Stochastic Calculus Notes 04 | Itô's Formula, Itô Processes and Geometric Brownian Motion

21 Oct 2026 • Ziyuan Zhang

Overview

Itô’s formula is the chain rule of stochastic calculus. Compared with the ordinary chain rule it has one extra term, \(\frac12 f''\,d[W,W]_t=\frac12f''\,dt\), and the reason is exactly \([W,W]_t=t>0\) from Part 02.

In practice you only need a second-order Taylor expansion plus the multiplication table \(dt^2=0,\quad dt\,dW_t=0,\quad (dW_t)^2=dt.\) This post derives the formula, extends it to Itô processes, and uses it to build the standard stock-price model, geometric Brownian motion.

1. Why the ordinary chain rule fails

For a smooth path \(g\), \(f(g_t)-f(g_0)=\int_0^tf'(g_u)\,g'_u\,du\). The naive stochastic analogue would be \(f(W_t)=f(W_0)+\int_0^tf'(W_u)\,dW_u\). Test it with \(f(x)=\frac12x^2\), which would predict \(\frac12W_t^2=\int_0^tW_u\,dW_u\). Compute directly with a partition of \([0,t]\):

\[\frac12W_t^2=\frac12\sum_j\big(W_{t_j}^2-W_{t_{j-1}}^2\big)=\sum_jW_{t_{j-1}}(W_{t_j}-W_{t_{j-1}})+\frac12\sum_j(W_{t_j}-W_{t_{j-1}})^2\ \xrightarrow{\Vert \Pi\Vert \to0}\ \int_0^tW_u\,dW_u+\frac12t.\]

(Check the algebra: \(W_{t_{j-1}}W_{t_j}-W_{t_{j-1}}^2+\frac12(W_{t_j}-W_{t_{j-1}})^2=\frac12W_{t_j}^2-\frac12W_{t_{j-1}}^2\).)

The extra \(\frac12t\) is the quadratic variation. So the chain rule needs a second-order correction.

2. Itô’s formula

For \(W\) a Brownian motion and \(f\) smooth:

\[f(W_t)-f(W_0)=\int_0^tf'(W_u)\,dW_u+\frac12\int_0^tf''(W_u)\,d[W,W]_u .\]

Shorthand: \(df(W_t)=f'(W_t)\,dW_t+\frac12f''(W_t)\,dt\). Note that “\(df(W_t)\)” by itself has no meaning — it is shorthand for the integral equation.

Heuristic derivation. Taylor to second order, \(f(x+y)-f(x)=f'(x)y+\frac12f''(x)y^2\), with \(x=W_t\), \(y=dW_t\) and \((dW_t)^2=d[W,W]_t\):

\[df(W_t)=f'(W_t)\,dW_t+\frac12f''(W_t)\,(dW_t)^2=f'(W_t)\,dW_t+\frac12f''(W_t)\,d[W,W]_t .\]

This is not rigorous — a two-term Taylor expansion is not exact, and plugging \(dW_t\) into it is formal — but it gives the right answer, and it is how everyone uses the formula.

Proof sketch (with the Taylor remainder made explicit)

Assume \(f,f',f''\) are continuous and bounded. Write the exact Taylor expansion with a remainder,

\[f(x+y)=f(x)+f'(x)y+\tfrac12f''(x)y^2+R(x,y)\,y^2,\qquad R(x,y)=\frac{f(x+y)-f(x)-f'(x)y-\frac12f''(x)y^2}{y^2}.\]

\(R\) is the “error per unit of \(y^2\)”. By L’Hôpital twice, \(\lim_{y\to0}R(x,y)=\lim_{y\to0}\frac12\big(f''(x+y)-f''(x)\big)=0\), i.e. the remainder is of smaller order than \(y^2\).

