Quadratic Variation of a Localized Ito Integral

The Energy Behind the Quadratic Variation

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. On a fixed interval $[0,T]$, an Ito integrand in $\mathcal{L}^2$ is a progressively measurable process $H_t$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Its Ito integral is $M_t=\int_0^t H_s\,dB_s$. A locally square-integrable integrand need only satisfy $\int_0^T H_s^2\,ds<\infty$ almost surely for every finite $T$; its integral is defined by stopping when accumulated energy reaches finite levels.

For a partition $\Pi=\{0=t_0<\cdots<t_m=t\}$, the quadratic variation sum of $M$ is $\sum_{i=1}^m(M_{t_i}-M_{t_{i-1}})^2$. Quadratic variation $[M]_t$ is the limit in probability of these sums as the mesh tends to zero. The formula below identifies that limit. Its right side, $\int_0^t H_s^2\,ds$, is the accumulated energy of the integrand.

Theorem: If $H$ is progressively measurable and locally square-integrable, then the continuous local martingale $M_t=\int_0^t H_s\,dB_s$ has quadratic variation $$ [M]_t=\int_0^t H_s^2\,ds. $$ More precisely, for each fixed $T<\infty$ and $t\in[0,T]$, the quadratic variation sums along any deterministic partitions of $[0,t]$ whose mesh tends to zero converge in probability to $\int_0^tH_s^2\,ds$. Equality of the continuous processes holds indistinguishably.

The proof has two stages. First establish the limit for an $\mathcal{L}^2$ integrand, using bounded-integrand estimates and approximation. Then stop a locally square-integrable integrand at finite energy levels and remove the stopping. In the proof, $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$ are the Ito multiplication rules; they encode quadratic variation, not the ordinary chain rule.

The Square-Integrable Case

Begin with a progressively measurable $H\in\mathcal{L}^2$ on $[0,T]$ that also satisfies $|H_s|\le C$ almost surely for a fixed finite $C$. For a partition of $[0,t]$, set $\Delta_i M=M_{t_i}-M_{t_{i-1}}$ and $A_i=\int_{t_{i-1}}^{t_i}H_s^2\,ds$. The Ito isometry gives $\mathbb{E}[(\Delta_i M)^2]=\mathbb{E}[A_i]$, and its conditional version gives $\mathbb{E}[(\Delta_i M)^2-A_i\mid\mathcal{F}_{t_{i-1}}]=0$. Thus the errors for successive intervals have conditional mean zero.

We need a fourth-moment bound for each increment. On an interval $[a,b]$, let $N_u=\int_a^uH_s\,dB_s$, where $a\le u\le b$. Ito's formula for $f(x)=x^4$ has $f'(x)=4x^3$, $f''(x)=12x^2$, and correction $\tfrac12 f''(x)H_s^2\,ds=6N_s^2H_s^2\,ds$. Stop $N$ when $|N|$ reaches $k$. The stopped stochastic integral in this formula is square-integrable and has expectation zero. The Ito isometry gives $\mathbb{E}[N_{s\wedge\rho_k}^2]\le C^2(s-a)$. Consequently, after taking expectations in the stopped formula and then using Fatou's lemma as $k\to\infty$,

\mathbb{E}[N_b^4]\le 6C^2\int_a^b C^2(s-a)\,ds =3C^4(b-a)^2.

Also $0\le A_i\le C^2(t_i-t_{i-1})$. Put $D_i=(\Delta_iM)^2-A_i$. The conditional-mean identity makes the $D_i$ orthogonal in $L^2$: for $i<j$, $D_i$ is $\mathcal{F}_{t_{j-1}}$-measurable, so $\mathbb{E}[D_iD_j]=\mathbb{E}[D_i\mathbb{E}(D_j\mid\mathcal{F}_{t_{j-1}})]=0$. Moreover, the fourth-moment bound and $(x-y)^2\le2x^2+2y^2$ give $\mathbb{E}[D_i^2]\le8C^4(t_i-t_{i-1})^2$. Therefore

\mathbb{E}\left[\left(\sum_i(\Delta_iM)^2-\int_0^tH_s^2\,ds\right)^2\right] =\sum_i\mathbb{E}[D_i^2] \le 8C^4t|\Pi|.

