Construction of the Itô Integral¶

1. Concept Definition¶

The Itô integral

\[ \int_0^t H_s \, dB_s \]

defines a stochastic integral with respect to Brownian motion. Unlike classical Riemann-Stieltjes integration, the pathwise construction fails because Brownian motion has infinite variation almost surely. Instead, the integral is defined first on simple adapted processes, then extended by continuity using the Itô isometry.

The construction exploits a special property of Brownian motion: although its total variation is infinite, its quadratic variation is finite. Because Brownian increments satisfy \((\Delta B)^2 \approx \Delta t\), squaring stochastic sums converts random fluctuations into deterministic time increments. This allows us to control the integral in the mean-square sense, leading to the Itô isometry

\[ \mathbb{E}\!\left[\left(\int_0^t H_s \, dB_s\right)^2\right] = \mathbb{E}\!\left[\int_0^t H_s^2 \, ds\right] \]

The construction proceeds in three steps:

Define the integral for simple adapted processes, where stochastic sums are well defined.
Establish the Itô isometry, which controls the integral in the \(L^2\) sense.
Extend to general \(L^2\)-adapted processes by density and continuity.

2. Why Riemann-Stieltjes Fails¶

Classical integration \(\int_0^t f(s)\, dg(s)\) for deterministic functions requires \(g\) to have bounded variation on \([0,t]\):

\[ \operatorname{Var}_{[0,t]}(g) := \sup_{\pi} \sum_{i=0}^{n-1} |g(t_{i+1}) - g(t_i)| < \infty \]

For Brownian motion \(B_t\), this condition fails almost surely.

Proof that Brownian motion has unbounded variation. Consider a uniform partition with \(n\) points: \(t_i = it/n\) and increments \(\Delta B_i = B_{t_{i+1}} - B_{t_i} \sim \mathcal{N}(0, t/n)\). Since \(\mathbb{E}[|\Delta B_i|] = \sqrt{2t/(\pi n)}\), the expected total variation satisfies

\[ \sum_{i=0}^{n-1} \mathbb{E}[|\Delta B_i|] = n \cdot \sqrt{\frac{2t}{\pi n}} = \sqrt{\frac{2nt}{\pi}} \to \infty \quad \text{as } n \to \infty \]

Since the expected total variation diverges, Brownian paths have unbounded variation almost surely. Pathwise Riemann-Stieltjes integration is therefore impossible.

Note

The expected total variation diverging implies almost sure divergence via the following argument: if \(\operatorname{Var}(B) < \infty\) a.s., then by the monotone convergence theorem the expectation would be finite—a contradiction.

However, Brownian motion has quadratic variation equal to \(t\), which suggests a different integration theory is possible.

3. Strategy: \(L^2\)-Approximation¶

The Itô integral is constructed via approximation in the mean-square sense rather than pathwise. The key steps are:

Define the integral for simple processes (piecewise constant, adapted integrands).
Show the integral satisfies the Itô isometry (an \(L^2\) bound).
Extend to general adapted processes in \(L^2\) by density and continuity.

flowchart TD
    A["Want to define ∫₀ᵗ H_s dB_s"] --> B["Pathwise Riemann-Stieltjes approach"]
    B --> C["Fails: Brownian motion has infinite variation"]

    C --> D["Step 1: Start with simple adapted processes"]
    D --> E["Define stochastic sum directly"]

    E --> F["Step 2: Prove Itô isometry"]
    F --> G["E[(∫₀ᵗ H_s dB_s)²] = E[∫₀ᵗ H_s² ds]"]

    G --> H["Step 3: Simple processes are dense in L²"]
    H --> I["Extend by L² limit"]

    I --> J["Itô integral for general adapted square-integrable processes"]

4. Step 1: Simple Processes¶

A simple process \(H_t\) has the form:

\[ H_t(\omega) = \sum_{i=0}^{n-1} H_i(\omega) \,\mathbf{1}_{(t_i, t_{i+1}]}(t) \]

where:

\(0 = t_0 < t_1 < \cdots < t_n = T\) is a partition of \([0,T]\)
Each \(H_i\) is \(\mathcal{F}_{t_i}\)-measurable (adapted)
\(\mathbb{E}[H_i^2] < \infty\) (square-integrable)

Simple processes correspond to piecewise constant trading strategies: the position is decided at time \(t_i\) using only information available up to \(t_i\), and held fixed until the next rebalancing date \(t_{i+1}\).

