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Construction of the Ito Integral for Simple Processes · Tutorial 83 of 600

The Ito Isometry for Simple Predictable Processes

See why adapted coefficients preserve the isometry: the diagonal terms give the integrated squared integrand, while distinct increments contribute zero cross terms.

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What You'll Learn

  • Define a simple predictable integrand on a finite time partition.
  • Expand its Ito integral as a sum of coefficient-weighted Brownian increments.
  • Evaluate diagonal terms using independence from the past.
  • Show why cross terms vanish even when coefficients are random.
  • Apply the isometry to deterministic and adapted random coefficients.
  • Identify when a simple integrand has zero integral in the square-integrable norm.

Why the Second Moment Matches an Integrated Square

The Ito isometry is an exact relation between the size of a simple stochastic integral and the size of its integrand. Its key feature is that it remains valid when the integrand’s coefficients are random, provided each coefficient is known at the beginning of the interval on which it is used. Expanding the integral reveals the reason: each increment contributes a diagonal term, while the cross terms between distinct increments have expectation zero.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) with respect to the filtration. In particular, for \(s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). Fix \(T>0\) and a partition \(0=t_0<t_1<\cdots<t_n=T\).

Definition: A simple predictable integrand on this partition is a process \(H_s=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(s)\), where each \(\xi_i\) is \(\mathcal{F}_{t_i}\)-measurable and \(\mathbb{E}[\xi_i^2]<\infty\). Its Ito integral is the finite sum \(\int_0^T H_s\,dB_s=\sum_{i=0}^{n-1}\xi_i(B_{t_{i+1}}-B_{t_i})\). Such \(H\) is progressively measurable and belongs to \(\mathcal{L}^2\), the class of progressively measurable processes satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\).

The values assigned at partition endpoints do not affect the time integral of \(H_s^2\). The convention \((t_i,t_{i+1}]\) makes the information requirement explicit: \(\xi_i\) is already measurable with respect to \(\mathcal{F}_{t_i}\) before the Brownian increment on that interval occurs.

The Ito Isometry for Simple Predictable Processes

Theorem: For every simple predictable integrand \(H_s=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(s)\) as defined above, $$ \mathbb{E}\left[\left(\int_0^T H_s\,dB_s\right)^2\right] = \mathbb{E}\int_0^T H_s^2\,ds = \sum_{i=0}^{n-1}(t_{i+1}-t_i)\mathbb{E}[\xi_i^2]. $$

Proof. Write \(\Delta B_i=B_{t_{i+1}}-B_{t_i}\) and \(\Delta t_i=t_{i+1}-t_i\). The integral is the finite sum \(\sum_i\xi_i\Delta B_i\). For a diagonal term, \(\xi_i\) is \(\mathcal{F}_{t_i}\)-measurable and \(\Delta B_i\) is independent of \(\mathcal{F}_{t_i}\), with mean zero and second moment \(\Delta t_i\). Therefore

$$ \mathbb{E}[(\xi_i\Delta B_i)^2] =\mathbb{E}[\xi_i^2]\mathbb{E}[(\Delta B_i)^2] =\Delta t_i\mathbb{E}[\xi_i^2]<\infty. $$

Now take two distinct indices \(i<j\). The random variable \(\xi_i\Delta B_i\xi_j\) is \(\mathcal{F}_{t_j}\)-measurable: \(\xi_i\) and \(\Delta B_i\) are known by time \(t_{i+1}\le t_j\), and \(\xi_j\) is known by time \(t_j\). The increment \(\Delta B_j\) is independent of \(\mathcal{F}_{t_j}\) and has conditional mean zero. The product is integrable: each \(\xi_k\Delta B_k\) is square-integrable by the diagonal calculation, so their product is integrable by Cauchy–Schwarz. Thus

$$ \mathbb{E}[\xi_i\Delta B_i\,\xi_j\Delta B_j] =\mathbb{E}\!\left[\xi_i\Delta B_i\xi_j\,\mathbb{E}[\Delta B_j\mid\mathcal{F}_{t_j}]\right] =0. $$

