Properties of the Itô Integral¶

1. Concept Definition¶

Having constructed the Itô integral \(\int_0^t H_s\, dB_s\) for adapted square-integrable processes, we now establish its fundamental properties. These properties—linearity, martingale structure, path continuity, and quadratic variation—form the foundation for stochastic calculus and distinguish the Itô integral from classical integration.

The four core properties are summarized below. Let \(I_t := \int_0^t H_s\, dB_s\) throughout.

Property	Statement
Linearity	\(\int (\alpha H + \beta K)\,dB = \alpha \int H\,dB + \beta \int K\,dB\)
Zero mean	\(\mathbb{E}[I_t] = 0\) for all \(t\)
Martingale	\(\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s\) for \(s \le t\)
Itô isometry	\(\mathbb{E}[I_t^2] = \mathbb{E}[\int_0^t H_s^2\,ds]\)
Continuity	\(t \mapsto I_t\) has continuous sample paths a.s.
Quadratic variation	\([I,I]_t = \int_0^t H_s^2\,ds\)

2. Explanation¶

Linearity¶

Theorem. Let \(H, K \in \mathcal{L}^2([0,T])\) and \(\alpha, \beta \in \mathbb{R}\). Then:

\[ \int_0^t (\alpha H_s + \beta K_s)\, dB_s = \alpha \int_0^t H_s\, dB_s + \beta \int_0^t K_s\, dB_s \]

Proof. Linearity holds by definition for simple processes. Since the integral extends by \(L^2\)-continuity and simple processes are dense, linearity passes to the limit. \(\square\)

Martingale property¶

Theorem. Let \(H \in \mathcal{L}^2([0,T])\). Then \(I_t = \int_0^t H_s\,dB_s\) is a continuous square-integrable martingale with respect to \(\{\mathcal{F}_t\}\).

Proof. We verify the three conditions.

Adaptedness. \(I_t\) is the \(L^2\)-limit of \(\mathcal{F}_t\)-measurable random variables, hence \(\mathcal{F}_t\)-measurable.

Integrability. By the Itô isometry:

\[ \mathbb{E}[I_t^2] = \mathbb{E}\!\left[\int_0^t H_s^2\, ds\right] \le \mathbb{E}\!\left[\int_0^T H_s^2\, ds\right] < \infty \]

So \(\mathbb{E}[|I_t|] \le \sqrt{\mathbb{E}[I_t^2]} < \infty\).

Martingale condition. For simple processes, \(\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s\) was verified directly in the construction using the independence and mean-zero property of future Brownian increments. For general \(H \in \mathcal{L}^2([0,T])\), approximate by simple processes \(H^{(n)} \to H\). Since conditional expectation is \(L^2\)-continuous:

\[ \mathbb{E}[I_t \mid \mathcal{F}_s] = \lim_{n \to \infty} \mathbb{E}[I_t^{(n)} \mid \mathcal{F}_s] = \lim_{n \to \infty} I_s^{(n)} = I_s \quad \square \]

Corollary (Zero mean). \(\mathbb{E}[I_t] = \mathbb{E}[I_0] = 0\) for all \(t\).

Path continuity¶

Theorem. Let \(H \in \mathcal{L}^2([0,T])\). Then \(I_t = \int_0^t H_s\,dB_s\) has a continuous modification.

Proof sketch. The proof uses the Burkholder-Davis-Gundy (BDG) inequality, which bounds moments of a continuous martingale's supremum by its quadratic variation:

\[ \mathbb{E}\!\left[\sup_{u \le T} |I_u|^p\right] \le C_p\, \mathbb{E}\!\left[\left(\int_0^T H_s^2\,ds\right)^{p/2}\right] \]

For \(p = 4\), this gives

\[ \mathbb{E}[|I_t - I_s|^4] \le C\,(t-s)^2\, \mathbb{E}\!\left[\sup_{u \le T} H_u^4\right] \]

under appropriate regularity on \(H\). By the Kolmogorov continuity criterion, since \(\mathbb{E}[|I_t - I_s|^4] \le K|t-s|^{1+\varepsilon}\) for some \(\varepsilon > 0\), \(I_t\) has a continuous modification. \(\square\)

Remark. Continuity is a special feature of integration with respect to Brownian motion. Stochastic integrals with respect to Poisson processes or general semimartingales may have jumps.