Definition for simple processes¶

For a simple process \(H_t = \sum_{i=0}^{n-1} H_i \mathbf{1}_{(t_i, t_{i+1}]}\), define:

\[ \boxed{ I_t(H) := \int_0^t H_s \, dB_s = \sum_{i=0}^{n-1} H_i (B_{t_{i+1} \wedge t} - B_{t_i \wedge t}) } \]

where \(a \wedge b := \min(a,b)\).

This is a left-point stochastic Riemann sum. At each time step \(t_i\), the process chooses a value \(H_i\) based only on past information, then multiplies by the next Brownian increment \(\Delta B_i = B_{t_{i+1}} - B_{t_i}\). Each term \(H_i \Delta B_i\) represents the contribution of the random fluctuation during \((t_i, t_{i+1}]\).

Key properties (verified directly from the definition):

Linearity: \(I_t(\alpha H + \beta K) = \alpha I_t(H) + \beta I_t(K)\)
Martingale property: \(\{I_t(H)\}_{t \ge 0}\) is a martingale
Mean zero: \(\mathbb{E}[I_t(H)] = 0\)

Martingale property for simple processes¶

We verify \(\mathbb{E}[I_t(H) \mid \mathcal{F}_s] = I_s(H)\) for \(s < t\). Consider a term \(H_i(B_{t_{i+1} \wedge t} - B_{t_i \wedge t})\):

If \(t_{i+1} \le s\): the term is \(\mathcal{F}_s\)-measurable and contributes to \(I_s(H)\).
If \(s \le t_i\): the increment lies entirely after time \(s\), so \(\mathbb{E}[H_i(B_{t_{i+1}} - B_{t_i}) \mid \mathcal{F}_s] = 0\) by independence and mean zero.
Boundary case \(t_i < s < t_{i+1}\): the increment splits into a past part \(B_s - B_{t_i}\) (which is \(\mathcal{F}_s\)-measurable) and a future part \(B_{t_{i+1} \wedge t} - B_s\) (whose conditional expectation is zero).

Combining these cases gives \(\mathbb{E}[I_t(H) \mid \mathcal{F}_s] = I_s(H)\). \(\square\)

5. Step 2: The Itô Isometry¶

Theorem (Itô Isometry for Simple Processes). Let \(H_t\) be a simple process. Then:

\[ \boxed{ \mathbb{E}\left[\left(\int_0^t H_s \, dB_s\right)^2\right] = \mathbb{E}\left[\int_0^t H_s^2 \, ds\right] } \]

Proof. Write \(H_t = \sum_{i=0}^{n-1} H_i \mathbf{1}_{(t_i, t_{i+1}]}\). Expanding the square:

\[ \left(\int_0^t H_s \, dB_s\right)^2 = \sum_{i} H_i^2 (\Delta B_i)^2 + 2 \sum_{i < j} H_i H_j \Delta B_i \Delta B_j \]

Cross terms vanish. For \(i < j\), condition on \(\mathcal{F}_{t_j}\):

\[ \mathbb{E}[H_i H_j \Delta B_i \Delta B_j] = \mathbb{E}\!\left[H_i H_j \Delta B_i \cdot \underbrace{\mathbb{E}[\Delta B_j \mid \mathcal{F}_{t_j}]}_{=\,0}\right] = 0 \]

since \(\Delta B_j = B_{t_{j+1}} - B_{t_j}\) has mean zero and is independent of \(\mathcal{F}_{t_j}\).

