H_s^2\,ds)<\infty\), implies that its stochastic exponential is a true martingale. For \(H_s=\theta_s\mathbf{1}_{\{s\le\tau_n\}}\), the exponential inside this expectation is at most \(e^{n/2}\). Hence Novikov’s condition holds, and \(\mathbb{E}[Z^{(n)}_t]=1\) for all \(t\le T\). Positivity and expectation one make \(Z^{(n)}_T\) a probability density. The Girsanov theorem applied to this true density says that \(B_t+\int_0^t H_s\,ds=B^{(n)}_t\) is Brownian motion under \(\mathbb{Q}_n\). \(\square\)
The result is local in the drift being changed: before \(\tau_n\), the shift uses the original \(\theta\); after \(\tau_n\), it uses zero. This distinction matters. The theorem does not say that an un-stopped density \(Z_T\) has expectation one, nor that there is a single measure under which the full, unstopped drift change is valid.
Worked Example: An Unbounded Deterministic Kernel
On \([0,1]\), take \(\theta_t=t^{-1/4}\) for \(t>0\), with its value at \(t=0\) chosen arbitrarily. This kernel is unbounded near zero, but its squared energy is finite. Find the stopping time at level \(n=1\), and verify the stopped density has expectation one.
First compute the accumulated energy for \(0\le t\le1\): $$ A_t=\int_0^t s^{-1/2}\,ds=2\sqrt{t}. $$ It reaches \(1\) when \(2\sqrt{t}=1\), so \(\tau_1=\tfrac14\). At that time the stochastic integral is centered normal with variance \(A_{\tau_1}=1\). If \(G=\int_0^{1/4}s^{-1/4}\,dB_s\), then \(G\sim N(0,1)\), and $$ Z^{(1)}_1=\exp(-G-\tfrac12). $$ For a standard normal \(G\), \(\mathbb{E}[e^{uG}]=e^{u^2/2}\). Substituting \(u=-1\) gives \(\mathbb{E}[Z^{(1)}_1]=e^{-1/2}e^{1/2}=1\).
Check: the stopped energy is exactly \(1\), so the Novikov bound is \(\mathbb{E}[\exp(\tfrac12 A_{\tau_1})]=e^{1/2}\), finite. The computation agrees with the theorem. The unboundedness of \(\theta\) at zero causes no problem because its squared energy over the stopped interval is finite.
Localized Drift Removal
Suppose a process \(X_t\), up to the same stopping time, satisfies $$ dX_t=\theta_t\,dt+dB_t. $$ Under \(\mathbb{Q}_n\), \(B^{(n)}\) is Brownian and, for \(t\le\tau_n\), \(dB_t=dB^{(n)}_t-\theta_t\,dt\). Substitution gives \(dX_t=dB^{(n)}_t\) before the stop. After \(\tau_n\), the density no longer changes the drift, so one must not claim that this equation holds there. This is the basic use in localization arguments: prove a statement on each stopped interval under \(\mathbb{Q}_n\), then separately justify any passage to a limit.
Worked Example: Removing Drift Only Before the Stop
Let \(\theta_t=B_t\), take \(T=1\), and define \(\tau_n=\inf\{t:\int_0^t B_s^2\,ds\ge n\}\). Set \(X_t=\int_0^t B_s\,ds+B_t\). Explain its dynamics under \(\mathbb{Q}_n\) before and after the stop.
The stopped energy satisfies \(\int_0^{1\wedge\tau_n}B_s^2\,ds\le n\), so the theorem applies even though \(B_t\) is not bounded. By the definition of \(X\), \(dX_t=B_t\,dt+dB_t\). Under \(\mathbb{Q}_n\), \(dB_t=dB^{(n)}_t-B_t\mathbf{1}_{\{t\le\tau_n\}}dt\). Thus $$ dX_t= \begin{cases} dB^{(n)}_t,&t\le\tau_n,\\ B_t\,dt+dB^{(n)}_t,&t>\tau_n. \end{cases} $$ At the single stopping instant the choice of endpoint convention does not affect the integrated equation.
Check: integrate to \(t\le\tau_n\): \(X_t=\int_0^t B_s\,ds+B_t=B^{(n)}_t\), since \(B^{(n)}_t=B_t+\int_0^tB_s\,ds\) before the stop. For \(t>\tau_n\), the additional drift integral after \(\tau_n\) remains; the stopped measure has not removed it.
Consistency Across Stopping Levels
The stopping times are nested: if \(m>n\), then \(\tau_n\le\tau_m\). Though the measures \(\mathbb{Q}_n\) may differ on all of \(\mathcal{F}_T\), they agree on information available by \(\tau_n\). Precisely, \(\mathcal{F}_{\tau_n}\) consists of events \(A\) such that \(A\cap\{\tau_n\le t\}\in\mathcal{F}_t\) for every \(t\).
