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The Ito Integral for L^2 Integrands · Tutorial 94 of 600

Constructing the Integral by L2 Limits

Learn how simple predictable approximations and the Ito isometry produce an L2 limit that serves as the integral of a general admissible process.

Advanced 12 min read

What You'll Learn

  • Define admissible integrands on a filtered probability space carrying Brownian motion
  • Use density of simple predictable processes to choose approximants
  • Apply the Ito isometry to prove their integrals form an L2 Cauchy sequence
  • Construct the integral at a fixed time as an equivalence class of random variables
  • Choose left-endpoint approximations for adapted random integrands
  • Identify why right-endpoint coefficients are not predictable

From Simple Sums to General Integrands

The Ito integral is first defined for simple predictable processes: on each time interval, the coefficient is known at the interval's left endpoint, before the Brownian increment is observed. A general square-integrable integrand need not be constant on any interval, so its integral is constructed by approximation. The key point is that the Ito isometry controls the mean-square difference of two simple integrals by the mean-square difference of their integrands.

Fix \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) adapted to the filtration, with increments independent of the past. In this tutorial every stochastic integral is an Ito integral with respect to this \(B_t\). The admissible integrands are progressively measurable processes \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); this class is denoted by \(\mathcal{L}^2\). Processes that agree outside a set of zero measure for the product measure \(d\mathbb{P}\,ds\) represent the same integrand.

Definition: A simple predictable process on \([0,T]\) has the form \(H_s=\sum_{k=0}^{m-1}\xi_k\mathbf{1}_{(t_k,t_{k+1}]}(s)\), where \(0=t_0<t_1<\cdots<t_m=T\), each \(\xi_k\) is \(\mathcal{F}_{t_k}\)-measurable, and \(\mathbb{E}[\xi_k^2]<\infty\). Its Ito integral at \(T\) is the random variable \(\sum_{k=0}^{m-1}\xi_k(B_{t_{k+1}}-B_{t_k})\).

The earlier results on simple predictable processes give the Ito isometry: if \(G\) is simple predictable, then \(\mathbb{E}\big[(\int_0^T G_s\,dB_s)^2\big]=\mathbb{E}\int_0^T G_s^2\,ds\). We will use this identity only for simple processes. We also use the density of simple predictable processes in \(\mathcal{L}^2\), established in the earlier material on that density. These two facts are enough to construct a limit; no pathwise Riemann-sum convergence is required.

The L2-Limit Construction

Theorem: Let \(H\in\mathcal{L}^2\) on \([0,T]\). Choose simple predictable processes \(H^n\) such that \(\mathbb{E}\int_0^T|H^n_s-H_s|^2\,ds\to0\). Then the simple Ito integrals \(\int_0^T H^n_s\,dB_s\) converge in \(L^2(\mathbb{P})\). Their limit, for this chosen approximating sequence, is the L2-limit construction of the Ito integral of \(H\) at time \(T\).

Proof. We must show that the integrals of the chosen approximants converge in \(L^2(\mathbb{P})\). For any \(m,n\), \(H^n-H^m\) is again simple predictable: refine the two finite partitions to their common partition, and on each refined interval subtract the two coefficients known at its left endpoint. Linearity of the simple integral and the Ito isometry therefore give

$$ \mathbb{E}\left[\left|\int_0^T H^n_s\,dB_s-\int_0^T H^m_s\,dB_s\right|^2\right] =\mathbb{E}\int_0^T|H^n_s-H^m_s|^2\,ds. $$

For real numbers \(x,y\), \(|x-y|^2\le 2|x|^2+2|y|^2\). Apply this with \(x=H^n_s-H_s\) and \(y=H^m_s-H_s\), then integrate and take expectations. The right-hand side is at most \(2\mathbb{E}\int_0^T|H^n_s-H_s|^2ds+2\mathbb{E}\int_0^T|H^m_s-H_s|^2ds\), which tends to zero as \(m,n\to\infty\). Thus the simple integrals form a Cauchy sequence in \(L^2(\mathbb{P})\). Since \(L^2(\mathbb{P})\) is complete, there is an \(L^2\) random-variable equivalence class to which they converge. This defines the claimed limit for the chosen sequence. \(\square\)

The theorem describes a construction from a selected approximating sequence. The assertion that the limit is unchanged if a different sequence is selected is a separate well-definedness property; it is not needed to show that this particular sequence has a limit. At a time \(t\in[0,T]\), the same construction applies on \([0,t]\): restrict the integrands to that interval, splitting a simple process's interval at \(t\) if necessary, and use the same argument.