Diagonal terms. For each \(i\), using independence of \(H_i\) (which is \(\mathcal{F}_{t_i}\)-measurable) from the future increment:

\[ \mathbb{E}[H_i^2 (\Delta B_i)^2] = \mathbb{E}[H_i^2] \cdot \mathbb{E}[(\Delta B_i)^2] = \mathbb{E}[H_i^2] \cdot (t_{i+1} \wedge t - t_i \wedge t) \]

Combining:

\[ \mathbb{E}\left[\left(\int_0^t H_s \, dB_s\right)^2\right] = \sum_{i} \mathbb{E}[H_i^2] \cdot (t_{i+1} \wedge t - t_i \wedge t) = \mathbb{E}\left[\int_0^t H_s^2 \, ds\right] \quad \square \]

Intuition from quadratic variation. The isometry holds because \((\Delta B_i)^2 \approx \Delta t_i\) (quadratic variation), so diagonal terms behave like \(\sum_i H_i^2 \Delta t_i \to \int H_s^2\,ds\), while cross terms vanish because disjoint Brownian increments are independent with mean zero.

flowchart LR
    A["Stochastic sum: Σ H_i ΔB_i"] --> B["Square the sum"]

    B --> C["Diagonal terms"]
    B --> D["Cross terms"]

    C --> E["Σ H_i² (ΔB_i)²"]
    E --> F["(ΔB_i)² ≈ Δt_i"]
    F --> G["Σ H_i² Δt_i → ∫₀ᵗ H_s² ds"]

    D --> I["Σ H_i H_j ΔB_i ΔB_j  (i≠j)"]
    I --> J["ΔB_j independent of past, mean zero"]
    J --> K["Cross terms vanish in expectation"]

    G --> M["E[(∫₀ᵗ H_s dB_s)²] = E[∫₀ᵗ H_s² ds]"]
    K --> M

6. Step 3: Extension to \(L^2\)-Adapted Processes¶

Let \(\mathcal{L}^2([0,T])\) denote the space of adapted processes \(H = \{H_t\}_{0 \le t \le T}\) satisfying:

\[ \mathbb{E}\left[\int_0^T H_t^2 \, dt\right] < \infty \]

This is a Hilbert space with inner product \(\langle H, K \rangle := \mathbb{E}[\int_0^T H_t K_t\, dt]\).

Theorem. The Itô integral extends uniquely from simple processes to all processes in \(\mathcal{L}^2([0,T])\), preserving linearity and the Itô isometry.

Proof sketch.

Density. Simple processes are dense in \(\mathcal{L}^2([0,T])\): for any \(H \in \mathcal{L}^2([0,T])\), there exist simple processes \(H^{(n)}\) with \(\mathbb{E}[\int_0^T (H_t - H_t^{(n)})^2\, dt] \to 0\).
Cauchy sequence. By the Itô isometry:

\[ \mathbb{E}\!\left[\left(\int_0^T H_s^{(n)} dB_s - \int_0^T H_s^{(m)} dB_s\right)^2\right] = \mathbb{E}\!\left[\int_0^T (H_s^{(n)} - H_s^{(m)})^2 ds\right] \to 0 \]

So \(\{\int_0^T H_s^{(n)} dB_s\}\) is Cauchy in \(L^2(\Omega)\).

Define the limit. Since \(L^2(\Omega)\) is complete, define:

\[ \int_0^T H_s \, dB_s := L^2\text{-}\lim_{n \to \infty} \int_0^T H_s^{(n)} dB_s \]

Well-defined. The limit is independent of the choice of approximating sequence. If \(\tilde{H}^{(n)}\) is any other approximating sequence converging to \(H\) in \(\mathcal{L}^2\), then by the isometry \(\|\int H^{(n)}\,dB - \int \tilde{H}^{(n)}\,dB\|_{L^2} = \|H^{(n)} - \tilde{H}^{(n)}\|_{\mathcal{L}^2} \to 0\), so both sequences have the same limit.
Isometry preserved. By continuity of the \(L^2\) norm, the Itô isometry holds for the extended integral. \(\square\)

7. The Itô Integral as a Process¶

For \(H \in \mathcal{L}^2([0,T])\), the Itô integral process is defined as:

\[ I_t := \int_0^t H_s \, dB_s, \quad 0 \le t \le T \]

Key properties (established rigorously in the next section):

\(I_t\) is a continuous adapted process
\(I_t\) is a martingale
\(I_t\) has quadratic variation \([I, I]_t = \int_0^t H_s^2\, ds\)

The continuity follows from the Burkholder-Davis-Gundy inequality and Kolmogorov's continuity criterion applied to the \(L^2\)-approximating sequence.