Proof. Both stopped exponentials are true martingales by the theorem. At \(T\wedge\tau_n\), the integrals defining \(Z^{(m)}\) and \(Z^{(n)}\) coincide, because \(\tau_n\le\tau_m\); hence \(Z^{(m)}_{T\wedge\tau_n}=Z^{(n)}_{T\wedge\tau_n}\). For an event \(A\) known by \(T\wedge\tau_n\), the martingale conditional-expectation identity gives $$ \mathbb{Q}_m(A)=\mathbb{E}[Z^{(m)}_T\mathbf{1}_A] =\mathbb{E}[Z^{(m)}_{T\wedge\tau_n}\mathbf{1}_A] =\mathbb{E}[Z^{(n)}_{T\wedge\tau_n}\mathbf{1}_A] =\mathbb{E}[Z^{(n)}_T\mathbf{1}_A] =\mathbb{Q}_n(A). $$ The first and last conditional-expectation steps use that \(A\) is measurable at the stopping time and that each density process is a true martingale. \(\square\)
Worked Example: Comparing Two Stopping Levels
For the kernel \(\theta_t=t^{-1/4}\) on \([0,1]\), compare the stopping times at levels \(n=1\) and \(n=4\), and identify the common density at the earlier stopping time.
Since \(A_t=2\sqrt t\), level \(1\) is reached at \(2\sqrt t=1\), giving \(\tau_1=\tfrac14\). Level \(4\) would be reached at \(2\sqrt t=4\), or \(t=4\), which lies beyond the horizon; hence \(\tau_4\ge1\) and the level-\(4\) process is not stopped before \(T=1\). At time \(\tau_1\), both exponentials equal $$ \exp\left(-\int_0^{1/4}s^{-1/4}\,dB_s-\frac12\int_0^{1/4}s^{-1/2}\,ds\right) =\exp(-G-\tfrac12), $$ where \(G\sim N(0,1)\). Its expectation is \(e^{-1/2}e^{1/2}=1\).
Check: the energy at \(\tau_1\) is \(2\sqrt{1/4}=1\), and the energy at \(T=1\) is \(2\), which is less than \(4\). Thus the level-\(4\) exponential runs unstopped through time \(1\), while its value at the earlier stopping time agrees exactly with the level-\(1\) density.
What Localization Does Not Establish
A frequent error is to infer a global change of measure merely because every stopped density has expectation one. The stopped variables are not automatically uniformly integrable as the stopping levels increase. In particular, almost-sure convergence of \(Z_{T\wedge\tau_n}\) to \(Z_T\), when it holds, does not by itself imply convergence of their expectations. Without such convergence, \(\mathbb{E}[Z_T]\) may be less than one, and \(Z_T\) then cannot be a probability density.
Common mistake. One might write, “Each \(\mathbb{E}[Z_{T\wedge\tau_n}]=1\), so taking \(n\to\infty\) gives \(\mathbb{E}[Z_T]=1\).” That step is not justified without an appropriate convergence theorem, such as uniform integrability of the stopped densities. The correction is to keep the conclusion at the level actually proved: each \(\mathbb{Q}_n\) is a probability measure, and the measures agree on the earlier stopped information. A global measure requires a separate argument establishing that the unstopped exponential is a true martingale.
Likewise, do not confuse a local martingale with a martingale. Positivity makes a stochastic exponential a supermartingale, but this only gives \(\mathbb{E}[Z_T]\le1\); it does not give equality. Novikov’s condition is a sufficient condition for the true-martingale property, not a necessary condition. Energy stopping gives a convenient sufficient bound at each level, while the global problem remains separate.
| Object | What is established | What is not automatic |
|---|---|---|
| \(Z^{(n)}_t=Z_{t\wedge\tau_n}\) | True martingale; expectation one | Equality with an unstopped density after the stop |
| \(\mathbb{Q}_n\) | Probability measure on \(\mathcal{F}_T\) | A single limiting measure for all \(n\) |
| Agreement on \(\mathcal{F}_{T\wedge\tau_n}\) | Consistency of the localized measures | Global equivalence or an unstopped drift change |
Check Your Understanding
Use the definitions and the stopped theorem to answer the following.
- Why does the bound \(\int_0^{T\wedge\tau_n}\theta_s^2\,ds\le n\) imply Novikov’s condition for the stopped kernel?
- With kernel sign \(-\theta\) in the density, what Brownian motion appears under the changed measure?
- For \(\theta_t=t^{-1/4}\), compute the energy at \(t=\tfrac1{16}\) and determine whether level \(1\) has been reached.
- In the adapted-kernel example, why does the drift \(B_t\,dt\) remain after \(\tau_n\)?
- What additional property of stopped densities would justify passing their expectation-one property to a limiting density?