This tends to zero with the mesh. In particular the quadratic variation sums converge in $L^2$ for bounded $H$.

For a general $H\in\mathcal{L}^2$, define $H^{(n)}=H\mathbf{1}_{\{|H|\le n\}}$, with integrals $M^{(n)}$. Then $\mathbb{E}\int_0^T(H-H^{(n)})^2\,ds\to0$ by dominated convergence. For any partition, write $Q(M)=\sum_i(\Delta_iM)^2$. The pathwise Cauchy–Schwarz inequality and the Ito isometry imply

\mathbb{E}|Q(M)-Q(M^{(n)})| \le \mathbb{E}Q(M-M^{(n)}) +2\sqrt{\mathbb{E}Q(M-M^{(n)})\,\mathbb{E}Q(M^{(n)})}.

Here $\mathbb{E}Q(M-M^{(n)})=\mathbb{E}\int_0^T(H-H^{(n)})^2\,ds$, while $\mathbb{E}Q(M^{(n)})=\mathbb{E}\int_0^T(H^{(n)})^2\,ds\le\mathbb{E}\int_0^TH^2\,ds$. These bounds do not depend on the partition, and the displayed expression tends to zero as $n\to\infty$. The target energies also converge in $L^1$, since $\mathbb{E}\int_0^T|H^2-(H^{(n)})^2|\,ds\to0$. Approximate first by $H^{(n)}$, apply the bounded case, and then let $n\to\infty$. This proves convergence in probability for every $H\in\mathcal{L}^2$.

Localization Extends the Formula

Now suppose only that $H$ is progressively measurable and has finite energy almost surely on every finite time interval. Define $A_t=\int_0^tH_s^2\,ds$ and the stopping time $\tau_n=\inf\{t\ge0:A_t\ge n\}$, for positive integers $n$. The stopped integrand $H_s\mathbf{1}_{\{s\le\tau_n\}}$ has energy at most $n$ on every finite interval, so it belongs to $\mathcal{L}^2$ there.

Proof. Fix $T<\infty$ and $t\in[0,T]$. Apply the square-integrable result to the stopped integrand. On the event $\{A_T<n\}$, the stopping time exceeds $T$, so its integral and energy agree with $M$ and $A$ throughout $[0,T]$. For any $\varepsilon>0$, the probability that the original quadratic variation sum at $t$ differs from $A_t$ by more than $\varepsilon$ is therefore at most $\mathbb{P}(A_T\ge n)$ plus the corresponding error probability for the stopped integral. For fixed $n$, the latter tends to zero as the partition mesh tends to zero. Then $\mathbb{P}(A_T\ge n)\to0$ as $n\to\infty$, because $A_T<\infty$ almost surely. This proves the asserted convergence in probability. At each fixed time the limit is unique up to almost-sure equality; applying this at rational times and using the continuous versions of quadratic variation and $A$ gives indistinguishable equality of the processes. $\square$

The argument is about quadratic variation, not an expectation identity for an unstopped local martingale. For $H\in\mathcal{L}^2$, the Ito isometry gives $\mathbb{E}[M_t^2]=\mathbb{E}\int_0^tH_s^2\,ds$; it also gives square integrability, and the conditional mean-zero property of future Ito increments verifies that $M$ is a martingale. For a merely locally square-integrable $H$, these expectations may be infinite, so the integral is guaranteed to be a local martingale, not necessarily a true martingale.

Worked Applications

Worked Example: A Deterministic Step Integrand

Let $H_s=3$ for $0\le s\le2$, and $H_s=-2$ for $2<s\le5$. Find the quadratic variation at every time through $T=5$.