8. Summary¶

The construction of the Itô integral proceeds in three steps:

Simple processes: define \(\int_0^t H_s\,dB_s\) directly for piecewise constant adapted integrands.
Itô isometry: establish the \(L^2\)-bound \(\mathbb{E}[(\int H\,dB)^2] = \mathbb{E}[\int H^2\,ds]\).
Completion: extend to all \(L^2\)-adapted processes via density and continuity.

The resulting integral is not defined pathwise—it is defined as a limit in \(L^2(\Omega)\) rather than pointwise in \(\omega\). This reflects the fundamental difference between stochastic and classical integration.

The Itô isometry is the central technical tool: it converts a stochastic problem (controlling the integral) into a deterministic problem (integrating \(H_s^2\) over time), which is tractable because Brownian motion has finite quadratic variation.

Advanced: extension to local integrands

The construction above requires \(\mathbb{E}[\int_0^T H_t^2\, dt] < \infty\). For many applications—including stochastic differential equations—we need to integrate processes that may not satisfy this global condition.

A process \(H\) is locally square-integrable if there exist stopping times \(\tau_n \uparrow \infty\) such that \(\mathbb{E}[\int_0^{\tau_n \wedge T} H_s^2\, ds] < \infty\) for all \(n\). For such \(H\), define:

\[ \int_0^t H_s \, dB_s := \lim_{n \to \infty} \int_0^{t \wedge \tau_n} H_s \, dB_s \]

The resulting process is a local martingale rather than a true martingale.

Advanced: predictable processes

In advanced treatments, the Itô integral is often constructed for predictable processes—those measurable with respect to the \(\sigma\)-algebra generated by left-continuous adapted processes. Predictability ensures that \(H_s\) uses only strictly past information, avoiding subtle measurability issues. For continuous adapted processes (which are progressively measurable), the distinction between adapted and predictable is minimal.

Exercises¶

Exercise 1. Let \(H_t = \mathbf{1}_{(a,b]}(t)\) for \(0 \le a < b \le T\). This is a simple process with a single nonzero piece. Compute \(\int_0^T H_s\, dB_s\) directly from the definition for simple processes and verify the Ito isometry for this integrand.

Solution to Exercise 1

The process \(H_t = \mathbf{1}_{(a,b]}(t)\) is a simple process with a single nonzero piece, \(H_0 = 1\) on \((a,b]\). By the definition for simple processes:

\[ \int_0^T H_s\, dB_s = 1 \cdot (B_{b \wedge T} - B_{a \wedge T}) = B_b - B_a \]

(assuming \(b \le T\)).

Verification of the Ito isometry. The left side is:

\[ \mathbb{E}\!\left[\left(\int_0^T H_s\, dB_s\right)^2\right] = \mathbb{E}[(B_b - B_a)^2] = b - a \]

since \(B_b - B_a \sim \mathcal{N}(0, b-a)\).

The right side is:

\[ \mathbb{E}\!\left[\int_0^T H_s^2\, ds\right] = \mathbb{E}\!\left[\int_0^T \mathbf{1}_{(a,b]}(s)\, ds\right] = b - a \]

Both sides equal \(b - a\), confirming the Ito isometry.

Exercise 2. Explain why the Riemann-Stieltjes integral \(\int_0^T f(s)\, dB_s(\omega)\) cannot be defined pathwise for a continuous function \(f\). In your answer, identify which property of Brownian motion causes the failure, and explain why finite variation of the integrator is essential for classical integration.