Now telescope over a partition, with \(x=W_{t_{j-1}}\), \(x+y=W_{t_j}\):

\[f(W_t)-f(W_0)=\underbrace{\sum_jf'(W_{t_{j-1}})\Delta W_j}_{(1)}+\underbrace{\tfrac12\sum_jf''(W_{t_{j-1}})(\Delta W_j)^2}_{(2)}+\underbrace{\sum_jR(W_{t_{j-1}},\Delta W_j)(\Delta W_j)^2}_{(3)}.\]

(1) \(\to\int_0^tf'(W_u)\,dW_u\) (a stochastic integral, Part 03).
(2) Split \((\Delta W_j)^2=\Delta t_j+\big((\Delta W_j)^2-\Delta t_j\big)\). The first part gives \(\int_0^tf''(W_u)\,du\). The second, \(\mathcal Z=\sum_jf''(W_{t_{j-1}})\big((\Delta W_j)^2-\Delta t_j\big)\), has mean \(0\) (because \(W_t^2-t\) is a martingale) and variance \(\le K(f,t)\,\Vert \Pi\Vert \to0\), so \(\mathcal Z\to0\).
(3) \(\vert(3)\vert\le\max_j\vert R(W_{t_{j-1}},\Delta W_j)\vert\cdot[W,W]^\Pi_t\to0\cdot t=0\).

Two examples.

\(f(x)=e^x\): \(e^{W_t}=1+\int_0^te^{W_u}\,dW_u+\frac12\int_0^te^{W_u}\,du\).
\(f(x)=\sin x\): \(\sin W_t=\int_0^t\cos W_u\,dW_u-\frac12\int_0^t\sin W_u\,du\). The \(du\) term changes sign, so \(\sin W_t\) is neither a sub-, super-, nor martingale.

Rule of thumb: the \(dW\) part is a martingale; the \(dt\) part is a drift. A function of \(W\) is a martingale exactly when its \(dt\) term vanishes.

3. Time-dependent Itô

If \(f(t,x)\) has continuous \(f_t,f_x,f_{xx}\):

\[df(t,W_t)=f_t(t,W_t)\,dt+f_x(t,W_t)\,dW_t+\frac12f_{xx}(t,W_t)\,d[W,W]_t .\]

Heuristic. Two-variable Taylor,

\[f(t+s,x+y)-f(t,x)=f_ts+f_xy+\tfrac12f_{tt}s^2+f_{tx}sy+\tfrac12f_{xx}y^2,\]

with \(s=dt\), \(y=dW_t\), and the multiplication table \(dt^2=0\), \(dt\,dW_t=0\), \((dW_t)^2=dt\). Why do the two mixed terms die? As \(\Vert \Pi\Vert \to0\):

\[\sum_i(t_i-t_{i-1})^2\le\max_i(t_i-t_{i-1})\cdot t\to0,\qquad \sum_i(t_i-t_{i-1})\vert W_{t_i}-W_{t_{i-1}}\vert\le\max_i\vert W_{t_i}-W_{t_{i-1}}\vert\cdot t\to0.\]

The exponential martingale (important)

\(f(t,x)=e^{\alpha x-\frac12\alpha^2t}\) has \(f_t=-\frac12\alpha^2f\), \(f_x=\alpha f\), \(f_{xx}=\alpha^2f\), so the \(dt\) terms cancel:

\[df(t,W_t)=\alpha f(t,W_t)\,dW_t\quad\Longrightarrow\quad e^{\alpha W_t-\frac12\alpha^2t}=1+\int_0^t\alpha\,e^{\alpha W_u-\frac12\alpha^2u}\,dW_u\]

is a martingale. The \(-\frac12\alpha^2t\) is exactly the “Itô correction” needed to kill the drift that \(e^{\alpha W_t}\) would otherwise have.

4. Itô processes

An Itô process has both a regular and a stochastic integral:

\[X_t=X_0+\underbrace{\int_0^t\Theta_u\,du}_{A_t:\ \text{finite-variation part}}+\underbrace{\int_0^t\Delta_u\,dW_u}_{M_t:\ \text{martingale part}},\qquad dX_t=\Theta_t\,dt+\Delta_t\,dW_t,\]

with \(\Theta,\Delta\) adapted so that the integrals exist and \(X_0\) constant. \(A\) is never a martingale unless \(\Theta\equiv0\).