Expanding the square of the finite sum now gives the sum of its diagonal terms plus twice the sum of its cross terms. Every cross term has expectation zero, so

$$ \mathbb{E}\left[\left(\sum_{i=0}^{n-1}\xi_i\Delta B_i\right)^2\right] =\sum_{i=0}^{n-1}\mathbb{E}[(\xi_i\Delta B_i)^2] =\sum_{i=0}^{n-1}\Delta t_i\mathbb{E}[\xi_i^2]. $$

On each interval \((t_i,t_{i+1}]\), \(H_s^2=\xi_i^2\). Integrating over the intervals and taking expectations gives \(\mathbb{E}\int_0^T H_s^2\,ds=\sum_i\Delta t_i\mathbb{E}[\xi_i^2]\). This proves both equalities. \(\square\)

The proof uses predictability in two distinct ways. For the diagonal term, the coefficient must be independent of the Brownian increment that follows it. For a cross term, the earlier increment and the later coefficient are already known by the start of the later interval; the later increment then has conditional mean zero.

Worked Applications

Worked Example: A Deterministic Step Integrand

On \([0,3]\), let \(H_s=1\) for \(0<s\le1\), and \(H_s=-2\) for \(1<s\le3\). Compute the second moment of the integral and compare it with \(\mathbb{E}\int_0^3H_s^2\,ds\).

The partition has intervals of lengths \(\Delta t_0=1\) and \(\Delta t_1=2\), and coefficients \(\xi_0=1\), \(\xi_1=-2\). Thus

$$ \int_0^3H_s\,dB_s=B_1-2(B_3-B_1). $$

Expanding the square and taking expectations, the cross term vanishes because the increments on \((0,1]\) and \((1,3]\) are independent and centered. The diagonal terms are \(\mathbb{E}[B_1^2]=1\) and \(4\mathbb{E}[(B_3-B_1)^2]=4\cdot2=8\). Hence the second moment is \(1+8=9\). On the other side, \(H_s^2=1\) for one unit of time and \(H_s^2=4\) for two units, so \(\mathbb{E}\int_0^3H_s^2\,ds=1\cdot1+4\cdot2=9\).

Check: applying the theorem’s sum directly gives \(1^2\cdot1+(-2)^2\cdot2=1+8=9\), matching the expansion.

Worked Example: A Coefficient Depending on the Past

On \([0,2]\), take \(H_s=1\) for \(0<s\le1\), and \(H_s=B_1\) for \(1<s\le2\). Find the second moment of its integral, taking care not to assume that terms involving \(B_1\) are independent.

The coefficient on the second interval is \(\mathcal{F}_1\)-measurable, and \(\mathbb{E}[B_1^2]=1\), so this is an admissible square-integrable simple integrand. Put \(D=B_2-B_1\). Then \(D\) is independent of \(B_1\), with \(\mathbb{E}[D]=0\) and \(\mathbb{E}[D^2]=1\). The integral is

$$ \int_0^2H_s\,dB_s=B_1+B_1D=B_1(1+D). $$

Since \(B_1\) and \(D\) are independent, \(\mathbb{E}[B_1^2(1+D)^2]=\mathbb{E}[B_1^2]\mathbb{E}[1+2D+D^2]\). Substituting the moments gives \(1\cdot(1+0+1)=2\). The isometry’s right side is \(1\cdot\mathbb{E}[1^2]+1\cdot\mathbb{E}[B_1^2]=1+1=2\).

Check: the two summands \(B_1\) and \(B_1D\) are centered and uncorrelated, since \(\mathbb{E}[B_1(B_1D)]=\mathbb{E}[B_1^2]\mathbb{E}[D]=0\). Each has second moment \(1\), giving \(2\). They are not independent: \(\mathbb{E}[(B_1D)^2\mid B_1]=B_1^2\), which is random rather than the constant \(\mathbb{E}[(B_1D)^2]=1\). The zero cross term follows from the centered later increment, not from independence of these two summands.

Worked Example: A Past-Dependent Indicator Coefficient

On \([0,2]\), let \(H_s=1\) for \(0<s\le1\), and \(H_s=\mathbf{1}_{\{B_1>0\}}\) for \(1<s\le2\). Calculate both sides of the isometry.

