From Integrands to Square-Integrable Random Variables
For a stochastic integral, the relevant size of an integrand is not its value at one time or along one path. It is the expected integrated square, \(\mathbb{E}\int_0^T H_s^2\,ds\). This quantity is also exactly the second moment of the terminal Itô integral. The Itô isometry therefore does more than give a useful estimate: it identifies a Hilbert space of integrands with a closed subspace of square-integrable terminal random variables.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Fix a finite horizon \(T>0\). In particular, for \(s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\), has mean zero, and has variance \(t-s\). All expectations below are under \(\mathbb{P}\). We consider predictable processes \(H\) on \([0,T]\times\Omega\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Predictable processes are progressively measurable, and this integrability condition makes their Itô integrals square-integrable.
The identification matters: changing \(H\) on a set of zero product measure does not change its norm or its Itô integral. The space \(L^2(\mathcal{F}_T)\) consists of \(\mathcal{F}_T\)-measurable random variables \(X\), identified when equal almost surely, with \(\mathbb{E}[X^2]<\infty\); its inner product is \(\mathbb{E}[XY]\). The Itô integral maps \(\mathcal{H}_T\) into this terminal-variable space.
The Itô Isometry
Proof. Begin with a simple predictable process \(H_s=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(s)\), where \(0=t_0<t_1<\cdots<t_n=T\), each \(\xi_i\) is square-integrable and \(\mathcal{F}_{t_i}\)-measurable, and \(\mathbf{1}_A\) denotes the indicator of an event \(A\). Its integral is \(\sum_i\xi_i(B_{t_{i+1}}-B_{t_i})\). Each summand has mean zero because the Brownian increment is independent of \(\mathcal{F}_{t_i}\) and centered. Its second moment is \(\mathbb{E}[\xi_i^2](t_{i+1}-t_i)\). If \(i<j\), the \(i\)-th summand is \(\mathcal{F}_{t_j}\)-measurable, while the conditional mean of the \(j\)-th summand given \(\mathcal{F}_{t_j}\) is zero. Thus the cross terms have expectation zero. Summing the diagonal terms gives $$ \mathbb{E}\left[\left(\int_0^T H_s\,dB_s\right)^2\right] =\sum_i\mathbb{E}[\xi_i^2](t_{i+1}-t_i) =\mathbb{E}\int_0^T H_s^2\,ds. $$
Simple predictable processes are dense in \(\mathcal{H}_T\): the predictable sigma-field is generated, up to the time-zero slice which has zero \(ds\,d\mathbb{P}\) measure, by sets \(A\times(s,t]\) with \(A\in\mathcal{F}_s\). Indicators of these generating sets and their finite linear combinations are dense among square-integrable measurable functions for the finite measure \(ds\,d\mathbb{P}\). Consequently, for general \(H\in\mathcal{H}_T\), choose simple predictable \(H^{(n)}\) converging to \(H\) in \(\mathcal{H}_T\). The simple-process identity applied to differences shows that their integrals are Cauchy in \(L^2(\mathcal{F}_T)\). The Itô integral of \(H\) is defined as that \(L^2\) limit. Taking limits in the identity proves the isometry and the mean-zero claim. Applying the isometry to \(H+G\), and subtracting the identities for \(H\) and \(G\), gives the stated inner-product identity. \(\square\)
The space \(\mathcal{H}_T\) is complete, so it is a Hilbert space. Indeed, it is the subspace of \(L^2(ds\,d\mathbb{P})\) consisting of predictable equivalence classes; an \(L^2\)-convergent sequence has a subsequence converging almost everywhere, whose limit has a predictable version. The Itô isometry then says that the map \(I_T:H\mapsto\int_0^T H_s\,dB_s\) preserves norms and inner products.
Worked Example: A Simple Predictable Integrand
On \([0,1]\), let \(H_s=2\) on \((0,\tfrac12]\) and \(H_s=-1\) on \((\tfrac12,1]\). Compute its norm and its terminal integral, then check the isometry directly.
