Worked Problems in Localization and Quadratic Variation

Localization as a Calculation Tool

For an Ito integral, the accumulated squared size of its integrand determines its quadratic variation. If that accumulated energy is unbounded, stopping when it first reaches a fixed level makes the stopped integral square-integrable. This gives a practical way to construct a localizing sequence and to calculate quadratic variations without assuming that the unstopped integral is a true martingale.

Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). An Ito integrand \(H\) is progressively measurable; on a finite interval \([0,T]\), it belongs to \(\mathcal{L}^2\) when \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). A progressively measurable \(H\) is locally square-integrable if \(\int_0^T H_s^2\,ds<\infty\) almost surely for every finite \(T\). For such an \(H\), its Ito integral is defined by localization, even if it does not belong to \(\mathcal{L}^2\) on the whole interval.

Definition: A sequence of stopping times \((\tau_n)\) is a localizing sequence on \([0,\infty)\) if \(\tau_n\) is nondecreasing and \(\tau_n\to\infty\) almost surely. On a finite time interval \([0,T]\), the corresponding requirement is that the stopping times increase to \(T\) almost surely. A continuous local martingale is a continuous adapted process that becomes a martingale when stopped at each time in a localizing sequence.

For a stopping time \(\tau\), the stopped integral and the integral with a stopped integrand agree: \(\left(\int_0^\cdot H_s\,dB_s\right)_{t\wedge\tau}=\int_0^t\mathbf{1}_{\{s\le\tau\}}H_s\,dB_s\). Here the indicator integrand is progressively measurable, and including or excluding the single endpoint \(s=\tau\) does not change the time integral. We use this identity together with the Ito isometry and the quadratic-variation formulas for Ito integrals established in the earlier material.

The Energy-Based Localizer

Theorem: Let \(H\) be progressively measurable and locally square-integrable, and set \(A_t=\int_0^t H_s^2\,ds\). Define \(\tau_n=\inf\{t\ge0:A_t\ge n\}\), with \(\inf\varnothing=\infty\). Then \((\tau_n)\) is a localizing sequence, and for each \(n\), the stopped integral \(M_{t\wedge\tau_n}\), where \(M_t=\int_0^t H_s\,dB_s\), is a square-integrable martingale. Its quadratic variation is \([M_{\cdot\wedge\tau_n}]_t=A_{t\wedge\tau_n}\). In particular, \([M]_t=A_t\).

Proof. The process \(A\) is adapted, continuous, nondecreasing, and finite on each compact time interval almost surely. Its first hitting time of the closed set \([n,\infty)\) is a stopping time. Since \(A\) is continuous, \(A_{t\wedge\tau_n}\le n\); if \(\tau_n<\infty\), then \(A_{\tau_n}=n\). The times are nondecreasing in \(n\). For any fixed finite \(T\), \(A_T<\infty\) almost surely, so for every integer \(n>A_T\) we have \(\tau_n>T\). Thus \(\tau_n\to\infty\) almost surely. By the stopping identity, \(M_{t\wedge\tau_n}\) is the Ito integral of \(\mathbf{1}_{\{s\le\tau_n\}}H_s\), whose expected squared energy on \([0,T]\) is at most \(n\). The Ito isometry gives \(\mathbb{E}[M_{T\wedge\tau_n}^2]\le n\), and the martingale property of an \(\mathcal{L}^2\) Ito integral gives \(\mathbb{E}[M_{t\wedge\tau_n}\mid\mathcal{F}_s]=M_{s\wedge\tau_n}\) for \(s\le t\). Hence the stopped process is a square-integrable martingale. The quadratic-variation formula for an Ito integral gives \([M_{\cdot\wedge\tau_n}]_t=\int_0^t\mathbf{1}_{\{s\le\tau_n\}}H_s^2\,ds=A_{t\wedge\tau_n}\). Finally, on every finite interval \([0,T]\), eventually \(\tau_n>T\), so the stopped processes agree with \(M\) on that interval; consequently \([M]_t=A_t\). \(\square\)

The conclusion about the stopped processes is stronger than a claim that \(M\) itself is a martingale. If \(\mathbb{E}\int_0^T H_s^2\,ds=\infty\), the Ito isometry does not put \(M_T\) in \(L^2\), and localization alone does not imply the integrability or conditional-expectation property required of a true martingale.

Worked Problems

Worked Example: A Singular Integrand Near a Finite Endpoint

On the interval \(0\le t<1\), take the deterministic integrand \(H_s=(1-s)^{-1}\). Find its energy, construct a localizing sequence approaching the endpoint, and compare the stopped integral and its quadratic variation.

The integrand is progressively measurable. For each \(t<1\), its energy is finite, and direct integration gives

A_t=\int_0^t\frac{1}{(1-s)^2}\,ds =\left[\frac{1}{1-s}\right]_{s=0}^{s=t} =\frac{1}{1-t}-1 =\frac{t}{1-t}.