In practice, the approximation is chosen to respect the information available at each time. For a process that varies with time, left-endpoint values are often useful: they are measurable at the start of the interval and can approximate the integrand in the \(\mathcal{L}^2\) norm. The norm, rather than pointwise closeness alone, is what the construction requires.

Worked Approximations

Worked Example: Approximating the Brownian Integrand \(H_s=B_s\)

On \([0,1]\), take \(H_s=B_s\). Brownian motion is progressively measurable, and \(\mathbb{E}\int_0^1B_s^2ds=\int_0^1s\,ds=\tfrac12<\infty\), so \(H\in\mathcal{L}^2\). For \(n\ge1\), define \(H^n_s=B_{k/n}\) on \((k/n,(k+1)/n]\), \(k=0,\ldots,n-1\). Each coefficient is measurable at the left endpoint, and \(\mathbb{E}[B_{k/n}^2]=k/n<\infty\), so these are simple predictable integrands.

On the \(k\)th interval, write \(s=k/n+u\), where \(0<u\le1/n\). The Brownian increment \(B_s-B_{k/n}\) has second moment \(u\). Hence the approximation error is

$$ \mathbb{E}\int_0^1|H^n_s-H_s|^2\,ds =\sum_{k=0}^{n-1}\int_0^{1/n}u\,du =n\cdot\frac{1}{2n^2} =\frac{1}{2n}. $$

It tends to zero, so the construction uses the L2 limit of the simple sums \(\sum_{k=0}^{n-1}B_{k/n}(B_{(k+1)/n}-B_{k/n})\). Check: for \(n=2\), the error is \(\tfrac{1}{2\cdot2}=\tfrac14\). Directly, the first interval contributes \(\int_0^{1/2}s\,ds=\tfrac18\); on the second, the error has second moment \(s-\tfrac12\), contributing \(\int_{1/2}^1(s-\tfrac12)ds=\tfrac18\). The total is \(\tfrac18+\tfrac18=\tfrac14\), as in the formula.

Worked Example: A Deterministic Integrand \(H_s=s\)

On \([0,1]\), let \(H_s=s\). It is progressively measurable and \(\mathbb{E}\int_0^1H_s^2ds=\int_0^1s^2ds=\tfrac13<\infty\). Use the simple predictable approximation \(H^n_s=k/n\) for \(s\in(k/n,(k+1)/n]\). Its coefficient is deterministic and therefore measurable at the left endpoint.

With \(s=k/n+u\), the error is \(s-k/n=u\). Thus

$$ \mathbb{E}\int_0^1|H^n_s-H_s|^2\,ds =\sum_{k=0}^{n-1}\int_0^{1/n}u^2\,du =n\cdot\frac{1}{3n^3} =\frac{1}{3n^2}. $$

The associated simple integral is \(\sum_{k=0}^{n-1}(k/n)(B_{(k+1)/n}-B_{k/n})\), and its sequence converges in \(L^2\). Check: when \(n=2\), the approximation error is \(\tfrac{1}{3\cdot 2^2}=\tfrac1{12}\). Direct integration gives \(\int_0^{1/2}s^2ds=\tfrac1{24}\) and \(\int_{1/2}^1(s-\tfrac12)^2ds=\tfrac1{24}\). Their sum is \(\tfrac1{12}\), matching the formula.

