Finite Energy Does Not Require Bounded Integrands
On a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration, the Ito integral is first defined for progressively measurable integrands \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). This is the class \(\mathcal{L}^2\) on \([0,T]\). But an adapted integrand can have finite accumulated squared magnitude on almost every path even when it is unbounded and its expected accumulated squared magnitude is infinite. Stopping at successive energy levels lets us define its integral on each finite time interval.
When finite pathwise energy holds, \(A_t\) has continuous, nondecreasing paths almost surely. For each positive integer \(n\), stop when \(A_t\) first reaches \(n\). The stopped integrand then has expected energy at most \(n\), so it belongs to \(\mathcal{L}^2\) on every finite horizon. The resulting stopped Ito integrals fit together to define the integral of \(H\).
Energy Stopping Times and the Localization Result
The inclusion of the endpoint \(s=\tau_n\) does not change a time integral, because a single time has Lebesgue measure zero. If \(A_t\) never reaches \(n\), then \(\tau_n=\infty\). If \(A_t\) is identically zero, every stopped integral below is zero; this degenerate case is included.
Proof. Continuity and monotonicity of \(A_t\) imply that, up to \(\tau_n\), the accumulated energy is at most \(n\). Thus, for every finite \(T\), \(\mathbb{E}\int_0^T H_s^2\mathbf{1}_{\{s\le\tau_n\}}\,ds\le n\). The stopped integrand is therefore in \(\mathcal{L}^2\). Define \(M^n_t=\int_0^t H_s\mathbf{1}_{\{s\le\tau_n\}}\,dB_s\). By the Ito isometry, \(\mathbb{E}[(M^n_t)^2]=\mathbb{E}\int_0^{t\wedge\tau_n}H_s^2\,ds\le n\), so \(M^n_t\) is integrable. The Ito integral of an \(\mathcal{L}^2\) integrand is a martingale: this holds for adapted step integrands by the zero conditional means of Brownian increments, and passes to \(\mathcal{L}^2\) limits by the Ito isometry and conditional-expectation contraction in \(L^1\). Hence \(\mathbb{E}[M^n_t\mid\mathcal{F}_u]=M^n_u\) for \(u\le t\). The stopping identity gives \(M^n_t=M_{t\wedge\tau_n}\), so these stops are martingales.
For each finite \(T\), \(A_T<\infty\) almost surely. On any such path, every integer \(n>A_T\) satisfies \(\tau_n>T\); consequently \(\tau_n\to\infty\) almost surely. The stopped integrals therefore define a continuous local martingale \(M\). The quadratic-variation formula for an Ito integral with an \(\mathcal{L}^2\) integrand gives \([M^n]_t=\int_0^{t\wedge\tau_n}H_s^2\,ds=A_{t\wedge\tau_n}\). This is the standard quadratic-variation result for Ito integrals. For each fixed finite \(t\), eventually \(\tau_n>t\), so the stopped process and its quadratic variation agree with the unstopped ones through time \(t\). Hence \([M]_t=A_t\). Applying this on all finite intervals gives the asserted identity. \(\square\)
This result concerns quadratic variation, not variance: it identifies a pathwise increasing process, while \(\mathbb{E}[A_t]\) may be infinite. In differential notation, the quadratic-variation identity is written \(dM_t\,dM_t=H_t^2\,dt\). The rules \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\) are shorthand for the corresponding limits of sums; they do not turn finite pathwise energy into finite expected energy.
Worked Examples: Unbounded Integrands and Their Energy
Worked Example: The Integrand \(H_t=B_t^2\)
On the filtered space carrying \(B_t\), take \(H_s=B_s^2\), a progressively measurable integrand. Brownian paths are continuous, so on every finite interval \(H_s\) is bounded along each path and \(A_T=\int_0^T B_s^4\,ds<\infty\) almost surely. Its expected energy is also finite. Since \(B_s\) is normal with mean zero and variance \(s\), \(\mathbb{E}[B_s^4]=3s^2\). Therefore, at \(T=1\),
The theorem gives \(M_t=\int_0^t B_s^2\,dB_s\) and \([M]_t=\int_0^t B_s^4\,ds\). In this example the expected energy is finite, so \(M\) is a square-integrable martingale and the Ito isometry gives \(\mathbb{E}[M_1^2]=1\).
For an independent check of the integral, apply Ito’s formula to \(f(x)=x^3\). Its derivatives are \(f'(x)=3x^2\) and \(f''(x)=6x\); the correction term is \(\tfrac12 f''(B_s)\,ds=3B_s\,ds\). Thus
Check: both terms in the expression for \(M_1\) have mean zero, since \(B_1\) and each \(B_s\) have symmetric distributions. Its second moment is \(1\) by the isometry calculation above, in agreement with the direct energy computation.
