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The Ito Integral for L^2 Integrands · Tutorial 95 of 600

Independence of the Approximating Sequence

See why the Ito isometry guarantees that every simple-process approximation to the same integrand has the same limiting Ito integral.

Advanced 12 min read

What You'll Learn

  • Define L2 convergence for progressively measurable Ito integrands on a finite time horizon.
  • Use the simple-process Ito isometry to compare approximating integrals.
  • Prove existence of the L2 limit from the Cauchy property.
  • Prove that two approximating sequences yield the same limit.
  • Understand why the integral depends on the integrand’s L2 equivalence class.
  • Identify a common error when calculating the isometry for centered random increments.

Why the Choice of Approximation Must Not Matter

The Ito integral for a general square-integrable integrand is constructed as an \(L^2\) limit of integrals of simple predictable processes. Density supplies such approximations, but a definition by limits is meaningful only if the result is independent of which approximating sequence is chosen. The key is the Ito isometry: it converts the \(L^2\) distance between simple integrands into the \(L^2\) distance between their integrals.

Work 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. Fix a finite horizon \(T>0\). The target integrands \(H\) in this tutorial are progressively measurable and satisfy \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). We identify two such integrands when they agree almost everywhere for the product measure \(dt\,d\mathbb{P}\).

Definition: A simple predictable process on \([0,T]\) has the form \(U_t=\sum_{i=0}^{m-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(t)\), where \(0=t_0<t_1<\cdots<t_m=T\), each \(\xi_i\) is \(\mathcal{F}_{t_i}\)-measurable, and \(\mathbb{E}[\xi_i^2]<\infty\). Its Ito integral is \(I(U)=\int_0^T U_t\,dB_t=\sum_{i=0}^{m-1}\xi_i(B_{t_{i+1}}-B_{t_i})\). The integrands and integrals used below are of this kind or are limits in \(L^2\) of such integrals.

For a random variable \(X\) with \(\mathbb{E}[X^2]<\infty\), write \(\|X\|_{L^2(\mathbb{P})}=(\mathbb{E}[X^2])^{1/2}\). For an integrand \(H\), write \(\|H\|_{\mathcal{L}^2}=(\mathbb{E}\int_0^T H_t^2\,dt)^{1/2}\). The simple-process Ito isometry, established for simple predictable processes, states that

$$ \mathbb{E}\left[\left(\int_0^T U_t\,dB_t\right)^2\right] = \mathbb{E}\int_0^T U_t^2\,dt. $$

In particular, if \(U\) and \(V\) are simple predictable processes, their difference is also simple predictable, so applying the isometry to \(U-V\) gives

$$ \|I(U)-I(V)\|_{L^2(\mathbb{P})} = \|U-V\|_{\mathcal{L}^2}. $$

This comparison is the entire mechanism behind independence: if two simple integrands are close in \(\mathcal{L}^2\), their integral random variables are close in \(L^2(\mathbb{P})\) by exactly the same amount.

The Limit Exists for Every Approximating Sequence

Proposition: Let \(H\) be a progressively measurable integrand with \(\|H\|_{\mathcal{L}^2}<\infty\), and let \((H^n)_{n\ge1}\) be simple predictable processes such that \(\|H^n-H\|_{\mathcal{L}^2}\to0\). Then \((I(H^n))_{n\ge1}\) is Cauchy in \(L^2(\mathbb{P})\) and has a limit in \(L^2(\mathbb{P})\).

Proof. For any \(n,m\), \(H^n-H^m\) is simple predictable. The simple-process isometry and the triangle inequality therefore give

$$ \|I(H^n)-I(H^m)\|_{L^2(\mathbb{P})} = \|H^n-H^m\|_{\mathcal{L}^2} \le \|H^n-H\|_{\mathcal{L}^2}+\|H^m-H\|_{\mathcal{L}^2}. $$

Both terms on the right tend to zero as \(n,m\to\infty\), so the integral sequence is Cauchy in \(L^2(\mathbb{P})\). To justify the existence of a limit, recall why \(L^2(\mathbb{P})\) is complete. Given any Cauchy sequence \((X_n)\) in that space, choose a subsequence \((X_{n_k})\) such that \(\|X_{n_{k+1}}-X_{n_k}\|_{L^2}\le2^{-k}\). Since \(\mathbb{P}\) is a probability measure, Cauchy–Schwarz gives \(\mathbb{E}|X_{n_{k+1}}-X_{n_k}|\le2^{-k}\). Tonelli’s theorem then implies that \(\sum_k|X_{n_{k+1}}-X_{n_k}|\) is finite almost surely, so the subsequence has a finite almost-sure limit \(X\). By Minkowski’s inequality, finite tails have \(L^2\) norm at most \(\sum_{j=k}^{\infty}2^{-j}\); Fatou’s lemma gives the same bound for the infinite tail. Thus \(X\) is in \(L^2\) and \(X_{n_k}\to X\) in \(L^2\). The original Cauchy sequence also converges to \(X\), by the triangle inequality. Applying completeness to \(I(H^n)\) proves the proposition. \(\square\)

