Itô Calculus

Setting

$L^2(\Omega)$ is a Hilbert space with $\langle X, Y \rangle = E[XY]$. A process $f(t, \omega)$ on $[0, \infty) \times \Omega$ is progressively measurable if for every $t \ge 0$ the map $(s, \omega) \mapsto f(s, \omega)$ on $[0, t] \times \Omega$ is $\mathcal{B}([0,t]) \otimes \mathcal{F}_t$-measurable. Adapted processes with left- or right-continuous paths are progressively measurable. $L^2_{prog}([0,T] \times \Omega)$ is the space of progressively measurable $f$ with $E\int_0^T f^2\, dt < \infty$.

The Itô Integral

For a simple adapted process, constant $\Delta(t_j)$ on each $[t_j, t_{j+1})$, and $t_k \le t < t_{k+1}$,

$$I(t) = \int_0^t \Delta(u)\, dW(u) = \sum_{j=0}^{k-1} \Delta(t_j)\big[W(t_{j+1}) - W(t_j)\big] + \Delta(t_k)\big[W(t) - W(t_k)\big]$$

For general $\Delta \in L^2_{prog}$, choose simple $\Delta_n$ with $\lim_{n \to \infty} E\int_0^T |\Delta_n(t) - \Delta(t)|^2\, dt = 0$ (simple, then bounded continuous, then bounded, then general), and define $\int_0^T \Delta\, dW$ as the $L^2$ limit of $\int_0^T \Delta_n\, dW$.

Properties

• $I(t)$ has continuous paths and is $\mathcal{F}_t$-measurable.
• It is linear in the integrand.
• It is a martingale, $E[I(t)|\mathcal{F}_s] = I(s)$, so $E[I(t)] = 0$.
• Itô isometry $E[I(t)^2] = E\int_0^t \Delta^2(u)\, du = \mathrm{Var}(I(t))$, i.e. the integral is an isometry from $L^2_{prog}$ into $L^2(\Omega)$.
• $[I, I](t) = \int_0^t \Delta^2(u)\, du$, i.e. $dI\, dI = \Delta^2\, dt$, and $[I, J](t) = \int_0^t \Delta\Gamma\, du$ for $J = \int \Gamma\, dW$.
• If $\Delta$ is deterministic, $\int_0^t \Delta\, dW \sim \mathcal{N}\big(0, \int_0^t \Delta^2(s)\, ds\big)$.

Itô Processes

$$X(t) = X(0) + \int_0^t \Delta(u)\, dW(u) + \int_0^t \Theta(u)\, du, \qquad dX = \underbrace{\Delta\, dW}_{\text{diffusion}} + \underbrace{\Theta\, dt}_{\text{drift}}$$

with $\Delta, \Theta$ adapted. Its quadratic variation is $[X, X](t) = \int_0^t \Delta^2(u)\, du$, i.e. $dX\, dX = \Delta^2\, dt$, since $dW\, dW = dt$ and $dW\, dt = dt\, dt = 0$.

Itô's Lemma

If $f(t, x)$ has continuous $f_t$, $f_x$, $f_{xx}$, then

$$f(T, X(T)) = f(0, X(0)) + \int_0^T f_t\, dt + \int_0^T f_x\, dX + \frac{1}{2}\int_0^T f_{xx}\, d[X, X](t)$$

or in differential form

$$df = f_t\, dt + f_x\, dX + \frac{1}{2} f_{xx}\, dX\, dX = \Big(f_t + \Theta f_x + \frac{1}{2}\Delta^2 f_{xx}\Big)\, dt + \Delta f_x\, dW$$

For $X = W$ this is $df = f_t\, dt + f_x\, dW + \frac{1}{2} f_{xx}\, dt$.

Multidimensional $df = f_t\, dt + \nabla f \cdot dX + \frac{1}{2}\, dX^T H_f\, dX$, with $H_f$ the Hessian in $x$.

Product rule $d(XY) = X\, dY + Y\, dX + dX\, dY$.

Examples

$\int_0^T W\, dW$ Approximate $W$ by the simple process $\Delta_n(t) = W(jT/n)$ on $[jT/n, (j+1)T/n)$. Writing $W_j = W(jT/n)$,

$$\sum_{j=0}^{n-1} W_j (W_{j+1} - W_j) = \frac{1}{2}W_n^2 - \frac{1}{2}\sum_{j=0}^{n-1} (W_{j+1} - W_j)^2 \;\longrightarrow\; \frac{1}{2}W(T)^2 - \frac{1}{2}T$$

since the sum of squared increments converges to $[W, W](T) = T$.

Others

$$\int_0^T t\, dW = TW(T) - \int_0^T W(t)\, dt, \qquad \int_0^T W^2\, dW = \frac{1}{3}W(T)^3 - \int_0^T W(t)\, dt$$

from the product rule and from Itô's lemma with $f(x) = x^3/3$.

Geometric Brownian motion For $dS = \mu S\, dt + \sigma S\, dW$ and $f(x) = \ln x$,

$$d(\log S) = \frac{dS}{S} - \frac{1}{2S^2} dS\, dS = \mu\, dt + \sigma\, dW - \frac{1}{2}\sigma^2\, dt = \Big(\mu - \frac{1}{2}\sigma^2\Big) dt + \sigma\, dW$$

Ornstein-Uhlenbeck For $dY = -aY\, dt + \sigma\, dW$, let $Z = Ye^{at}$. Then $dZ = \sigma e^{at}\, dW$, so

$$Y(t) = e^{-at}Y(0) + \sigma\int_0^t e^{-a(t-s)}\, dW(s)$$