In particular, at \(t=\tfrac12\), the energy is \(\tfrac{1/2}{1-1/2}=1\). For each positive integer \(n\), solve \(A_t=n\): \(t/(1-t)=n\) implies \(t=n(1-t)\), hence \(t(1+n)=n\), so \(\tau_n=n/(n+1)\). These times increase to \(1\), the endpoint of the interval. The stopped integrand has energy at most \(n\), since \(A_{t\wedge\tau_n}\le n\).

Let \(M_t=\int_0^t(1-s)^{-1}\,dB_s\), defined for \(t<1\). The stopping identity gives \(M_{t\wedge\tau_n}=\int_0^t\mathbf{1}_{\{s\le\tau_n\}}(1-s)^{-1}\,dB_s\). For each \(T<1\), this stopped integral is square-integrable on \([0,T]\): its expected squared energy is the deterministic quantity \(A_{T\wedge\tau_n}\le n\). Its quadratic variation is \([M_{\cdot\wedge\tau_n}]_t=A_{t\wedge\tau_n}\). For example, at \(t=\tfrac12\) and \(n=1\), the stopped quadratic variation is \(A_{1/2}=1\), and for \(t\ge\tfrac12\) it remains \(1\).

Check: Substitution of \(\tau_n=n/(n+1)\) into the energy gives \(A_{\tau_n}=\big(n/(n+1)\big)\big/\big(1/(n+1)\big)=n\), as required. Also, the integral diverges at the endpoint because \(A_t=t/(1-t)\to\infty\) as \(t\uparrow1\); the sequence therefore approaches \(1\), not a time beyond the domain.

Worked Example: Localizing a Brownian Integrand

Take \(H_s=B_s\) on \([0,\infty)\). Construct an energy-based localizer and find the quadratic variation of the stopped integral and its covariation with \(B\).

The paths of \(B\) are continuous, so \(\int_0^T B_s^2\,ds<\infty\) almost surely for every finite \(T\). The integrand is progressively measurable, and its accumulated energy is \(A_t=\int_0^t B_s^2\,ds\). Set \(\tau_n=\inf\{t\ge0:A_t\ge n\}\). This is a stopping time because \(A\) is continuous and adapted. For every finite \(T\), the random variable \(A_T\) is finite almost surely; hence, once \(n>A_T\), \(\tau_n>T\). Thus \(\tau_n\to\infty\) almost surely.

Write \(M_t=\int_0^t B_s\,dB_s\), defined by localization. Then

M_{t\wedge\tau_n} =\int_0^t\mathbf{1}_{\{s\le\tau_n\}}B_s\,dB_s, \qquad \mathbb{E}\int_0^T\mathbf{1}_{\{s\le\tau_n\}}B_s^2\,ds =\mathbb{E}[A_{T\wedge\tau_n}] \le n.

The Ito isometry gives \(\mathbb{E}[M_{T\wedge\tau_n}^2]\le n\), while the conditional-expectation property of an \(\mathcal{L}^2\) Ito integral gives \(\mathbb{E}[M_{t\wedge\tau_n}\mid\mathcal{F}_s]=M_{s\wedge\tau_n}\) for \(s\le t\). Therefore each stopped process is a true square-integrable martingale. Its quadratic variation is \([M_{\cdot\wedge\tau_n}]_t=\int_0^{t\wedge\tau_n}B_s^2\,ds=A_{t\wedge\tau_n}\).

For the covariation, the covariation formula for Ito integrals, applied to the integrands \(B_s\) and \(1\), gives \([M_{\cdot\wedge\tau_n},B]_t=\int_0^{t\wedge\tau_n}B_s\,ds\). The integrand \(1\) defines \(B_t=\int_0^t1\,dB_s\); on every finite interval it is in \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^T1\,ds=T\). Removing the stopping gives \([M,B]_t=\int_0^tB_s\,ds\). These are pathwise time integrals of the indicated continuous paths.

Check: At \(t=0\), both the quadratic variation and covariation are \(0\), matching their integral formulas. The quadratic variation is nondecreasing because its integrand \(B_s^2\) is nonnegative; the covariation need not be nondecreasing because \(B_s\) can have either sign.

Worked Example: Two Deterministic Integrands and Their Covariation

On \([0,\infty)\), let \(H_s=(1+s)^{-1/2}\) and \(K_s=(1+s)^{-1}\). Compute the quadratic variations of their Ito integrals and their mutual covariation. Find an energy-based localizer for the first integral.

