Properties of the Itô Integral¶

1. Concept Definition¶

The left-endpoint sampling rule \(\sum H_{t_k}(B_{t_{k+1}}-B_{t_k})\) is more than a technical choice -- it freezes \(H_{t_k}\) before the increment \(\Delta B\) arrives, making each summand a fair conditional bet. From this single feature, four properties of \(I_t = \int_0^t H_s\,dB_s\) follow automatically: the sum of fair bets is a martingale (zero mean, conditional expectation preserved), the variance bookkeeping collapses to the Itô isometry, the \(L^2\)-Cauchy structure used in construction gives continuous sample paths, and quadratic variation reduces to \(\int_0^t H_s^2\,ds\). These four facts are what make the Itô integral a usable object of calculus.

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 \(\{\mathcal{F}_t\}\).

Recall (see § Construction of the Itô Integral): the martingale identity \(\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s\) is verified directly for simple processes using independence and mean-zero of future Brownian increments, then extended to general \(H \in \mathcal{L}^2([0,T])\) by the \(L^2\)-continuity of conditional expectation applied to an approximating sequence \(H^{(n)} \to H\). Adaptedness follows because \(I_t\) is the \(L^2\)-limit of \(\mathcal{F}_t\)-measurable random variables, and square-integrability follows from the Itô isometry: \(\mathbb{E}[I_t^2] \le \mathbb{E}[\int_0^T H_s^2\,ds] < \infty\).

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:

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

Recall (see § Quadratic Variation): the identity is the canonical Itô-integral quadratic-variation formula. Its proof reduces partition sums \(\sum_i (I_{t_{i+1}} - I_{t_i})^2 \approx \sum_i H_{t_i}^2 (\Delta B_i)^2\) to \(\int_0^t H_s^2\,ds\) via \((\Delta B)^2 \sim \Delta t\).

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.

Recall (see § Itô Isometry): the isometry is the Hilbert-space identity that makes the \(L^2\)-extension of the integral well-defined. The polarized form

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

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

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 — Hₛ = σ¶

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 — Hₛ = 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 — Hₛ = Bₛ¶

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

Recall (see § Itô's Lemma): this closed form follows by applying Itô's formula 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 \(H_s = 3(B_s^2 - s) = 3B_s^2 - 3s\).