Examples: \(\Theta\equiv0,\ \Delta\equiv1,\ X_0=x\) gives \(X_t=x+W_t\), Brownian motion started at \(x\); \(\Delta\equiv0\) gives a smooth process \(X_t=X_0+\int_0^t\Theta_u\,du\).

Quadratic variation. Only the \(dW\) part contributes:

\[d[X,X]_t=(dX_t)^2=\Theta_t^2\,dt^2+2\Theta_t\Delta_t\,dt\,dW_t+\Delta_t^2\,(dW_t)^2=\Delta_t^2\,dt,\qquad [X,X]_t=\int_0^t\Delta_u^2\,du.\]

Integrating against an Itô process. For adapted \(\Gamma\),

\[\int_0^t\Gamma_u\,dX_u:=\int_0^t\Gamma_u\Theta_u\,du+\int_0^t\Gamma_u\Delta_u\,dW_u,\]

provided \(\int_0^t\vert\Gamma_u\Theta_u\vert du<\infty\) and \(E\int_0^t\Gamma_u^2\Delta_u^2du<\infty\). If \(X\) is a share price, this is the P/L of trading \(\Gamma\) shares. The result \(Y_t=\int_0^t\Gamma_u\,dX_u\) is again an Itô process (a martingale only if \(\Theta\equiv0\)), with \([Y,Y]_t=\int_0^t\Gamma_u^2\Delta_u^2\,du\).

5. Itô’s formula for Itô processes

For nice \(f(t,x)\) (smooth, bounded derivatives):

\[df(t,X_t)=f_t(t,X_t)\,dt+f_x(t,X_t)\,dX_t+\frac12f_{xx}(t,X_t)\,d[X,X]_t .\]

Plugging in \(dX_t=\Theta_tdt+\Delta_tdW_t\) and \(d[X,X]_t=\Delta_t^2dt\):

\[df(t,X_t)=\underbrace{\Big(f_t+f_x\Theta_t+\tfrac12f_{xx}\Delta_t^2\Big)dt}_{\text{finite variation}}+\underbrace{f_x\,\Delta_t\,dW_t}_{\text{martingale}} .\]

So a function of an Itô process is again an Itô process — but the martingale property is not preserved in general (e.g. \(W\) is a martingale, \(W^2\) is a sub-martingale).

How to remember it: pretend the second-order Taylor expansion is exact, then apply the multiplication table, using \(dt\,dX_t=dt(\Theta_tdt+\Delta_tdW_t)=0\) and \((dX_t)^2=\Delta_t^2dt\).

Discounting (used all the time in Parts 05–06). If \(dY_t=\Theta_tdt+\Delta_tdW_t\), apply the formula to \(f(t,y)=e^{-rt}y\) (\(f_t=-rf\), \(f_y=e^{-rt}\), \(f_{yy}=0\)):

\[d(e^{-rt}Y_t)=e^{-rt}(\Theta_t-rY_t)\,dt+e^{-rt}\Delta_t\,dW_t .\]

6. Geometric Brownian motion

Let \(\mu=\{\mu_t\}\), \(\sigma=\{\sigma_t\}\) be adapted with \(\int_0^t\vert\mu_s\vert ds<\infty\) and \(E\int_0^t\sigma_s^2ds<\infty\). A (generalized) geometric Brownian motion, written \(S\sim GBM(\mu,\sigma)\), is

\[S_t=S_0\exp\Big\{\int_0^t\Big(\mu_u-\tfrac12\sigma_u^2\Big)du+\int_0^t\sigma_u\,dW_u\Big\},\qquad S_0>0.\]