Both deterministic integrands are progressively measurable and square-integrable on every finite interval. Their energies and cross-integrand integral are

\begin{aligned} A^H_t&=\int_0^t\frac{1}{1+s}\,ds=\log(1+t),\\ A^K_t&=\int_0^t\frac{1}{(1+s)^2}\,ds =\left[-\frac{1}{1+s}\right]_{s=0}^{s=t} =1-\frac{1}{1+t}=\frac{t}{1+t},\\ C_t&=\int_0^t H_sK_s\,ds =\int_0^t(1+s)^{-3/2}\,ds =\left[-2(1+s)^{-1/2}\right]_{s=0}^{s=t} =2\left(1-\frac{1}{\sqrt{1+t}}\right). \end{aligned}

Define \(M_t=\int_0^t H_s\,dB_s\) and \(N_t=\int_0^t K_s\,dB_s\), with both integrals defined by localization on the infinite time interval. The first energy reaches \(n\) at \(\tau_n=e^n-1\), because \(\log(1+\tau_n)=n\). These times increase to infinity. The stopped integral \(M_{t\wedge\tau_n}\) is square-integrable on every finite interval, with expected squared energy \(\int_0^{t\wedge\tau_n}(1+s)^{-1}\,ds\le n\). The process \(N\) is itself square-integrable on each finite interval because \(A^K_t=t/(1+t)\le1\).

The quadratic-variation and covariation formulas now give \([M]_t=\log(1+t)\), \([N]_t=t/(1+t)\), and \([M,N]_t=2(1-(1+t)^{-1/2})\). For the stopped version, replace \(t\) by \(t\wedge\tau_n\) in each formula. At the stopping time, these values are respectively \(n\), \(1-e^{-n}\), and \(2(1-e^{-n/2})\).

Check: At \(t=0\), each displayed expression is \(0\). For \(t>0\), the two quadratic variations are nonnegative, and the covariation is positive because \(1+t>1\). At \(t=\tau_n=e^n-1\), direct substitution gives \(\log(e^n)=n\), \((e^n-1)/e^n=1-e^{-n}\), and \(2(1-1/\sqrt{e^n})=2(1-e^{-n/2})\), confirming all three stopped values.

Why the Covariation Formula Works

Proposition: If \(H\) and \(K\) are progressively measurable and locally square-integrable, and \(M_t=\int_0^tH_s\,dB_s\), \(N_t=\int_0^tK_s\,dB_s\), then \([M]_t=\int_0^tH_s^2\,ds\) and \([M,N]_t=\int_0^tH_sK_s\,ds\), almost surely for all \(t\).

Proof. Choose a common localizing sequence by stopping when \(\int_0^t(H_s^2+K_s^2)\,ds\) first reaches \(n\). The energy argument in the theorem shows that the two integrals, stopped at these times, are square-integrable martingales. The quadratic-variation formula for a localized Ito integral gives the stated formula for each stopped \(M\) and \(N\). For the mutual covariation, polarization gives \([M,N]=\tfrac14([M+N]-[M-N])\). The integrals \(M+N\) and \(M-N\) have integrands \(H+K\) and \(H-K\), respectively. Therefore their quadratic variations have difference \(\int_0^t((H_s+K_s)^2-(H_s-K_s)^2)\,ds=4\int_0^tH_sK_s\,ds\), using the identity \((x+y)^2-(x-y)^2=4xy\). This proves the covariation formula for the stopped processes. On each finite interval the common localizers eventually exceed the interval almost surely, so the stopped identities give the asserted identities for \(M\) and \(N\). \(\square\)

Common Mistakes and Practical Checks

Do not integrate the energy incorrectly. For the first example, the incorrect calculation \(\left[\frac{1}{1-s}\right]_{s=0}^{s=t}-1=\frac{t}{1-t}\) subtracts \(1\) twice: the bracket already equals \(\frac{1}{1-t}-1=\frac{t}{1-t}\). At \(t=\tfrac12\), the integral is \(1\), whereas the incorrect middle expression gives \(0\). Writing out both endpoint evaluations prevents this error.

A stopped martingale is not automatically an unstopped martingale. The energy-based argument verifies integrability and conditional expectation for each stopped integral. Without a separate integrability argument, it only establishes that the original process is a local martingale. In particular, localization does not preserve expectations automatically.

Covariation uses the product of integrands. The quadratic variation of \(\int H\,dB\) uses \(H^2\), while the covariation of \(\int H\,dB\) and \(\int K\,dB\) uses \(HK\). Replacing the latter by \(H^2\) is valid only when the two integrands agree.

Check Your Understanding

For each question, state the integrability or localization condition that justifies the calculation.

For \(H_s=(1-s)^{-1}\) on \(0\le s<1\), compute \(A_t\) and solve \(A_{\tau_n}=n\).
For a locally square-integrable \(H\), explain why stopping when \(\int_0^tH_s^2\,ds\) first reaches \(n\) gives a square-integrable stopped Ito integral.
If \(M_t=\int_0^tH_s\,dB_s\) and \(N_t=\int_0^tK_s\,dB_s\), write their quadratic variations and mutual covariation.
For \(H_s=(1+s)^{-1/2}\) and \(K_s=(1+s)^{-1}\), calculate \([M]_t\), \([N]_t\), and \([M,N]_t\).
Why does proving that every stopped integral is a true martingale not, by itself, prove that the unstopped integral is a true martingale?