Why Stop According to Energy?
An integrand can have finite square-integral on every bounded time interval for almost every outcome, while its expected square-integral is infinite. The first property is enough to describe local square-integrability, but it does not put the integrand in the usual \(L^2\) class required to define an Ito integral directly on a fixed interval. The remedy is to stop when the accumulated square-integral reaches a prescribed level. The stopped integrand then has a deterministic bound on its energy, and hence finite expected energy.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Assume the filtration satisfies the usual conditions. Let \(H\) be a predictable process and suppose it is locally square-integrable in the pathwise sense: for every finite \(T\), \(\int_0^T H_s^2\,ds<\infty\) almost surely. All time integrals defining energy below are Lebesgue integrals. For an Ito integral on \([0,T]\), the required class is the progressively measurable processes \(G\) satisfying \(\mathbb{E}\int_0^T G_s^2\,ds<\infty\); this is the class \(\mathcal{L}^2\).
Outside a single null set, \(A_t\) is finite on every bounded interval, continuous, nondecreasing, and starts at \(A_0=0\). Continuity follows because the indefinite integral of a locally integrable function is continuous. The choice of \(\{s\le\tau_n\}\), rather than \(\{s<\tau_n\}\), does not change any time integral: the two indicators differ at most at the single time \(s=\tau_n\).
The Hitting Times Are Stopping Times
Proof. Fix \(t\ge0\). Since \(A\) is nondecreasing and continuous, it reaches or exceeds \(n\) at some time in \([0,t]\) if and only if \(A_t\ge n\). Therefore \(\{\tau_n\le t\}=\{A_t\ge n\}\). The energy \(A_t\) is \(\mathcal{F}_t\)-measurable: it is the time integral up to \(t\) of the adapted measurable process \(H_s^2\). Thus \(\{\tau_n\le t\}\in\mathcal{F}_t\) for every \(t\), which is the definition of a stopping time. If the threshold is never reached, the event on the left is empty for each finite \(t\), consistent with \(\tau_n=\infty\). \(\square\)
The stopped indicator is also suitable for integration. The process \(s\mapsto\mathbf{1}_{\{s\le\tau_n\}}\) is adapted and left-continuous, hence predictable. Its product with the predictable process \(H\) is predictable, and in particular progressively measurable. The important issue is whether this product has finite expected square-integral; the next result gives a pathwise bound first.
The Energy Bound and Localization
Proof. Because \(A\) is continuous and nondecreasing, it cannot exceed \(n\) before \(\tau_n\). If \(\tau_n\le T\), continuity gives \(A_{\tau_n}=n\); if \(\tau_n>T\), then \(A_T<n\). (The same bound holds if the threshold is never reached.) Thus \(A_{T\wedge\tau_n}\le n\). By the definition of \(H^{(n)}\) and the fact that changing an integrand at one time does not change its Lebesgue integral, \(\int_0^T(H^{(n)}_s)^2\,ds=\int_0^{T\wedge\tau_n}H_s^2\,ds=A_{T\wedge\tau_n}\). Taking expectations gives the claimed expected-energy bound. Progressive measurability and this finite expectation are exactly the conditions for \(H^{(n)}\in\mathcal{L}^2\) on \([0,T]\). If \(n\) increases, reaching the higher energy threshold cannot occur before reaching the lower one, so the stopping times are nondecreasing. Finally, on the probability-one event where \(A_k<\infty\) for every positive integer \(k\), fix any finite \(T\). For every integer \(n>A_T\), the threshold has not been reached by time \(T\), so \(\tau_n>T\). This holds for every finite \(T\), and proves \(\tau_n\to\infty\) almost surely. The case \(H=0\) is included: then \(A_t=0\), every \(\tau_n=\infty\), and the stopped energy is zero. \(\square\)
This theorem uses only the continuity and monotonicity of accumulated energy. It does not assert that the original \(H\) has finite expected energy. Instead, each stopped version has a deterministic upper bound \(n\), while the stopping times move arbitrarily far out as the threshold increases. Thus, on any fixed bounded interval, sufficiently large thresholds leave the integrand unstopped almost surely.
Worked Example: A Deterministic Integrand with a Known Hitting Time
Set \(H_s=2\) for \(0\le s<1\) and \(H_s=1\) for \(s\ge1\). Find the energy hitting time for threshold \(n=3\), then compute the stopped energy through \(T=2\).