Independence of the Approximating Sequence

Theorem: Suppose \(H\) is progressively measurable and \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). If \((H^n)\) and \((G^n)\) are any two sequences of simple predictable processes satisfying \(\|H^n-H\|_{\mathcal{L}^2}\to0\) and \(\|G^n-H\|_{\mathcal{L}^2}\to0\), then \(I(H^n)\) and \(I(G^n)\) converge in \(L^2(\mathbb{P})\) to the same random variable.

Proof. The proposition shows that each integral sequence has an \(L^2\) limit; call the limits \(X\) and \(Y\), respectively. For each \(n\), the difference \(H^n-G^n\) is simple predictable, so the isometry applies. The triangle inequality in \(\mathcal{L}^2\) yields

$$ \|I(H^n)-I(G^n)\|_{L^2(\mathbb{P})} = \|H^n-G^n\|_{\mathcal{L}^2} \le \|H^n-H\|_{\mathcal{L}^2}+\|G^n-H\|_{\mathcal{L}^2} \longrightarrow 0. $$

Also, convergence of each sequence implies \(I(H^n)-I(G^n)\to X-Y\) in \(L^2\), since

$$ \|(I(H^n)-I(G^n))-(X-Y)\|_{L^2} \le \|I(H^n)-X\|_{L^2}+\|I(G^n)-Y\|_{L^2} \longrightarrow0. $$

The same sequence of differences therefore converges in \(L^2\) both to zero and to \(X-Y\). Limits in a normed space are unique: the triangle inequality gives \(\|X-Y\|_{L^2}\le\|X-Y-(I(H^n)-I(G^n))\|_{L^2}+\|I(H^n)-I(G^n)\|_{L^2}\to0\). Hence \(X=Y\) almost surely. \(\square\)

We can therefore define \(\int_0^T H_t\,dB_t\) to be the \(L^2(\mathbb{P})\) limit of the integrals of any simple predictable sequence converging to \(H\) in \(\mathcal{L}^2\). Density of simple predictable processes ensures that at least one such sequence exists; the theorem ensures that its choice does not affect the result. This definition assigns an equivalence class of random variables equal almost surely, rather than a particular pointwise representative.

Worked Comparisons

Worked Example: Two Grids for a Deterministic Integrand

On \([0,1]\), let \(H_t=t\). For each positive integer \(n\), define \(A^n_t=i/n\) on \((i/n,(i+1)/n]\), for \(i=0,\ldots,n-1\). Compare the integral approximations from \(A^n\) and \(A^{2n}\).

On each interval of length \(1/n\), the error \(H_t-A^n_t\) increases from \(0\) to \(1/n\). Thus

$$ \|A^n-H\|_{\mathcal{L}^2}^2 = n\int_0^{1/n}u^2\,du = n\frac{(1/n)^3}{3} = \frac{1}{3n^2}. $$

For \(A^{2n}\), the same calculation with interval length \(1/(2n)\) gives \(\|A^{2n}-H\|_{\mathcal{L}^2}^2=1/(12n^2)\). On the first half of each coarse interval, the two processes agree. On the second half, their difference is \(1/(2n)\). There are \(n\) such half-intervals, each of length \(1/(2n)\), so

$$ \|I(A^n)-I(A^{2n})\|_{L^2}^2 = \|A^n-A^{2n}\|_{\mathcal{L}^2}^2 = n\left(\frac{1}{2n}\right)^2\frac{1}{2n} = \frac{1}{8n^2}. $$

Check: at \(n=2\), the two approximation errors squared are \(1/12\) and \(1/48\), and the squared distance between the integral approximations is \(1/32\). The formula gives \(1/(8\cdot2^2)=1/32\), as required. Both grids approximate the same integrand, and their integral difference tends to zero.

Worked Example: Approximating Brownian Motion on Nested Grids

On \([0,1]\), take \(H_t=B_t\), and set \(A^n_t=B_{i/n}\) on \((i/n,(i+1)/n]\). These coefficients are measurable at the left endpoints, as required for simple predictable processes. Compare \(A^n\) with the finer approximation \(A^{2n}\).

