The Ito integral and isometry

Definition Fix a filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t \ge 0}, \mathbb{P})$, a horizon $T > 0$ and a standard Brownian motion $W$ with respect to $(\mathcal{F}_t)$. A simple predictable integrand on $[0, T]$ is a process

with a partition $0 = t_0 < t_1 < \dots < t_n = T$ and coefficients such that each $H_k$ is $\mathcal{F}_{t_k}$-measurable and $\mathbb{E}[H_k^2] < \infty$. Its stochastic integral is defined to be the finite sum

and the integral up to a time $t \le T$ is the same sum with every partition time replaced by its minimum with $t$. Each product $H_k \Delta W_k$ is square integrable, since $\Delta W_k$ is independent of $\mathcal{F}_{t_k}$ and so $\mathbb{E}[H_k^2 \Delta W_k^2] = \mathbb{E}[H_k^2]\,(t_{k+1} - t_k)$. Each product also has mean zero, because the known factor comes out of the conditional expectation,

Definition For a simple predictable integrand $H$ on $[0, T]$ with square-integrable coefficients,

The proof expands the square of the sum. A cross term with $j < k$ has mean zero, because conditioning on $\mathcal{F}_{t_k}$ fixes $H_j \Delta W_j H_k$ and leaves the factor $\mathbb{E}[\Delta W_k \mid \mathcal{F}_{t_k}] = 0$, and the square integrability of every product $H_i \Delta W_i$ makes these cross terms integrable. A diagonal term gives

and these terms add up to $\mathbb{E}\int_0^T H_t^2 \, dt$. The name says that the map from $H$ to its integral preserves norms, the norm $\bigl(\mathbb{E}\int_0^T H_t^2 \, dt\bigr)^{1/2}$ of integrands and the norm $\bigl(\mathbb{E}[Y^2]\bigr)^{1/2}$ of random variables $Y$. Applied to a difference $H - K$, the isometry turns closeness of two integrands into closeness of their integrals, and after the extension in the next two cards it holds for every square-integrable integrand. Applying it to $H + K$ and $H - K$ and subtracting gives the covariance form

For integrals against two Brownian motions $W$ and $B$ of the same filtration with correlation $\rho$, meaning $[W, B]_t = \rho t$, the right side gets the factor $\rho$. Since the integral has mean zero, its second moment is its variance.

Definition A process $H$ on $[0, T]$ is a square-integrable integrand when it is predictable, meaning measurable with respect to the predictable sigma algebra on $[0, T] \times \Omega$, and

Two such processes $H$ and $K$ are identified when $\lVert H - K \rVert = 0$. Every continuous adapted process is predictable, so for example $H_t = W_t$ qualifies, with $\lVert W \rVert^2 = \int_0^T t \, dt = T^2/2$ by Fubini’s theorem. The simple predictable integrands with square-integrable coefficients belong to this class and are dense in it, meaning that for each square-integrable integrand $H$ there are simple ones $H^{(n)}$ with $\lVert H^{(n)} - H \rVert \to 0$. This density is an approximation result stated here without proof. The weaker condition that $\int_0^T H_t^2 \, dt < \infty$ almost surely holds for every continuous adapted process. It is enough for the local construction described in the next card, but not for the isometry.

Definition Let $H$ be a square-integrable integrand on $[0, T]$, write $I(K)$ for the integral over $[0, T]$ of a simple predictable integrand $K$, and choose simple predictable integrands $H^{(n)}$ with $\lVert H^{(n)} - H \rVert \to 0$. The isometry applied to $H^{(n)} - H^{(m)}$ gives

so the integrals form a Cauchy sequence in $L^2(\mathbb{P})$. That space is complete, so the sequence has a mean-square limit, and the Itô integral is defined as

with the limit taken in $L^2(\mathbb{P})$. Two approximating sequences give the same limit, because interleaving them gives a third approximating sequence whose integrals must also converge, so the definition does not depend on the choice. The integral is linear in $H$, has mean zero and satisfies the isometry for every square-integrable $H$, and the integral up to $t \le T$ is the integral of $H \mathbf{1}_{[0, t]}$. Its values are random variables defined up to sets of probability zero, and the process $t \mapsto \int_0^t H_s \, dW_s$ has a version with continuous paths, a result stated here. If $H$ is only predictable with $\int_0^T H_t^2 \, dt < \infty$ almost surely, the integral is defined by localization. With the stopping times $\tau_n = \inf\{t \ge 0 : \int_0^t H_s^2 \, ds \ge n\} \wedge T$, each stopped integrand $H \mathbf{1}_{[0, \tau_n]}$ is square integrable, the stopped integrals agree where they overlap, and they join into one process. That process is a local martingale, and the expectation identities of the square-integrable construction can fail for it. General localization is beyond this lesson.

Definition Let $H$ be a square-integrable integrand on $[0, T]$ and $M_t = \int_0^t H_s \, dW_s$ for $0 \le t \le T$. Then $M$ is a square-integrable martingale with respect to $(\mathcal{F}_t)$ under $\mathbb{P}$, with $M_0 = 0$, so

and in particular $\mathbb{E}[M_t] = 0$. For a simple integrand, refine the partition so that $s$ and $t$ are partition times. Then $M_t - M_s$ is a sum of terms $H_k \Delta W_k$ with $t_k \ge s$, and the tower rule gives

For a general $H$ the property passes to the limit, because $M^{(n)}_t \to M_t$ in $L^2(\mathbb{P})$ for each $t$ and conditional expectation does not increase mean-square distances. The same argument gives the conditional isometry

If $H$ is deterministic with $\int_0^T H_t^2 \, dt < \infty$, then $M_t$ is normal with mean $0$ and variance $\int_0^t H_s^2 \, ds$, since it is a mean-square limit of sums of independent normal variables. A random predictable $H$ keeps the martingale property but in general loses the normal law. When only $\int_0^T H_t^2 \, dt < \infty$ almost surely, $M$ is in general only a local martingale.

Definition The Itô integral evaluates the integrand at the left endpoint of each interval. For a continuous adapted process $H$, the sums $\sum_k H_{t_k} (W_{t_{k+1}} - W_{t_k})$ over partitions of $[0, t]$ converge in probability to $\int_0^t H_s \, dW_s$ as the mesh tends to zero, a result stated here. For $H = W$ the limit can be computed. Each increment of the square satisfies

and summing over a partition of $[0, t]$ telescopes, since $W_0 = 0$, to

The quadratic sum converges to $t$ in mean square, because the squared increments are independent with $\operatorname{Var}\bigl((\Delta W_k)^2\bigr) = 2 (t_{k+1} - t_k)^2$, so its mean-square distance from $t$ is $2 \sum_k (t_{k+1} - t_k)^2$, at most twice $t$ times the mesh. Hence

Right-endpoint sums $\sum_k W_{t_{k+1}} \Delta W_k$ exceed the left-endpoint sums by exactly $\sum_k (\Delta W_k)^2$, so they converge to $\tfrac12 (W_t^2 + t)$, and the gap does not shrink as the grid is refined. Sums that use the average $\tfrac12 (W_{t_k} + W_{t_{k+1}})$ telescope to exactly $\tfrac12 W_t^2$ for every partition, the value of the Stratonovich integral, whose theory is beyond this lesson.