Quadratic variation¶

Theorem. Let \(H \in \mathcal{L}^2([0,T])\). Then the quadratic variation of \(I_t = \int_0^t H_s\,dB_s\) is:

\[ \boxed{ [I, I]_t = \int_0^t H_s^2 \, ds } \]

Proof (heuristic). Consider a partition \(\pi: 0 = t_0 < t_1 < \cdots < t_n = t\). Each increment is:

\[ I_{t_{i+1}} - I_{t_i} = \int_{t_i}^{t_{i+1}} H_s\, dB_s \approx H_{t_i}(B_{t_{i+1}} - B_{t_i}) \]

for small intervals where \(H_s \approx H_{t_i}\). Therefore:

\[ \sum_{i} (I_{t_{i+1}} - I_{t_i})^2 \approx \sum_{i} H_{t_i}^2 (B_{t_{i+1}} - B_{t_i})^2 \to \int_0^t H_s^2\,ds \]

where the last step uses the quadratic variation of Brownian motion: \(\sum_i (\Delta B_i)^2 \to t\). The convergence holds in probability. \(\square\)

Corollary. The process \(M_t := I_t^2 - \int_0^t H_s^2\,ds\) is a martingale. This follows from the Doob-Meyer decomposition: \(I_t^2\) is a submartingale with compensator \(\int_0^t H_s^2\,ds\).

Itô isometry (restated)¶

Theorem (Itô Isometry). For \(H \in \mathcal{L}^2([0,T])\):

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

Since \(\mathbb{E}[I_t] = 0\), the left side equals \(\operatorname{Var}(I_t)\), so the isometry is a variance formula.

Generalization (polarization). For two processes \(H, K \in \mathcal{L}^2([0,T])\):

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

This follows by polarization: \(\langle H, K \rangle_{L^2} = \tfrac{1}{4}(\|H+K\|^2 - \|H-K\|^2)\).

3. Diagram¶

The six properties form an interconnected structure. The martingale property and Itô isometry are the two central pillars from which the others follow.

flowchart TD
    A["Itô integral I_t = ∫₀ᵗ H_s dB_s"]

    A --> B["Linearity: α∫H dB + β∫K dB"]
    A --> C["Martingale property: E[I_t | F_s] = I_s"]
    A --> D["Itô isometry: E[I_t²] = E[∫ H_s² ds]"]

    C --> E["Zero mean: E[I_t] = 0"]
    C --> F["Continuity: paths continuous a.s."]
    D --> G["Quadratic variation: [I,I]_t = ∫₀ᵗ H_s² ds"]
    D --> H["Variance formula: Var(I_t) = E[∫ H_s² ds]"]

4. Examples¶

Example 1: Constant integrand — \(H_s = \sigma\)¶

Let \(\sigma > 0\) be a constant. Then \(I_t = \sigma B_t\).

Martingale: \(\mathbb{E}[\sigma B_t \mid \mathcal{F}_s] = \sigma B_s\). ✓

Itô isometry: \(\mathbb{E}[(\sigma B_t)^2] = \sigma^2 t\) and \(\mathbb{E}[\int_0^t \sigma^2\, ds] = \sigma^2 t\). ✓

Quadratic variation: \([I,I]_t = \sigma^2 t\). ✓

Example 2: Deterministic integrand — \(H_s = s\)¶

\[ I_t = \int_0^t s\, dB_s \]

Since \(H_s = s\) is deterministic, \(I_t\) is Gaussian with mean zero and variance

\[ \operatorname{Var}(I_t) = \mathbb{E}\!\left[\int_0^t s^2\, ds\right] = \frac{t^3}{3} \]

So \(I_t \sim \mathcal{N}(0,\, t^3/3)\).