The energy is the integral of the squared coefficient on each interval. For $0\le t\le2$, $A_t=3^2t=9t$. For $2<t\le5$, the first interval contributes $3^2\cdot2=18$, and the second contributes $(-2)^2(t-2)=4(t-2)$. Thus $A_t=18+4(t-2)=4t+10$ on the second interval. The theorem gives $[M]_t=A_t$.

Check: at $t=2$, both expressions give $9\cdot2=18$. At $t=5$, the accumulated energy is $9\cdot2+4\cdot3=18+12=30$; the second formula gives $4\cdot5+10=30$ as well.

Worked Example: A Time-Varying Deterministic Integrand

Take $H_s=s$ on $[0,2]$. The quadratic variation is

[M]_t=\int_0^t s^2\,ds=\frac{t^3}{3},\qquad 0\le t\le2.

For an independent check of the energy at $t=2$, $[M]_2=2^3/3=8/3$. The Ito isometry applies because $\mathbb{E}\int_0^2s^2\,ds=8/3<\infty$, and gives $\mathbb{E}[M_2^2]=8/3$, matching the computed energy. Integration by parts also gives $M_t=tB_t-\int_0^tB_s\,ds$, since $t$ has finite variation and hence zero quadratic covariation with $B$.

Worked Example: An Integrand Chosen Using the Past

On $[0,4]$, let $H_s=\mathbf{1}_{\{B_1>0\}}\mathbf{1}_{(1,3]}(s)$. This process is progressively measurable: after time $1$, the random factor is already known. The event $\{B_1>0\}$ has probability $1/2$, since $B_1$ has a continuous centered normal distribution. Its energy is

[M]_t=\mathbf{1}_{\{B_1>0\}}\int_0^t\mathbf{1}_{(1,3]}(s)\,ds = \begin{cases} 0,&0\le t\le1,\\ (t-1)\mathbf{1}_{\{B_1>0\}},&1<t\le3,\\ 2\mathbf{1}_{\{B_1>0\}},&3<t\le4. \end{cases}

The expected terminal energy is $2\cdot\tfrac12=1$, so the integrand is in $\mathcal{L}^2$. Check: at $t=4$, the pathwise energy is $2$ on $\{B_1>0\}$ and $0$ otherwise; averaging these two values with probabilities $1/2$ and $1/2$ gives $1$, as calculated.

Interpretation and a Common Mistake

The formula says that the diffusion amplitude $H_s$ contributes its square to quadratic variation at time $s$. It is the stochastic-integral counterpart of $dB_t\,dB_t=dt$: formally, $dM_t=H_t\,dB_t$ yields $dM_t\,dM_t=H_t^2\,dt$. The assertion is rigorous as a limit of squared increments. It does not say that each individual squared increment equals the corresponding energy increment.

Common mistake: for a local integrand, one might write $\mathbb{E}[M_t^2]=\mathbb{E}\int_0^tH_s^2\,ds$ without checking finiteness, then use the Ito isometry globally. That equality is justified for $\mathcal{L}^2$ integrands, not automatically for locally square-integrable ones. The localization proof instead applies the isometry to each stopped integral, whose energy is bounded, and then removes the stopping in probability. Quadratic variation remains $\int_0^tH_s^2\,ds$ even when that random variable has infinite expectation.

Check Your Understanding

Use the theorem and its hypotheses to answer the following.

If $H_s=-4$ on $[0,1]$, what is $[M]_t$ for $0\le t\le1$, and why is the sign irrelevant?
For $H_s=s$ on $[0,3]$, compute the accumulated energy at $t=3$ and state whether $H\in\mathcal{L}^2$.
Why does finite energy almost surely not by itself imply that the integral is a true martingale?
In the random-indicator example, which information makes the integrand adapted after time $1$?
What convergence of partition sums is used to define quadratic variation in the theorem?