From Simple Integrands to the Full L2 Class
For a simple predictable integrand, the Ito integral is a finite sum of adapted coefficients multiplied by Brownian increments. The Ito isometry says that the expected square of this sum equals the expected time integral of the squared integrand. The extension to general square-integrable predictable integrands rests on a precise point: approximation must preserve this equality in the limit on both sides.
Fix \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration. In particular, \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\) for \(0\le s<t\). All expectations below are under \(\mathbb{P}\). Integrands are predictable processes; in particular, they are progressively measurable and adapted.
For a simple process, its Ito integral at time \(T\) is the finite sum \(\int_0^T H_s\,dB_s=\sum_{k=0}^{m-1}\xi_k(B_{t_{k+1}}-B_{t_k})\). The isometry for simple predictable processes, established in the earlier construction, states that \(\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds\). Simple predictable processes are dense in the square-integrable predictable class: for every such \(H\), there is a sequence of simple predictable \(H^n\) with \(\mathbb{E}\int_0^T|H^n_s-H_s|^2\,ds\to0\). These are the two facts that the extension needs.
Why the Two Norms Have the Same Limit
Write \(\|H\|_{L^2(\mathbb{P}\times dt)}=(\mathbb{E}\int_0^T H_s^2\,ds)^{1/2}\) for the integrand norm, and \(\|X\|_{L^2(\mathbb{P})}=(\mathbb{E}[X^2])^{1/2}\) for the random-variable norm. The simple-process isometry says these norms agree after integration. The following elementary observation explains why convergence in either norm also gives convergence of the norms themselves.
Proof. The triangle inequality gives \(\|x_n\|\le\|x_n-x\|+\|x\|\), so \(\|x_n\|-\|x\|\le\|x_n-x\|\). Interchanging \(x_n\) and \(x\) gives \(\|x\|-\|x_n\|\le\|x_n-x\|\). Therefore \(\big|\|x_n\|-\|x\|\big|\le\|x_n-x\|\), whose right side tends to zero. \(\square\)
Proof. Set \(I_n=\int_0^T H^n_s\,dB_s\). For any \(n,m\), \(H^n-H^m\) is simple predictable, so the simple-process isometry gives
Since \(H^n\to H\) in \(L^2(\mathbb{P}\times dt)\), the sequence \(H^n\) is Cauchy in that norm. The displayed equality therefore says that \(I_n\) is Cauchy in \(L^2(\mathbb{P})\). The space \(L^2(\mathbb{P})\) is complete, so there is an \(I\in L^2(\mathbb{P})\) such that \(I_n\to I\) in that space. Define \(\int_0^T H_s\,dB_s=I\).
For each \(n\), the simple-process isometry also gives \(\|I_n\|_{L^2(\mathbb{P})}=\|H^n\|_{L^2(\mathbb{P}\times dt)}\). The lemma applies to both convergences: \(\|I_n\|_{L^2(\mathbb{P})}\to\|I\|_{L^2(\mathbb{P})}\) and \(\|H^n\|_{L^2(\mathbb{P}\times dt)}\to\|H\|_{L^2(\mathbb{P}\times dt)}\). Since the two norms are equal at every \(n\), their limits are equal. Squaring this equality gives the asserted isometry.
To check that the definition does not depend on the chosen approximating sequence, let \(\widetilde H^n\) be another sequence of simple predictable processes converging to \(H\), and set \(\widetilde I_n=\int_0^T\widetilde H^n_s\,dB_s\). The same simple isometry yields \(\mathbb{E}[(I_n-\widetilde I_n)^2]=\mathbb{E}\int_0^T(H^n_s-\widetilde H^n_s)^2\,ds\). The right side tends to zero because \(\|H^n-\widetilde H^n\|_{L^2(\mathbb{P}\times dt)}\le\|H^n-H\|_{L^2(\mathbb{P}\times dt)}+\|\widetilde H^n-H\|_{L^2(\mathbb{P}\times dt)}\to0\). Thus both sequences of integrals have the same \(L^2(\mathbb{P})\) limit. The construction is consequently well-defined on equivalence classes of integrands, rather than depending on a particular representative or approximation. \(\square\)
Worked Applications of the Extension
Worked Example: A Deterministic Integrand \(H_t=t\)
On \([0,1]\), take \(H_t=t\). This deterministic process is predictable, and \(\mathbb{E}\int_0^1H_t^2\,dt=\int_0^1t^2\,dt=\tfrac13<\infty\). Approximate it using \(H^n_t=k/n\) on each interval \((k/n,(k+1)/n]\), for \(k=0,\ldots,n-1\). Each \(H^n\) is simple predictable.
On each interval, write \(t=k/n+u\), where \(0<u\le1/n\). The squared approximation error there is \(u^2\), so
The norm squared of the simple integrand is \(\mathbb{E}\int_0^1(H^n_t)^2\,dt=\frac{1}{n^3}\sum_{k=0}^{n-1}k^2=\frac{(n-1)(2n-1)}{6n^2}\), which tends to \(\tfrac13\). The theorem therefore gives \(\mathbb{E}[(\int_0^1t\,dB_t)^2]=\tfrac13\). Check: for \(n=2\), the approximant is zero on \((0,\tfrac12]\) and \(\tfrac12\) on \((\tfrac12,1]\). Its integrand norm squared is \(\tfrac12(\tfrac12)^2=\tfrac18\), matching \(\frac{(2-1)(4-1)}{6\cdot 2^2}=\tfrac18\).