Quadratic variation: \([I,I]_t = \int_0^t s^2\, ds = t^3/3\). For this deterministic integrand the quadratic variation is the same deterministic value \(t^3/3\) on every path.

Example 3: Random integrand — \(H_s = B_s\)¶

\[ I_t = \int_0^t B_s\, dB_s = \frac{B_t^2 - t}{2} \]

(from Itô's formula applied to \(f(x) = x^2/2\)).

Martingale: \(\frac{B_t^2 - t}{2}\) is indeed a martingale since \(B_t^2 - t\) is the well-known martingale. ✓

Itô isometry: \(\mathbb{E}[(B_t^2-t)^2/4] = \operatorname{Var}(B_t^2)/4 + (t^2 - 2t^2 + t^2)/4\). More directly:

\[ \mathbb{E}\!\left[\int_0^t B_s^2\, ds\right] = \int_0^t \mathbb{E}[B_s^2]\, ds = \int_0^t s\, ds = \frac{t^2}{2} \]

And \(\mathbb{E}[(B_t^2-t)^2/4] = t^2/2\). ✓

Non-Gaussianity: since \(B_t^2 - t\) is a centered chi-squared random variable (scaled), \(I_t = (B_t^2-t)/2\) is not Gaussian. Random integrands generally produce non-Gaussian integrals.

Example 4: Quadratic variation in action¶

Let \(H_s = \sigma(s, B_s)\) for some function \(\sigma\). Then the quadratic variation of the Itô process

\[ X_t = x + \int_0^t \mu_s\, ds + \int_0^t \sigma_s\, dB_s \]

is \([X,X]_t = \int_0^t \sigma_s^2\, ds\). The drift term \(\int_0^t \mu_s\, ds\) contributes zero quadratic variation (it has finite variation). This shows that quadratic variation detects only the stochastic component of an Itô process.

5. Summary¶

The Itô integral \(I_t = \int_0^t H_s\,dB_s\) satisfies six fundamental properties:

Linearity: \(\int (\alpha H + \beta K)\,dB = \alpha \int H\,dB + \beta \int K\,dB\)
Zero mean: \(\mathbb{E}[I_t] = 0\) for all \(t\)
Martingale: \(\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s\)
Itô isometry: \(\mathbb{E}[I_t^2] = \mathbb{E}[\int_0^t H_s^2\, ds]\)
Continuity: \(t \mapsto I_t\) has continuous sample paths a.s.
Quadratic variation: \([I,I]_t = \int_0^t H_s^2\,ds\)

These properties reveal the deep connection between stochastic integration and martingale theory. They form the foundation for:

Itô's formula — the stochastic chain rule
Stochastic differential equations
Change of measure (Girsanov's theorem)
Mathematical finance — option pricing and hedging

In the next section, we introduce Itô processes, which combine ordinary and stochastic integration into the general class \(dX_t = \mu_t\,dt + \sigma_t\,dB_t\).

Advanced: martingale representation theorem

Every square-integrable martingale \(M_t\) adapted to the Brownian filtration \(\mathcal{F}_t = \sigma(B_s: s \le t)\) can be represented as:

\[ M_t = M_0 + \int_0^t H_s\, dB_s \]

for some adapted process \(H \in \mathcal{L}^2([0,T])\). This result—the martingale representation theorem—says that in a Brownian filtration, all randomness comes from Brownian motion, and every martingale is a reweighting of Brownian increments via stochastic integration. It is fundamental in option pricing (every hedging strategy can be represented as a stochastic integral) and filtering theory.

Advanced: local martingales

For processes not globally in \(\mathcal{L}^2\), the Itô integral is defined as a local martingale: there exist stopping times \(\tau_n \uparrow \infty\) such that each stopped process \(I_{t \wedge \tau_n}\) is a true martingale. Local martingales need not have constant expectation. For example, \(\int_0^t e^{B_s}\,dB_s\) is a local martingale but its expectation is not zero in general, since \(e^{B_s}\) is not globally square-integrable.