The squared norm is $$ \|H\|_{\mathcal{H}_1}^2 =\int_0^{1/2}4\,ds+\int_{1/2}^1 1\,ds =4\cdot\tfrac12+1\cdot\tfrac12 =\tfrac52. $$ The terminal integral is \(2B_{1/2}-(B_1-B_{1/2})\). The two increments are independent, centered normal random variables, each with variance \(\tfrac12\). Therefore the integral has mean zero and variance \(4\cdot\tfrac12+1\cdot\tfrac12=\tfrac52\), agreeing with the norm. This direct computation also confirms the cross term vanishes.
Worked Example: The Integrand \(H_s=B_s\)
Take \(T=2\) and \(H_s=B_s\). Brownian paths are continuous and adapted, so this is a predictable integrand. Its expected integrated square is $$ \mathbb{E}\int_0^2 B_s^2\,ds =\int_0^2\mathbb{E}[B_s^2]\,ds =\int_0^2s\,ds =2, $$ so \(H\in\mathcal{H}_2\).
To identify the integral, apply Itô's formula to \(f(x)=x^2\): \(f'(x)=2x\), \(f''(x)=2\), and the correction term is \(\tfrac12 f''(B_s)\,ds=ds\). Hence $$ B_2^2-B_0^2=2\int_0^2 B_s\,dB_s+\int_0^2 ds, \qquad \int_0^2 B_s\,dB_s=\frac{B_2^2-2}{2}. $$
Check: Since \(B_2\) is normal with mean zero and variance \(2\), \(\mathbb{E}[B_2^2]=2\) and \(\mathbb{E}[B_2^4]=3\cdot2^2=12\). The displayed integral has mean \((2-2)/2=0\). Its second moment is $$ \mathbb{E}\left[\left(\frac{B_2^2-2}{2}\right)^2\right] =\frac{12-4\cdot2+4}{4} =2, $$ which independently matches \(\mathbb{E}\int_0^2 B_s^2\,ds=2\).
Why the Range Is Closed
Proof. Linearity of the integral makes \(\mathcal{I}_T\) a linear subspace. To prove it is closed, suppose \(X_n\in\mathcal{I}_T\) and \(X_n\to X\) in \(L^2(\mathcal{F}_T)\). Choose \(H_n\in\mathcal{H}_T\) such that \(X_n=\int_0^T H_n\,dB\). For any \(n,m\), the isometry gives $$ \|H_n-H_m\|_{\mathcal{H}_T}^2 =\mathbb{E}[(X_n-X_m)^2] =\|X_n-X_m\|_{L^2(\mathcal{F}_T)}^2. $$ Since \((X_n)\) converges in \(L^2\), it is Cauchy there; hence \((H_n)\) is Cauchy in the complete space \(\mathcal{H}_T\). Let \(H\in\mathcal{H}_T\) be its limit. Another application of the isometry gives \(\int_0^T H_n\,dB\to\int_0^T H\,dB\) in \(L^2\). But the same sequence converges to \(X\), and limits in \(L^2\) are unique up to almost-sure equality. Thus \(X=\int_0^T H\,dB\) in \(L^2\), so \(X\in\mathcal{I}_T\). Finally, the mean-zero property follows from the isometry construction, and for any constant \(c\), \(\mathbb{E}[cX]=c\mathbb{E}[X]=0\). Therefore the range is orthogonal to constants. \(\square\)
This conclusion is specifically that the terminal integrals form a closed subspace; it does not by itself say that every centered element of \(L^2(\mathcal{F}_T)\) belongs to that subspace. A representation theorem requires an additional argument about the filtration. The isometry provides the Hilbert-space structure needed for that question without assuming its answer.