It is the most common stock-price model. With constant \(\mu,\sigma\),

\[S_t=S_0\exp\Big\{\big(\mu-\tfrac12\sigma^2\big)t+\sigma W_t\Big\}=S_0e^{\mu t}\cdot\underbrace{e^{\sigma W_t-\frac12\sigma^2t}}_{\text{exponential martingale}} .\]

Dynamics. Let \(X_t=\int_0^t(\mu_u-\frac12\sigma_u^2)du+\int_0^t\sigma_udW_u\), so \(S_t=f(X_t)\) with \(f(x)=S_0e^x\) (\(f=f'=f''\)):

\[dS_t=S_t\Big(\big(\mu_t-\tfrac12\sigma_t^2\big)dt+\sigma_tdW_t\Big)+\tfrac12S_t\sigma_t^2dt\quad\Longrightarrow\quad\frac{dS_t}{S_t}=\mu_t\,dt+\sigma_t\,dW_t .\]

\(dS_t/S_t\) is the instantaneous return over \([t,t+dt]\). Since \(\mu_t,\sigma_t\) are \(\mathcal F_t\)-measurable and \(dW_t\perp\!\!\!\perp\mathcal F_t\),

\[E\Big[\frac{dS_t}{S_t}\,\Big\vert\,\mathcal F_t\Big]=\mu_t\,dt,\qquad \mathrm{Var}\Big[\frac{dS_t}{S_t}\,\Big\vert\,\mathcal F_t\Big]=\sigma_t^2\,dt,\]

so \(\mu\) and \(\sigma\) are the drift and volatility of returns. The \(-\frac12\sigma^2\) in the exponent is the Itô correction again: the log-price grows more slowly than the expected return.

7. Application: integrals of deterministic functions are Gaussian

Claim. If \(f\) is deterministic (non-random), then

\[\int_0^tf_u\,dW_u\sim N\Big(0,\ \int_0^tf_u^2\,du\Big).\]

Proof via the exponential martingale

Let \(S\sim GBM(0,\alpha f)\) with \(S_0=1\) and \(\alpha\in\mathbb R\). With zero drift, \(S\) is a martingale, so

\[1=E[S_t]=E\Big[e^{\alpha\int_0^tf_udW_u-\frac12\alpha^2\int_0^tf_u^2du}\Big]=e^{-\frac12\alpha^2\int_0^tf_u^2du}\,E\Big[e^{\alpha\int_0^tf_udW_u}\Big],\]

where we pulled out the non-random factor. Hence \(E\big[e^{\alpha\int_0^tf_udW_u}\big]=e^{\frac12\alpha^2\int_0^tf_u^2du}\), which is the moment generating function of \(N\big(0,\int_0^tf_u^2du\big)\) (for \(X\sim N(0,\sigma^2)\), \(E e^{\alpha X}=e^{\frac12\alpha^2\sigma^2}\)). The mgf determines the distribution.

The variance \(\int_0^tf_u^2du\) is exactly what the Itô isometry predicts; the new information is that the distribution is normal. We will use this for the Ornstein–Uhlenbeck process in Part 07.

Series contents

01 · Probability Spaces, Filtrations, Conditional Expectation and Martingales
02 · Random Walks, Quadratic Variation and Brownian Motion
03 · Stochastic Integration
04 · Itô’s Formula, Itô Processes and Geometric Brownian Motion (this post)
05 · Black–Scholes: Replication, the PDE and the Greeks
06 · Risk-Neutral Pricing, Change of Measure and Girsanov
07 · SDEs, Feynman–Kac and Stopping Times

← Previous: Stochastic Integration　·　Next: Black–Scholes: Replication, the PDE and the Greeks →

This series collects my notes from QST MF 790 Stochastic Calculus at Boston University’s Questrom School of Business, in 7 parts. The content follows the order of the course; wording, intuition and a few corrections are my own. Throughout, \(N(\mu,\sigma^2)\) uses the variance as its second argument. Corrections are welcome. Chinese version of this post: 中文版.