Exercises¶

Exercise 1. Let \(H_s = \cos(s)\) and \(K_s = \sin(s)\) on \([0, \pi]\). Using linearity and the Ito isometry, compute the variance of

\[ \int_0^\pi \bigl(3\cos(s) + 2\sin(s)\bigr)\, dB_s \]

Solution to Exercise 1

By linearity: \(\int_0^\pi (3\cos(s) + 2\sin(s))\, dB_s = 3\int_0^\pi \cos(s)\, dB_s + 2\int_0^\pi \sin(s)\, dB_s\).

Since both integrands are deterministic, the Ito isometry gives:

\[ \operatorname{Var}\!\left(\int_0^\pi (3\cos(s) + 2\sin(s))\, dB_s\right) = \int_0^\pi (3\cos(s) + 2\sin(s))^2\, ds \]

Expanding the square:

\[ = \int_0^\pi \left(9\cos^2(s) + 12\cos(s)\sin(s) + 4\sin^2(s)\right)\, ds \]

Evaluate each integral:

\(\int_0^\pi \cos^2(s)\, ds = \pi/2\)
\(\int_0^\pi \sin^2(s)\, ds = \pi/2\)
\(\int_0^\pi \cos(s)\sin(s)\, ds = \int_0^\pi \frac{1}{2}\sin(2s)\, ds = \left[-\frac{1}{4}\cos(2s)\right]_0^\pi = -\frac{1}{4}(1 - 1) = 0\)

Therefore:

\[ \operatorname{Var} = 9 \cdot \frac{\pi}{2} + 12 \cdot 0 + 4 \cdot \frac{\pi}{2} = \frac{13\pi}{2} \]

Exercise 2. Show directly from the martingale property that \(\mathbb{E}[I_t \cdot I_s] = \mathbb{E}[I_s^2]\) for \(s \le t\), where \(I_t = \int_0^t H_u\, dB_u\). Hint: Write \(I_t = I_s + (I_t - I_s)\) and use the fact that \(I_t - I_s\) is independent of \(\mathcal{F}_s\).

Solution to Exercise 2

Write \(I_t = I_s + (I_t - I_s)\) where \(I_t - I_s = \int_s^t H_u\, dB_u\). Then:

\[ \mathbb{E}[I_t \cdot I_s] = \mathbb{E}[(I_s + (I_t - I_s)) \cdot I_s] = \mathbb{E}[I_s^2] + \mathbb{E}[(I_t - I_s) \cdot I_s] \]

For the cross term, note that \(I_t - I_s = \int_s^t H_u\, dB_u\) depends only on Brownian increments after time \(s\), while \(I_s = \int_0^s H_u\, dB_u\) is \(\mathcal{F}_s\)-measurable. By the martingale property:

\[ \mathbb{E}[(I_t - I_s) \cdot I_s] = \mathbb{E}\!\left[\mathbb{E}[(I_t - I_s) \cdot I_s \mid \mathcal{F}_s]\right] = \mathbb{E}\!\left[I_s \cdot \mathbb{E}[I_t - I_s \mid \mathcal{F}_s]\right] \]

Since \(I_t\) is a martingale, \(\mathbb{E}[I_t - I_s \mid \mathcal{F}_s] = 0\). Therefore:

\[ \mathbb{E}[(I_t - I_s) \cdot I_s] = 0 \]

and so \(\mathbb{E}[I_t \cdot I_s] = \mathbb{E}[I_s^2]\).

Exercise 3. Let \(I_t = \int_0^t B_s\, dB_s = \frac{1}{2}(B_t^2 - t)\). Verify each of the six fundamental properties (linearity, zero mean, martingale, Ito isometry, continuity, quadratic variation) directly for this specific integral.

Solution to Exercise 3

Let \(I_t = \int_0^t B_s\, dB_s = \frac{1}{2}(B_t^2 - t)\).