Worked Example: A Random Predictable Step
On \([0,2]\), set \(A=\{B_1>0\}\) and define \(H_s=1\) on \((0,1]\) and \(H_s=\mathbf{1}_A\) on \((1,2]\). The event \(A\) has probability \(\tfrac12\) and belongs to \(\mathcal{F}_1\), so the coefficient used on \((1,2]\) is known at its left endpoint. Thus \(H\) is predictable. Its squared norm is $$ \mathbb{E}\int_0^2 H_s^2\,ds =1+\mathbb{E}[\mathbf{1}_A](2-1) =1+\tfrac12 =\tfrac32. $$ The integral is \(X=B_1+\mathbf{1}_A(B_2-B_1)\). Its mean is zero: the first term is centered, and the increment \(B_2-B_1\) is independent of \(\mathcal{F}_1\) and centered.
Check: Write \(\Delta=B_2-B_1\). Since \(B_1\) is \(\mathcal{F}_1\)-measurable and \(\mathbb{E}[\Delta\mid\mathcal{F}_1]=0\), the cross term has expectation \(2\mathbb{E}[B_1\mathbf{1}_A\Delta]=0\). Also, independence gives \(\mathbb{E}[\mathbf{1}_A\Delta^2]=\mathbb{P}(A)\mathbb{E}[\Delta^2]=\tfrac12\cdot1=\tfrac12\). Therefore $$ \mathbb{E}[X^2]=\mathbb{E}[B_1^2]+\mathbb{E}[\mathbf{1}_A\Delta^2]=1+\tfrac12=\tfrac32, $$ again matching the integrand norm.
Martingale Property and a Common Pitfall
For \(H\in\mathcal{H}_T\), define \(M_t=\int_0^t H_s\,dB_s\) for \(0\le t\le T\). This process is a true square-integrable martingale, not merely a local martingale. In fact, for \(0\le s\le t\le T\), the integral over \((s,t]\) has conditional expectation zero given \(\mathcal{F}_s\). For simple predictable processes this follows from the independent centered Brownian increments; approximation in \(L^2\), together with the contraction property of conditional expectation in \(L^2\), gives the same result for every \(H\in\mathcal{H}_T\). Moreover, the isometry gives \(\mathbb{E}[M_t^2]=\mathbb{E}\int_0^t H_u^2\,du<\infty\). Both integrability and the conditional-expectation identity are therefore verified.
Common mistake: using future information in an integrand. A tempting but invalid calculation is to set \(H_s=\mathbf{1}_{\{B_2>0\}}\) for \(0<s\le1\), note that \(\mathbb{E}\int_0^1H_s^2\,ds=\tfrac12\), and apply the Itô isometry. The coefficient depends on \(B_2\), which is not known at times \(s\le1\); this process is not predictable. The Itô integral in the stated class and the isometry do not apply to it. A random coefficient on an interval must be measurable with respect to the information available at the interval's left endpoint, as in the preceding example.
Keep the two spaces distinct: \(\mathcal{H}_T\) measures integrands through expected integrated squared size, while \(L^2(\mathcal{F}_T)\) measures terminal random variables through their second moments. The isometry preserves these sizes exactly and, by completeness, ensures that limits of terminal integrals remain terminal integrals.
Check Your Understanding
Use the definitions and the isometry to answer the following.
- For a deterministic integrand \(H_s=s\) on \([0,2]\), compute \(\|H\|_{\mathcal{H}_2}^2\) and the second moment of its terminal integral.
- Why does identifying integrands that agree \(ds\,d\mathbb{P}\)-almost everywhere fit the definition of the Itô integral?
- In the random-step example, what measurability property makes \(\mathbf{1}_{\{B_1>0\}}\) an admissible coefficient on \((1,2]\)?
- Which part of the closed-range proof uses completeness of \(\mathcal{H}_T\), and which equality makes \((H_n)\) Cauchy?
- What additional fact about the filtration would be needed to conclude that every centered \(L^2(\mathcal{F}_T)\) variable is a stochastic integral?