Solution to Exercise 2

The Riemann-Stieltjes integral \(\int_0^T f(s)\, dg(s)\) is defined as the limit of sums \(\sum_k f(t_k^*)(g(t_{k+1}) - g(t_k))\) as the partition mesh tends to zero. A sufficient condition for this limit to exist is that \(g\) has bounded variation on \([0,T]\).

For Brownian motion \(B_t(\omega)\), the total variation is almost surely infinite on every interval \([0,T]\). This is because \(\mathbb{E}[\sum |B_{t_{k+1}} - B_{t_k}|] = n\sqrt{2T/(\pi n)} = \sqrt{2nT/\pi} \to \infty\), and the expected total variation diverging implies almost sure divergence.

When the integrator has infinite variation, the Riemann-Stieltjes sums may not converge, or they may converge to different limits depending on the choice of evaluation points \(t_k^*\). Finite variation is essential because it ensures that the oscillation of \(g\) on each subinterval goes to zero fast enough that the choice of evaluation point within each subinterval becomes irrelevant in the limit. Without this, the sum \(\sum f(t_k^*) \Delta g_k\) is sensitive to the exact location of \(t_k^*\) within \([t_k, t_{k+1}]\), as demonstrated by the different values of the Ito and Stratonovich integrals.

Exercise 3. Let \(H_t^{(n)} = \sum_{k=0}^{n-1} B_{t_k}\, \mathbf{1}_{(t_k, t_{k+1}]}(t)\) be a simple process approximating \(H_t = B_t\) on a uniform partition of \([0,1]\). Compute

\[ \mathbb{E}\!\left[\int_0^1 (B_t - H_t^{(n)})^2\, dt\right] \]

and show that it tends to zero as \(n \to \infty\), verifying that simple processes are dense in \(\mathcal{L}^2([0,1])\) for this particular integrand.

Solution to Exercise 3

On the uniform partition of \([0,1]\) with \(n\) points, \(t_k = k/n\) and \(\Delta t = 1/n\). The simple process approximation is \(H_t^{(n)} = B_{t_k}\) on \((t_k, t_{k+1}]\). We need:

\[ \mathbb{E}\!\left[\int_0^1 (B_t - H_t^{(n)})^2\, dt\right] = \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} \mathbb{E}[(B_t - B_{t_k})^2]\, dt \]

Since \(B_t - B_{t_k}\) is an increment of Brownian motion with \(\mathbb{E}[(B_t - B_{t_k})^2] = t - t_k\):

\[ = \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} (t - t_k)\, dt = \sum_{k=0}^{n-1} \frac{(\Delta t)^2}{2} = n \cdot \frac{1}{2n^2} = \frac{1}{2n} \]

As \(n \to \infty\), \(\frac{1}{2n} \to 0\). This confirms that simple processes are dense in \(\mathcal{L}^2([0,1])\) for the integrand \(H_t = B_t\).

Exercise 4. Using the Ito isometry for simple processes, show that if \(H\) and \(K\) are simple processes, then

\[ \mathbb{E}\!\left[\int_0^T H_s\, dB_s \cdot \int_0^T K_s\, dB_s\right] = \mathbb{E}\!\left[\int_0^T H_s K_s\, ds\right] \]

Hint: Use the polarization identity \(\langle X, Y \rangle = \frac{1}{4}(\|X+Y\|^2 - \|X-Y\|^2)\).

Solution to Exercise 4

By the polarization identity, for any random variables \(X, Y\) in \(L^2\):

\[ \mathbb{E}[XY] = \frac{1}{4}\left(\mathbb{E}[(X+Y)^2] - \mathbb{E}[(X-Y)^2]\right) \]

Apply this with \(X = \int_0^T H_s\, dB_s\) and \(Y = \int_0^T K_s\, dB_s\). By linearity:

\[ X + Y = \int_0^T (H_s + K_s)\, dB_s, \qquad X - Y = \int_0^T (H_s - K_s)\, dB_s \]

By the Ito isometry:

\[ \mathbb{E}[(X+Y)^2] = \mathbb{E}\!\left[\int_0^T (H_s + K_s)^2\, ds\right] \]

\[ \mathbb{E}[(X-Y)^2] = \mathbb{E}\!\left[\int_0^T (H_s - K_s)^2\, ds\right] \]