Linearity: This is a single integral, so linearity is trivially satisfied.

Zero mean: \(\mathbb{E}[I_t] = \frac{1}{2}(\mathbb{E}[B_t^2] - t) = \frac{1}{2}(t - t) = 0\). ✓

Martingale: For \(s \le t\):

\[ \mathbb{E}[I_t \mid \mathcal{F}_s] = \frac{1}{2}\mathbb{E}[B_t^2 - t \mid \mathcal{F}_s] \]

Write \(B_t = B_s + (B_t - B_s)\), so \(B_t^2 = B_s^2 + 2B_s(B_t - B_s) + (B_t - B_s)^2\). Taking conditional expectation:

\[ \mathbb{E}[B_t^2 \mid \mathcal{F}_s] = B_s^2 + 0 + (t - s) \]

Therefore \(\mathbb{E}[I_t \mid \mathcal{F}_s] = \frac{1}{2}(B_s^2 + t - s - t) = \frac{1}{2}(B_s^2 - s) = I_s\). ✓

Ito isometry: \(\mathbb{E}[I_t^2] = \mathbb{E}[(B_t^2 - t)^2/4]\). Since \(\mathbb{E}[B_t^4] = 3t^2\):

\[ \mathbb{E}[(B_t^2 - t)^2] = \mathbb{E}[B_t^4] - 2t\,\mathbb{E}[B_t^2] + t^2 = 3t^2 - 2t^2 + t^2 = 2t^2 \]

So \(\mathbb{E}[I_t^2] = t^2/2\). The right side of the isometry is \(\mathbb{E}[\int_0^t B_s^2\, ds] = \int_0^t s\, ds = t^2/2\). ✓

Continuity: \(I_t = \frac{1}{2}(B_t^2 - t)\) is continuous since \(B_t\) has continuous paths. ✓

Quadratic variation: \([I,I]_t = \int_0^t B_s^2\, ds\). This can be verified by noting that \(I_t^2 - [I,I]_t = (B_t^2 - t)^2/4 - \int_0^t B_s^2\, ds\) should be a martingale, which follows from the Doob-Meyer decomposition. ✓

Exercise 4. Using the polarization identity

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

compute \(\mathbb{E}[I_t \cdot J_t]\) where \(I_t = \int_0^t s\, dB_s\) and \(J_t = \int_0^t s^2\, dB_s\).

Solution to Exercise 4

Using the polarization identity with \(H_s = s\) and \(K_s = s^2\):

\[ \mathbb{E}[I_t \cdot J_t] = \mathbb{E}\!\left[\int_0^t s\, dB_s \cdot \int_0^t s^2\, dB_s\right] = \mathbb{E}\!\left[\int_0^t s \cdot s^2\, ds\right] = \int_0^t s^3\, ds = \frac{t^4}{4} \]

(The expectation can be moved inside the ordinary integral because the integrand is deterministic.)

Exercise 5. Let \(I_t = \int_0^t \sigma(s)\, dB_s\) for a deterministic function \(\sigma(s)\). Show that \(I_t\) is Gaussian and compute its distribution. Then verify that \(I_t^2 - \int_0^t \sigma(s)^2\, ds\) is a martingale by computing its expectation.

Solution to Exercise 5

Since \(\sigma(s)\) is deterministic, \(I_t = \int_0^t \sigma(s)\, dB_s\) is the \(L^2\)-limit of sums \(\sum_k \sigma(t_k)(B_{t_{k+1}} - B_{t_k})\). Each such sum is a finite linear combination of independent Gaussian random variables (the Brownian increments), and hence Gaussian. The \(L^2\)-limit of Gaussian random variables is Gaussian.

Distribution: \(\mathbb{E}[I_t] = 0\) (martingale property) and by the Ito isometry:

\[ \operatorname{Var}(I_t) = \int_0^t \sigma(s)^2\, ds \]

So \(I_t \sim \mathcal{N}\!\left(0,\; \int_0^t \sigma(s)^2\, ds\right)\).