Worked Example: Approximating \(H_s=B_s^2\)

On \([0,1]\), set \(H_s=B_s^2\). It is progressively measurable. Since \(B_s\) is centered normal with variance \(s\), \(\mathbb{E}[B_s^4]=3s^2\); consequently \(\mathbb{E}\int_0^1H_s^2ds=\int_0^1 3s^2ds=1<\infty\). Choose \(H^n_s=B_{k/n}^2\) on \((k/n,(k+1)/n]\). The coefficients are \(\mathcal{F}_{k/n}\)-measurable and square-integrable because \(\mathbb{E}[B_{k/n}^4]=3(k/n)^2<\infty\).

Put \(s=k/n+u\), \(0<u\le1/n\). Then \(B_s=B_{k/n}+\Delta B\), where \(\Delta B\) is independent of \(\mathcal{F}_{k/n}\), has mean zero, variance \(u\), and fourth moment \(3u^2\). Expanding the squared error and taking expectations gives

$$ \begin{aligned} \mathbb{E}\big[(B_s^2-B_{k/n}^2)^2\big] &=\mathbb{E}\big[(2B_{k/n}\Delta B+(\Delta B)^2)^2\big]\\ &=4(k/n)u+3u^2. \end{aligned} $$

Here the cross term has expectation zero because \(\mathbb{E}[\Delta B^3]=0\) and \(\Delta B\) is independent of \(B_{k/n}\). Integrating over all intervals yields

$$ \mathbb{E}\int_0^1|H^n_s-H_s|^2ds =\sum_{k=0}^{n-1}\left(\frac{2k}{n^3}+\frac{1}{n^3}\right) =\frac{n(n-1)+n}{n^3} =\frac1n. $$

This tends to zero, so the corresponding simple Ito integrals converge in \(L^2\). Check: for \(n=2\), the first interval contributes \(\int_0^{1/2}3u^2du=\tfrac18\). The second contributes \(\int_0^{1/2}(2u+3u^2)du=\tfrac14+\tfrac18=\tfrac38\). The total is \(\tfrac12=1/2\), as the formula \(1/n\) gives.

Why Predictability Matters

The approximation method is not simply “replace the process by nearby values.” Coefficients must be known at the beginning of the interval whose Brownian increment they multiply. This is the predictable structure that makes the simple Ito integral and its isometry applicable. In the example \(H_s=B_s\), replacing the left endpoint \(B_{k/n}\) by the right endpoint \(B_{(k+1)/n}\) gives coefficients that depend on the increment being integrated over. That is not an admissible simple predictable approximation.

For a concrete illustration, on the single interval \((0,1]\) the right-endpoint choice would give coefficient \(B_1\). It is not \(\mathcal{F}_0\)-measurable (Brownian motion starts at \(B_0=0\), while \(B_1\) is nonconstant). Treating the resulting product as an Ito simple integral would give \(B_1(B_1-B_0)=B_1^2\), whose expectation is \(1\). By contrast, the admissible left-endpoint coefficient is \(B_0=0\), giving the simple integral \(0\) on this one-interval partition. The right-endpoint calculation is not a competing construction: it violates the predictability requirement and cannot be justified by the simple-process Ito isometry.

A second common mistake is to require the approximants to be pointwise close at every time. The construction only requires convergence in \(\mathbb{E}\int_0^T|H^n_s-H_s|^2ds\). This allows random errors and does not require uniform pathwise approximation. Once this norm tends to zero, the isometry proves the integral sequence is Cauchy; completeness of \(L^2(\mathbb{P})\) supplies its limit.

Check Your Understanding

In each question, assume the filtered probability space carries a Brownian motion adapted to the filtration, and that all integrals are Ito integrals.

  1. State the measurability and integrability conditions on the coefficient \(\xi_k\) in a simple predictable process.
  2. If \(\mathbb{E}\int_0^T|H^n_s-H_s|^2ds\to0\), explain how the Ito isometry makes the simple integrals a Cauchy sequence in \(L^2(\mathbb{P})\).
  3. For \(H_s=B_s\) on \([0,1]\), compute the approximation error for \(n=4\) using left endpoints.
  4. For the deterministic integrand \(H_s=s\) on \([0,1]\), give the squared L2 approximation error for \(n\) equal subintervals.
  5. Explain why using \(B_{(k+1)/n}\) as the coefficient on \((k/n,(k+1)/n]\) fails the simple predictable integrand condition.