Subtracting:

\[ \mathbb{E}[(X+Y)^2] - \mathbb{E}[(X-Y)^2] = \mathbb{E}\!\left[\int_0^T 4 H_s K_s\, ds\right] \]

since \((H+K)^2 - (H-K)^2 = 4HK\). Dividing by 4:

\[ \mathbb{E}\!\left[\int_0^T H_s\, dB_s \cdot \int_0^T K_s\, dB_s\right] = \mathbb{E}\!\left[\int_0^T H_s K_s\, ds\right] \]

Exercise 5. Verify the martingale property of the Ito integral for the simple process \(H_t = c \cdot \mathbf{1}_{(0, T/2]}(t)\), where \(c\) is a constant. Specifically, show that for \(s < T/2 < t\), \(\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s\).

Solution to Exercise 5

The simple process is \(H_t = c \cdot \mathbf{1}_{(0, T/2]}(t)\), so:

\[ I_t = \int_0^t c \cdot \mathbf{1}_{(0, T/2]}(s)\, dB_s \]

For \(t \le T/2\): \(I_t = c(B_t - B_0) = cB_t\).

For \(t > T/2\): \(I_t = c(B_{T/2} - B_0) = cB_{T/2}\).

Now take \(s < T/2 < t\). We need to show \(\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s\).

Since \(t > T/2\), \(I_t = cB_{T/2}\). Since \(s < T/2\), \(I_s = cB_s\). So we need:

\[ \mathbb{E}[cB_{T/2} \mid \mathcal{F}_s] = cB_s \]

Write \(B_{T/2} = B_s + (B_{T/2} - B_s)\). Since \(B_{T/2} - B_s\) is independent of \(\mathcal{F}_s\) with mean zero:

\[ \mathbb{E}[cB_{T/2} \mid \mathcal{F}_s] = c\left(B_s + \mathbb{E}[B_{T/2} - B_s \mid \mathcal{F}_s]\right) = cB_s = I_s \]

This confirms \(\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s\).

Exercise 6. The construction extends the integral from simple processes to \(\mathcal{L}^2([0,T])\) using completeness of \(L^2(\Omega)\). Suppose \(H^{(n)}\) and \(\tilde{H}^{(n)}\) are two sequences of simple processes both converging to \(H\) in \(\mathcal{L}^2([0,T])\). Use the Ito isometry to prove that

\[ \int_0^T H_s^{(n)}\, dB_s \quad \text{and} \quad \int_0^T \tilde{H}_s^{(n)}\, dB_s \]

converge to the same limit in \(L^2(\Omega)\), so the extended integral is well defined.

Solution to Exercise 6

By linearity of the Ito integral for simple processes:

\[ \int_0^T H_s^{(n)}\, dB_s - \int_0^T \tilde{H}_s^{(n)}\, dB_s = \int_0^T (H_s^{(n)} - \tilde{H}_s^{(n)})\, dB_s \]

By the Ito isometry:

\[ \mathbb{E}\!\left[\left(\int_0^T (H_s^{(n)} - \tilde{H}_s^{(n)})\, dB_s\right)^2\right] = \mathbb{E}\!\left[\int_0^T (H_s^{(n)} - \tilde{H}_s^{(n)})^2\, ds\right] \]

By the triangle inequality in \(\mathcal{L}^2([0,T])\):

\[ \|H^{(n)} - \tilde{H}^{(n)}\|_{\mathcal{L}^2} \le \|H^{(n)} - H\|_{\mathcal{L}^2} + \|H - \tilde{H}^{(n)}\|_{\mathcal{L}^2} \]

Both terms tend to zero by assumption. Therefore:

\[ \mathbb{E}\!\left[\left(\int_0^T H_s^{(n)}\, dB_s - \int_0^T \tilde{H}_s^{(n)}\, dB_s\right)^2\right] \to 0 \]

This means the two sequences of integrals converge to the same limit in \(L^2(\Omega)\), so the extended integral is well defined (independent of the approximating sequence).