Itô's Lemma: Derivation and Applications

Setup

We work on $(\Omega, \mathcal{F}, \mathbb{P})$ with filtration $\mathbb{F} = (\mathcal{F}_t)_{t \geq 0}$ satisfying the usual conditions. Let $(W_t)_{t \geq 0}$ be a standard $\mathbb{P}$-Brownian motion.

An Itô process is a continuous adapted process of the form: $X_t = X_0 + \int_0^t \mu_s \, ds + \int_0^t \sigma_s \, dW_s,$ where $\mu$ (drift) and $\sigma$ (diffusion) are $\mathbb{F}$-progressively measurable and satisfy: $\int_0^T |\mu_s| \, ds < \infty \quad \text{and} \quad \int_0^T \sigma_s^2 \, ds < \infty \quad \text{a.s.}$

In differential notation: $dX_t = \mu_t \, dt + \sigma_t \, dW_t$.

The Itô Integral

The stochastic integral $\int_0^T \sigma_s \, dW_s$ is defined as the $L^2$ limit of non-anticipating Riemann sums: $\int_0^T \sigma_s \, dW_s = L^2\text{-}\!\lim_{|\pi| \to 0} \sum_{i} \sigma_{t_{i-1}} \bigl(W_{t_i} - W_{t_{i-1}}\bigr).$

The requirement to evaluate $\sigma$ at the left endpoint $t_{i-1}$ — not the midpoint — is what makes the integral Itô. Left-endpoint evaluation ensures $\sigma_{t_{i-1}}$ is $\mathcal{F}_{t_{i-1}}$-measurable (non-anticipating). The choice of evaluation point matters: different conventions yield different integrals (see Itô vs Stratonovich below).

Itô isometry. For square-integrable adapted $\sigma$: $\mathbb{E}\!\left[\left(\int_0^T \sigma_s \, dW_s\right)^2\right] = \mathbb{E}\!\left[\int_0^T \sigma_s^2 \, ds\right].$

This is the key $L^2$ norm identity. It follows from expanding the square and using independence of non-overlapping Brownian increments: cross terms $\mathbb{E}[\sigma_{t_{i-1}}(W_{t_i} - W_{t_{i-1}}) \cdot \sigma_{t_{j-1}}(W_{t_j} - W_{t_{j-1}})] = 0$ for $i \neq j$.

A square-integrable adapted integrand produces a martingale: $\mathbb{E}\!\left[\int_0^t \sigma_s \, dW_s \,\Big|\, \mathcal{F}_r\right] = \int_0^r \sigma_s \, dW_s, \quad r \leq t.$

Itô's Lemma: Statement

Let $f: \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}$ be $C^{1,2}$ (once continuously differentiable in $t$, twice in $x$). Define $Y_t = f(t, X_t)$. Then $Y$ is again an Itô process and: $dY_t = \frac{\partial f}{\partial t}(t, X_t) \, dt + \frac{\partial f}{\partial x}(t, X_t) \, dX_t + \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(t, X_t) \, (dX_t)^2,$ where $(dX_t)^2$ is evaluated using the Itô multiplication table: $dt \cdot dt = 0, \qquad dt \cdot dW_t = 0, \qquad dW_t \cdot dW_t = dt.$

Substituting $dX_t = \mu_t \, dt + \sigma_t \, dW_t$ and $(dX_t)^2 = \sigma_t^2 \, dt$: $\boxed{dY_t = \left(\frac{\partial f}{\partial t} + \mu_t \frac{\partial f}{\partial x} + \frac{1}{2}\sigma_t^2 \frac{\partial^2 f}{\partial x^2}\right) dt + \sigma_t \frac{\partial f}{\partial x} \, dW_t.}$

The term $\frac{1}{2}\sigma_t^2 f_{xx}$ is the Itô correction. It has no analogue in classical calculus and arises solely from the non-zero quadratic variation of Brownian motion.

Derivation

Apply Taylor's theorem to $f(t + dt, X_{t+dt})$ around $(t, X_t)$: $df = f_t \, dt + f_x \, dX + \frac{1}{2}f_{xx}(dX)^2 + \frac{1}{2}f_{tt}(dt)^2 + f_{xt} \, dt \, dX + \cdots$

Substitute $dX = \mu \, dt + \sigma \, dW$ and expand $(dX)^2$: $(dX)^2 = \mu^2(dt)^2 + 2\mu\sigma \, dt \, dW + \sigma^2(dW)^2.$

Apply the Itô multiplication table. The key step: $(dW)^2 = dt$ is the quadratic variation result $[W]_t = t$ in differential form. The other products vanish as $o(dt)$: $(dX)^2 = \sigma^2 \, dt + O\!\left((dt)^{3/2}\right).$

Similarly, $(dt)^2 = 0$ and $dt \cdot dW = 0$ in the $L^2$ sense. Retaining only terms of order $dt$: $df = f_t \, dt + f_x(\mu \, dt + \sigma \, dW) + \frac{1}{2}f_{xx}\sigma^2 \, dt.$

Collecting drift and diffusion terms: $df = \left(f_t + \mu f_x + \frac{1}{2}\sigma^2 f_{xx}\right) dt + \sigma f_x \, dW.$

This is Itô's lemma. The heuristic argument is exact in identifying the correct terms; the rigorous version replaces the Taylor remainder analysis with an $L^2$ convergence argument using the Itô isometry.

Application: Solving the GBM SDE

Let $S_t$ satisfy the geometric Brownian motion SDE: $dS_t = \mu S_t \, dt + \sigma S_t \, dW_t.$

Apply Itô's lemma to $f(S_t) = \ln S_t$, with $f_x = 1/x$, $f_{xx} = -1/x^2$, $f_t = 0$: $d(\ln S_t) = \frac{1}{S_t} \, dS_t - \frac{1}{2} \cdot \frac{1}{S_t^2} \cdot \sigma^2 S_t^2 \, dt = \left(\mu - \frac{\sigma^2}{2}\right) dt + \sigma \, dW_t.$

Integrating from 0 to $T$: $\ln S_T - \ln S_0 = \left(\mu - \frac{\sigma^2}{2}\right)T + \sigma W_T.$

Therefore: $S_T = S_0 \exp\!\left(\left(\mu - \frac{\sigma^2}{2}\right)T + \sigma W_T\right).$

The Role of the Itô Correction

The term $-\sigma^2/2$ is not a typo. It is the Itô correction from $f_{xx} = -1/S^2$. Without it — if one naively wrote $\ln S_T = \ln S_0 + \mu T + \sigma W_T$ — the expectation would be wrong: $\mathbb{E}[S_T] = S_0 e^{\mu T}, \quad \text{correct from the SDE:} \quad d\mathbb{E}[S_t] = \mu \mathbb{E}[S_t] \, dt.$

But $\mathbb{E}[e^{(\mu + \sigma^2/2)T + \sigma W_T}] = e^{(\mu + \sigma^2/2)T} \cdot e^{\sigma^2 T/2} = e^{(\mu + \sigma^2)T} \neq e^{\mu T}$. The Itô correction $-\sigma^2/2$ is what reconciles the arithmetic mean growth $\mu$ of the SDE with the geometric mean growth $\mu - \sigma^2/2$ of the log process.

Itô vs Stratonovich

The Stratonovich integral evaluates the integrand at the midpoint: $\int_0^T \sigma_s \circ dW_s = L^2\text{-}\!\lim_{|\pi| \to 0} \sum_i \frac{\sigma_{t_i} + \sigma_{t_{i-1}}}{2}\bigl(W_{t_i} - W_{t_{i-1}}\bigr).$