Worked Example: An Exponential Brownian Integrand
Set \(T=\ln 2\) and \(H_s=e^{B_s}\), which is progressively measurable. Continuity of \(B_s\) makes \(H_s\) bounded along each path on \([0,T]\), so its energy is finite almost surely. Using the normal moment-generating function, \(\mathbb{E}[e^{2B_s}]=e^{2s}\), and hence
Thus \(H\) is in \(\mathcal{L}^2\) on this interval. Its Ito integral \(M_t=\int_0^t e^{B_s}\,dB_s\) is a square-integrable martingale, and \([M]_t=\int_0^t e^{2B_s}\,ds\). To check the integral itself, use \(f(x)=e^x\), with \(f'(x)=e^x\) and \(f''(x)=e^x\). Ito’s correction is \(\tfrac12 e^{B_s}\,ds\), giving
Check: \(\mathbb{E}[e^{B_T}]=e^{T/2}=\sqrt{2}\), and \(\mathbb{E}\int_0^T e^{B_s}\,ds=\int_0^T e^{s/2}\,ds=2(\sqrt{2}-1)\). Substitution gives \(\mathbb{E}[M_T]=\sqrt{2}-1-(\sqrt{2}-1)=0\), as required for this integrable martingale starting at zero.
Worked Example: Finite Energy Without a True Martingale
Let \(\xi\) be an \(\mathcal{F}_0\)-measurable random variable, independent of the Brownian motion, with \(\mathbb{P}(\xi>x)=x^{-1/2}\) for \(x\ge1\). It is finite almost surely, but \(\mathbb{E}[\xi]=\infty\). Take \(H_s=\xi\). This integrand is progressively measurable and has energy \(A_t=t\xi^2<\infty\) almost surely for every finite \(t\). The energy stopping times are explicit: \(\tau_n=n/\xi^2\). The stopped integral is \(M^n_t=\xi B_{t\wedge\tau_n}\), and its expected squared value satisfies
Each stopped integral is therefore a square-integrable martingale, with quadratic variation \(A_{t\wedge\tau_n}\); the unstopped local integral has \([M]_t=t\xi^2\). But for every \(t>0\), independence gives \(\mathbb{E}|M_t|=\mathbb{E}[\xi]\mathbb{E}|B_t|=\infty\). Thus \(M_t=\xi B_t\) is not integrable and cannot be a true martingale. The localization theorem still applies.
Check: the stopped energy is \(\min(t\xi^2,n)\), so it is bounded above by \(n\) for every outcome. This verifies the integrability needed for each stopped Ito integral without falsely asserting that the unstopped integral is integrable.
Why Expected Energy and Pathwise Energy Must Not Be Confused
A common mistake is to require \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) before concluding that any Ito integral can be defined. That condition gives a square-integrable martingale directly, but it is stronger than the finite-pathwise-energy condition used for localization. In the heavy-tailed example, \(A_T=T\xi^2\) is finite almost surely, while its expectation is infinite. Stopping still produces integrands in \(\mathcal{L}^2\) and defines a continuous local martingale.
The converse mistake is to infer that the unstopped integral is a true martingale merely because each stopped integral is one. A local martingale need not be integrable at each time; the example \(M_t=\xi B_t\) fails precisely this requirement. A useful sufficient condition is finite expected energy: if \(H\in\mathcal{L}^2\) on \([0,T]\), then its Ito integral is a square-integrable martingale and \(\mathbb{E}[M_t^2]=\mathbb{E}[A_t]\). Finite energy almost surely alone does not supply that conclusion.
In applications, the energy stopping times provide two pieces of information at once: they control the integrand enough to construct the integral, and their increasing limits recover the full process on every finite interval. The quadratic variation records exactly how much energy has accumulated, even when no finite expectation is available.
Check Your Understanding
Use the energy stopping times and distinguish pathwise statements from expectation statements.
- For \(H_s=B_s\) on \([0,T]\), compute \(\mathbb{E}[A_T]\) and state the quadratic variation of its Ito integral.
- Explain why \(\tau_n=\inf\{t:A_t\ge n\}\) is a stopping time when \(A_t\) is continuous and adapted.
- For \(H_s=\xi\) in the heavy-tailed example, show that \(\mathbb{E}[A_T]=\infty\) even though \(A_T<\infty\) almost surely.
- State the integrability and conditional-expectation requirements that distinguish a martingale from a local martingale.
- For an integrand with \(A_t=0\) for every \(t\), identify its localized Ito integral and its quadratic variation.