Worked Example: The Brownian Integrand \(H_t=B_t\)
On \([0,1]\), \(H_t=B_t\) is adapted and continuous, hence predictable, and \(\mathbb{E}\int_0^1B_t^2\,dt=\int_0^1t\,dt=\tfrac12<\infty\). Let \(H^n_t=B_{k/n}\) on \((k/n,(k+1)/n]\). The coefficient \(B_{k/n}\) is \(\mathcal{F}_{k/n}\)-measurable, so this is simple predictable.
For \(t\) in this interval, Brownian increments have mean zero and variance \(t-k/n\). Thus \(\mathbb{E}|B_t-B_{k/n}|^2=t-k/n\), and integrating over all \(n\) intervals gives
The simple integrand norm squared is \(\mathbb{E}\int_0^1(H^n_t)^2\,dt=\frac1n\sum_{k=0}^{n-1}\mathbb{E}[B_{k/n}^2]=\frac1{n^2}\sum_{k=0}^{n-1}k=\frac{n-1}{2n}\), tending to \(\tfrac12\). Hence the extended integral has second moment \(\tfrac12\). Check: for \(n=2\), the simple integral is \(B_{1/2}(B_1-B_{1/2})\). Its factors are independent, each has second moment \(\tfrac12\), so the second moment of their product is \(\tfrac12\cdot\tfrac12=\tfrac14\), exactly the simple integrand norm squared \(\frac{2-1}{2\cdot2}=\tfrac14\).
Worked Example: A Brownian Integrand with Drift \(H_t=B_t+t\)
On \([0,1]\), \(H_t=B_t+t\) is predictable. Since \(\mathbb{E}[B_t]=0\) and \(\mathbb{E}[B_t^2]=t\), its squared norm is \(\mathbb{E}\int_0^1(B_t+t)^2\,dt=\int_0^1(t+t^2)\,dt=\tfrac12+\tfrac13=\tfrac56\). Use the simple predictable approximation \(H^n_t=B_{k/n}+k/n\) on \((k/n,(k+1)/n]\).
For \(t=k/n+u\), the error is \((B_t-B_{k/n})+u\). Its expected square is \(u+u^2\): the increment has mean zero, so the cross term with \(u\) has expectation zero. Therefore
On interval \(k\), \(\mathbb{E}[(B_{k/n}+k/n)^2]=k/n+(k/n)^2\). Consequently \(\mathbb{E}\int_0^1(H^n_t)^2\,dt=\frac{n-1}{2n}+\frac{(n-1)(2n-1)}{6n^2}\), which tends to \(\tfrac12+\tfrac13=\tfrac56\). The extended integral therefore has second moment \(\tfrac56\). Check: at \(n=2\), only the second interval contributes, giving integrand norm squared \(\tfrac12\mathbb{E}[(B_{1/2}+\tfrac12)^2]=\tfrac12(\tfrac12+\tfrac14)=\tfrac38\). The formula gives \(\tfrac14+\tfrac18=\tfrac38\) as well.
What the Extension Does—and Does Not—Say
The isometry identifies the full second moment of the integral with the squared integrand norm. It does not assert that the integral is an ordinary pathwise Riemann integral, nor does it turn the Ito integral into a Stratonovich integral. The construction here is an \(L^2(\mathbb{P})\) limit of simple Ito integrals; the equality follows by taking limits in the simple-process isometry. No Ito formula or ordinary chain rule is needed.
A frequent mistake is to confuse zero mean with zero second moment. For example, with the \(n=2\) approximation to \(B_t\) on \([0,1]\), the simple integral is \(I=B_{1/2}(B_1-B_{1/2})\). Its mean is zero, but its second moment is \(\tfrac14\). The incorrect calculation \(\mathbb{E}[I^2]=(\mathbb{E}[I])^2=0\) uses a false identity: in general, \(\mathbb{E}[I^2]=\operatorname{Var}(I)+(\mathbb{E}[I])^2\), not just \((\mathbb{E}[I])^2\). Here the independent factors each have second moment \(\tfrac12\), giving \(\mathbb{E}[I^2]=\tfrac12\cdot\tfrac12=\tfrac14\), in agreement with the simple-process isometry.
Finally, the isometry concerns equality of norms, not pointwise equality between an integral and its integrand. Its role in the extension is exact: convergence of the approximating integrands makes their simple integrals Cauchy, and the same equality of norms survives in the limit.
Check Your Understanding
Use the simple-process isometry and the limiting argument to answer the following.
- State the integrability condition required of a predictable integrand on \([0,T]\) for the extended Ito isometry.
- Why does the simple-process isometry imply that the integrals of an approximating sequence are Cauchy in \(L^2(\mathbb{P})\)?
- For \(H_t=t\) on \([0,1]\), compute the integrand norm squared and the limiting second moment of its Ito integral.
- For \(H_t=B_t\) on \([0,1]\), what is the squared \(L^2(\mathbb{P}\times dt)\) approximation error for the left-endpoint approximation with \(n\) equal intervals?
- Explain why two different simple predictable approximations of the same integrand produce the same limiting Ito integral.