The indicator is \(\mathcal{F}_1\)-measurable and bounded, so its square has finite expectation. The integral equals \(B_1+\mathbf{1}_{\{B_1>0\}}D\), where \(D=B_2-B_1\). On expanding its square, the first diagonal term has expectation \(\mathbb{E}[B_1^2]=1\). The cross term has expectation \(2\mathbb{E}[B_1\mathbf{1}_{\{B_1>0\}}D]=0\), because \(D\) is independent of \(\mathcal{F}_1\) and centered. The remaining term has expectation \(\mathbb{E}[\mathbf{1}_{\{B_1>0\}}D^2]=\mathbb{P}(B_1>0)\mathbb{E}[D^2]=\tfrac12\cdot1=\tfrac12\). Therefore the integral’s second moment is \(1+0+\tfrac12=\tfrac32\).

For the right side, the first interval contributes \(1\cdot\mathbb{E}[1^2]=1\). The second contributes \(1\cdot\mathbb{E}[\mathbf{1}_{\{B_1>0\}}^2]=\mathbb{P}(B_1>0)=\tfrac12\), since \(B_1\sim N(0,1)\) has a symmetric continuous distribution. Thus \(\mathbb{E}\int_0^2H_s^2\,ds=1+\tfrac12=\tfrac32\).

Check: the theorem gives \(1\cdot1+1\cdot\tfrac12=\tfrac32\), in agreement with the direct expansion.

A Consequence: The Norm Detects a Nonzero Integrand

Proposition: For a simple predictable integrand on a partition with strictly positive interval lengths, \(\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=0\) if and only if \(H=0\) almost everywhere with respect to the product of time measure and \(\mathbb{P}\).

Proof. By the isometry, the second moment is \(\sum_i\Delta t_i\mathbb{E}[\xi_i^2]\), a finite sum of nonnegative terms. If the second moment is zero, every term is zero. Since \(\Delta t_i>0\), this implies \(\mathbb{E}[\xi_i^2]=0\), hence \(\xi_i=0\) almost surely for every \(i\). On each interval the integrand equals that coefficient, so \(H=0\) almost everywhere for the product measure. Conversely, if \(H=0\) almost everywhere for that measure, then \(\sum_i\Delta t_i\mathbb{E}[\xi_i^2]=\mathbb{E}\int_0^T H_s^2\,ds=0\), and the isometry gives a zero second moment. \(\square\)

This proposition concerns equality up to sets of zero time-probability measure, not necessarily equality at every time and outcome. Together, the theorem and proposition say that the second-moment size of a simple Ito integral exactly records the square-integrable size of its integrand. The identity is also a central reason that the integral can later be extended from simple processes to a larger class of integrands.

Common Pitfall: Random Coefficients Are Not Future Increments

A common error is to treat every random term in an expanded sum as independent of every other term. In the second example, \(B_1\) and \(B_1D\) are not independent, even though their cross expectation is zero. The correct argument uses that \(D\) is independent of the information at time \(1\) and has mean zero. More generally, when \(i<j\), the coefficient-weighted earlier increment and the later coefficient are known at time \(t_j\); it is the later increment \(\Delta B_j\) whose conditional mean zero eliminates the cross term.

Predictability is essential to this calculation. A coefficient chosen using the increment on the same interval need not be independent of that increment, so the diagonal calculation \(\mathbb{E}[\xi_i^2(\Delta B_i)^2]=\Delta t_i\mathbb{E}[\xi_i^2]\) may fail. For simple predictable processes, the coefficient is fixed using only information available at the interval’s left endpoint, exactly the condition used in the proof.

Check Your Understanding

For each question, identify where predictability or the centered-increment property enters the calculation.

  1. For \(H_s=3\) on \((0,2]\), compute both sides of the isometry.
  2. For \(H_s=1\) on \((0,1]\) and \(H_s=2\) on \((1,3]\), expand the integral’s square and identify its cross term.
  3. In the second worked example, explain why \(\mathbb{E}[B_1(B_1D)]=0\) but \(B_1\) and \(B_1D\) are not independent.
  4. What measurability condition on \(\xi_i\) permits the diagonal-term calculation on \((t_i,t_{i+1}]\)?
  5. For a simple integrand on strictly positive-length intervals, what does a zero second moment of its integral imply about the integrand?

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