For \(t=i/n+u\), Brownian increments have mean zero and second moment \(u\). Therefore

$$ \|A^n-H\|_{\mathcal{L}^2}^2 = \sum_{i=0}^{n-1}\int_0^{1/n}\mathbb{E}[(B_{i/n+u}-B_{i/n})^2]\,du = n\int_0^{1/n}u\,du = \frac{1}{2n}. $$

For the comparison of the two grids, the processes agree on the first half of every coarse interval. On the second half, their difference is \(B_{i/n+1/(2n)}-B_{i/n}\), whose second moment is \(1/(2n)\). Integrating this expectation over a half-interval of length \(1/(2n)\), then summing over all \(n\) coarse intervals, gives

$$ \|I(A^n)-I(A^{2n})\|_{L^2}^2 = n\frac{1}{2n}\frac{1}{2n} = \frac{1}{4n}. $$

Check: for \(n=2\), the error squared for the coarse integrand is \(1/4\), the error squared for \(A^4\) is \(1/8\), and the squared distance between their integrals is \(1/8\). The comparison formula gives \(1/(4\cdot2)=1/8\). Both approximation errors vanish as the grid is refined, so the two integral sequences have the same limit.

Worked Example: Different Approximations to the Zero Integrand

On \([0,1]\), let \(H_t=0\). Compare the zero simple process \(G^n_t=0\) with \(U^n_t=\mathbf{1}_{(0,1/n]}(t)\). Both are simple predictable, and \(U^n\) converges to \(H\) in \(\mathcal{L}^2\).

The simple integral is \(I(U^n)=B_{1/n}-B_0=B_{1/n}\), while \(I(G^n)=0\). Since \(B_0=0\) and \(\mathbb{E}[B_{1/n}^2]=1/n\),

$$ \|U^n-H\|_{\mathcal{L}^2}^2=\int_0^{1/n}1\,dt=\frac1n, \qquad \|I(U^n)-I(G^n)\|_{L^2}^2 = \mathbb{E}[B_{1/n}^2] = \frac1n. $$

Check: at \(n=4\), both squared quantities equal \(1/4\). As \(n\to\infty\), the integral approximation converges to zero in \(L^2\), just like the identically zero approximation. A nonzero integral at a finite stage does not contradict independence; what matters is the common limit.

What the Isometry Protects

The integral is defined for the \(\mathcal{L}^2\) equivalence class of \(H\). If \(H\) and \(G\) agree \(dt\,d\mathbb{P}\)-almost everywhere, then \(\|H-G\|_{\mathcal{L}^2}=0\); approximations to these two representatives are therefore also close to each other in the limit. The isometry makes their integral limits the same almost surely. No choice of values on a time-probability null set can change the defined integral.

Common mistake: taking the mean before the square. In the Brownian-grid example, each increment \(B_{i/n+u}-B_{i/n}\) has mean zero. It would be wrong to infer that the squared error is zero by writing \(\mathbb{E}[B_{i/n+u}-B_{i/n}]=0\) and using that value in place of the second moment. The isometry uses \(\mathbb{E}[(B_{i/n+u}-B_{i/n})^2]=u\), not the square of the mean. For example, at \(u=1/4\), the mean is \(0\), but the second moment is \(1/4\). It is this nonzero second moment that produces the \(\mathcal{L}^2\) error and controls the integral approximation.

Check Your Understanding

Use the simple-process isometry and the distinction between first and second moments.

  1. Suppose \(\|H^n-H\|_{\mathcal{L}^2}\to0\). Write the inequality that proves \(I(H^n)\) is Cauchy.
  2. Why must two \(L^2\) limits \(X\) and \(Y\) of integral approximations be equal almost surely if the \(L^2\) distance between the approximations tends to zero?
  3. For \(H_t=t\) on \([0,1]\), what is \(\|A^n-H\|_{\mathcal{L}^2}^2\) for the left-endpoint grid used above?
  4. For \(H_t=B_t\), why is \(\mathbb{E}[(B_{s+u}-B_s)^2]=u\) even though \(\mathbb{E}[B_{s+u}-B_s]=0\)?
  5. For \(U^n_t=\mathbf{1}_{(0,1/n]}(t)\), calculate \(\|U^n\|_{\mathcal{L}^2}^2\) and \(\mathbb{E}[I(U^n)^2]\). What does their equality illustrate?

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