Martingale verification for \(I_t^2 - \int_0^t \sigma(s)^2\, ds\): We compute:

\[ \mathbb{E}\!\left[I_t^2 - \int_0^t \sigma(s)^2\, ds\right] = \operatorname{Var}(I_t) - \int_0^t \sigma(s)^2\, ds = 0 \]

for all \(t\). This is consistent with \(I_t^2 - \int_0^t \sigma(s)^2\, ds\) being a martingale (constant expectation equal to zero).

Exercise 6. The quadratic variation of \(I_t = \int_0^t H_s\, dB_s\) is \([I,I]_t = \int_0^t H_s^2\, ds\). Consider the Ito process \(X_t = \mu t + \int_0^t \sigma_s\, dB_s\). Show that \([X,X]_t = \int_0^t \sigma_s^2\, ds\), i.e., the drift contributes nothing to the quadratic variation.

Solution to Exercise 6

The Ito process \(X_t = \mu t + \int_0^t \sigma_s\, dB_s\) has increments:

\[ X_{t_{i+1}} - X_{t_i} = \mu(t_{i+1} - t_i) + \int_{t_i}^{t_{i+1}} \sigma_s\, dB_s \]

The quadratic variation sum is:

\[ \sum_i (X_{t_{i+1}} - X_{t_i})^2 = \sum_i \left(\mu \Delta t_i + \int_{t_i}^{t_{i+1}} \sigma_s\, dB_s\right)^2 \]

Expanding:

\[ = \mu^2 \sum_i (\Delta t_i)^2 + 2\mu \sum_i \Delta t_i \int_{t_i}^{t_{i+1}} \sigma_s\, dB_s + \sum_i \left(\int_{t_i}^{t_{i+1}} \sigma_s\, dB_s\right)^2 \]

As the mesh tends to zero, the first term vanishes (same argument as for smooth functions: sum of \((\Delta t)^2\) goes to zero). The cross term also vanishes. The third term converges to \(\int_0^t \sigma_s^2\, ds\) (this is the quadratic variation of the Ito integral).

Therefore \([X,X]_t = \int_0^t \sigma_s^2\, ds\), and the drift \(\mu t\) contributes nothing to the quadratic variation because its increments \(\mu \Delta t_i\) are of order \(\Delta t_i\), making squared increments of order \((\Delta t_i)^2\), which sum to zero.

Exercise 7. The martingale representation theorem states that every square-integrable martingale in the Brownian filtration is an Ito integral. The process \(M_t = B_t^3 - 3tB_t\) is a martingale. Find the integrand \(H_s\) such that \(M_t = \int_0^t H_s\, dB_s\). Hint: Apply Ito's formula to \(f(t, x) = x^3 - 3tx\).

Solution to Exercise 7

Apply Ito's formula to \(f(t, x) = x^3 - 3tx\). The partial derivatives are:

\[ f_t = -3x, \qquad f_x = 3x^2 - 3t, \qquad f_{xx} = 6x \]

With \(X_t = B_t\) (so \(dX_t = dB_t\), \(\mu_t = 0\), \(\sigma_t = 1\)):

\[ dM_t = f_t\, dt + f_x\, dB_t + \frac{1}{2}f_{xx}\, dt \]

\[ = (-3B_t)\, dt + (3B_t^2 - 3t)\, dB_t + \frac{1}{2}(6B_t)\, dt \]

\[ = -3B_t\, dt + (3B_t^2 - 3t)\, dB_t + 3B_t\, dt \]

\[ = (3B_t^2 - 3t)\, dB_t \]

The drift terms cancel, confirming that \(M_t\) is a martingale. Integrating:

\[ M_t = \int_0^t (3B_s^2 - 3s)\, dB_s = 3\int_0^t (B_s^2 - s)\, dB_s \]

Therefore \(H_s = 3(B_s^2 - s) = 3B_s^2 - 3s\).