For \(0\le t\le1\), \(A_t=\int_0^t4\,ds=4t\). For \(t\ge1\), the energy accumulated before time \(1\) is \(4\), and the subsequent rate is \(1\), so \(A_t=4+(t-1)=t+3\). The first time \(A_t\ge3\) is \(4t=3\), giving \(\tau_3=\tfrac34\). At \(T=2\), the stopped energy is \(\int_0^2H_s^2\mathbf{1}_{\{s\le3/4\}}\,ds=\int_0^{3/4}4\,ds=3\). Since this integrand is deterministic, its expected stopped energy is also \(3\).
Check: At \(t=\tfrac34\), \(A_t=4\cdot\tfrac34=3\); for every \(t<\tfrac34\), \(A_t=4t<3\). The hitting time and the stopped energy therefore agree with the threshold exactly.
Worked Example: A Random Initial Amplitude
Let \(\xi\) be \(\mathcal{F}_0\)-measurable, independent of the Brownian increments, with \(\mathbb{P}(\xi=1)=\mathbb{P}(\xi=2)=\tfrac12\), and let \(H_s=\xi\). Use threshold \(n=2\) and compute the expected stopped energy through \(T=1\).
For either value of \(\xi\), \(A_t=\xi^2t\), so \(\tau_2=2/\xi^2\). If \(\xi=1\), then \(\tau_2=2\) and the energy by time \(1\) is \(1^2\cdot1=1\). If \(\xi=2\), then \(\tau_2=\tfrac12\) and the stopped energy by time \(1\) is \(2^2\cdot\tfrac12=2\). Therefore \(\mathbb{E}\int_0^1(H^{(2)}_s)^2\,ds=\tfrac12(1)+\tfrac12(2)=\tfrac32\).
Check: In both outcomes the stopped energy is at most the threshold \(2\), as the theorem requires. Without stopping, the expected energy through time \(1\) is \(\mathbb{E}[\xi^2]=\tfrac12(1)+\tfrac12(4)=\tfrac52\), so truncation has reduced the expected energy.
Worked Example: Finite Pathwise Energy but Infinite Expected Energy
Let \(\xi\) have density \(f(x)=x^{-2}\) for \(x\ge1\), and zero otherwise. Take \(\xi\) to be \(\mathcal{F}_0\)-measurable and independent of the Brownian motion, and set \(H_s=\xi\). Use threshold \(n=1\) and \(T=1\).
The density integrates to \(\int_1^\infty x^{-2}\,dx=1\), and \(\xi\) is finite almost surely. Thus \(A_t=\xi^2t\) is finite for every finite \(t\), almost surely. The original expected energy through time \(1\), however, is \(\mathbb{E}[\xi^2]=\int_1^\infty x^2x^{-2}\,dx=\int_1^\infty1\,dx=\infty\). For every outcome, the threshold is reached at \(\tau_1=1/\xi^2\le1\). Hence the stopped energy through time \(1\) is \(\int_0^1\xi^2\mathbf{1}_{\{s\le1/\xi^2\}}\,ds=\xi^2(1/\xi^2)=1\), and its expectation is exactly \(1\).
Check: The original expected energy is infinite, but each sample path has finite energy on \([0,1]\); the stopped energy is exactly the threshold for every outcome. This illustrates why pathwise local square-integrability and membership in \(\mathcal{L}^2\) are different conditions.
A Common Miscalculation
A common mistake is to infer \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) from the fact that \(\int_0^T H_s^2\,ds<\infty\) almost surely. The heavy-tailed example disproves that inference: for \(T=1\), the pathwise energy is \(\xi^2<\infty\) almost surely, while its expectation is \(\int_1^\infty1\,dx=\infty\). The correction is not to take expectations of a merely finite random variable and assume they remain finite. Instead, stop at an energy threshold and use the pathwise inequality \(\int_0^T(H^{(n)}_s)^2\,ds\le n\), which directly gives the expectation bound.
Energy stopping is useful because it converts a pathwise condition into the integrability condition needed for Ito integration on bounded intervals. It also gives a sequence of stopping times tending to infinity, which is the basic form of localization. The bound is uniform over the time horizon \(T\), but it depends on the threshold \(n\); it does not claim a uniform bound as \(n\) itself tends to infinity.
Check Your Understanding
For each question, distinguish the pathwise energy statement from its expectation.
- For \(H_s=3\), find \(A_t\) and the hitting time \(\tau_5\).
- Why does continuity of \(A_t\) ensure that a finite hitting time satisfies \(A_{\tau_n}=n\)?
- Explain why \(\int_0^T(H^{(n)}_s)^2\,ds\le n\) implies \(H^{(n)}\in\mathcal{L}^2\) on \([0,T]\).
- If \(A_t=0\) for every \(t\), what are \(\tau_n\) and the stopped energy?
- In the random-amplitude example, what is the unstopped expected energy through time \(1\), and why is it larger than the stopped expected energy?