Quant Academy Math · Probability · Quant Finance
Phase 9 - Lesson 9.1

The Itô Integral

Why ordinary calculus fails for Brownian paths, and how to build an integral against them anyway.

⏱ 55 min● Advanced🔗 Prereqs: 8.x Brownian motion; 7.x conditional expectation
↖ Phase 9 hub
Builds on: Phase 8 built Brownian motion, filtrations, and martingales.
Leads to: 9.2 differentiates functions of the Itô integral; 9.6 uses it to price.

Learning Objectives

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Key Vocabulary

Quadratic variation
The limit of summed squared increments; for Brownian motion on [0,t] it equals t, not 0.
Adapted / non-anticipating
An integrand whose value at time t uses only information up to t (measurable w.r.t. the filtration).
Simple process
A step-function integrand, constant on a partition, for which the integral is defined by hand.
Itô isometry
The identity equating the L² norm of the integral to the L² norm of the integrand in time.
Itô integral
The L²-limit of integrals of simple processes; evaluated at the LEFT endpoint of each subinterval.
Martingale
A process whose conditional expected future value equals its present value; the Itô integral is one.

Intuition & Motivation

Intuition
You already integrate against smooth functions with Riemann sums. Brownian motion \(W_t\) is too rough for that: over \([0,t]\) its path wiggles so much that its total length (first variation) is infinite, so the usual Riemann–Stieltjes recipe diverges. The saving grace is that its squared increments do not vanish - they accumulate to \(t\). Itô's construction embraces this: we always sample the integrand at the left endpoint of each step (so we never peek into the future), take an \(L^2\) limit, and get a well-defined, mean-zero martingale.

Why ordinary calculus breaks

Fix a partition \(0=t_0\lt t_1\lt \cdots\lt t_n=t\). The first variation \(\sum_i |W_{t_{i+1}}-W_{t_i}|\) diverges as the mesh shrinks, so a Riemann–Stieltjes integral \(\int g\,dW\) defined path-by-path does not exist. But the quadratic variation converges:

\[[W,W]_t=\lim_{\|\Pi\|\to0}\sum_i\big(W_{t_{i+1}}-W_{t_i}\big)^2 = t\quad\text{(in }L^2\text{, a.s.)}\] (9.1)

Each squared increment has mean \(t_{i+1}-t_i\) and variance \(2(t_{i+1}-t_i)^2\); the means sum to \(t\) while the variances vanish. This single fact - quadratic variation accumulates - is the engine of the whole theory.

Definition - Itô integral of a simple process
Let \(\Delta\) be adapted and constant on \([t_i,t_{i+1})\), i.e. \(\Delta_s=\Delta_{t_i}\). Define \(I_t=\int_0^t\Delta_s\,dW_s := \sum_{i}\Delta_{t_i}\,(W_{t_{i+1}\wedge t}-W_{t_i\wedge t}).\) The integrand is sampled at the LEFT endpoint \(t_i\), using only information available then.
Theorem - Itô isometry & martingale property
For adapted \(\Delta\) with \(\E\int_0^t\Delta_s^2\,ds\lt \infty\), the process \(I_t=\int_0^t\Delta_s\,dW_s\) is a martingale with \(\E[I_t]=0\) and
\[\E\!\left[\Big(\int_0^t\Delta_s\,dW_s\Big)^{\!2}\right]=\E\!\int_0^t\Delta_s^2\,ds\qquad(\text{It\^o isometry}).\] (9.2)

The isometry is what lets us extend the integral from step functions to all square-integrable adapted integrands: approximate \(\Delta\) by simple processes in \(L^2\), and (9.2) guarantees the integrals converge in \(L^2\) to a unique limit - the Itô integral.

Worked Example - The signature identity: \(\int_0^T W_s\,dW_s\)
1
Telescope \(W_T^2=\sum_i(W_{t_{i+1}}^2-W_{t_i}^2)\). Use \(b^2-a^2=2a(b-a)+(b-a)^2\) with \(a=W_{t_i},\,b=W_{t_{i+1}}\).
2
So \(W_T^2=2\sum_i W_{t_i}(W_{t_{i+1}}-W_{t_i})+\sum_i(W_{t_{i+1}}-W_{t_i})^2.\) The first sum is the Itô (left-endpoint) sum; the second is the quadratic variation.
3
Take \(\|\Pi\|\to0\): the first sum \(\to 2\int_0^T W_s\,dW_s\), the second \(\to T\). Hence \(W_T^2=2\int_0^T W_s\,dW_s+T.\)
4
Solve: \(\displaystyle\int_0^T W_s\,dW_s=\tfrac12 W_T^2-\tfrac12 T.\) The extra \(-\tfrac12 T\) (absent in ordinary calculus, where \(\int x\,dx=\tfrac12x^2\)) is the fingerprint of quadratic variation.
5
Check the martingale property: \(\E[\tfrac12 W_T^2-\tfrac12 T]=\tfrac12 T-\tfrac12 T=0=I_0.\) Consistent with (9.2).

Interactive: watch quadratic variation accumulate

Common Mistakes to Avoid
  • Evaluating the integrand at the right endpoint or midpoint - that gives the Stratonovich integral, a different answer, and destroys the martingale property.
  • Writing \(\int_0^T W\,dW=\tfrac12 W_T^2\). You must subtract \(\tfrac12 T\).
  • Assuming \((dW)^2\) is negligible. It is order \(dt\), the whole point.
  • Using an anticipating integrand (peeking at the future); then the isometry and martingale property fail.
Quant Practitioner Tips
  • Whenever you see a squared Brownian increment, replace it with \(dt\): \((dW)^2\to dt\).
  • Use the isometry to get variances for free: \(\Var\big(\int_0^t\Delta\,dW\big)=\E\int_0^t\Delta^2\,ds\).
  • ‘Adapted & left-endpoint’ is the modeling assumption that trading strategies cannot see the future - why Itô (not Stratonovich) is used in finance.

Knowledge Check

Q1 Medium
Why can't the integral against a Brownian path be defined pathwise as a Riemann–Stieltjes integral?
Brownian paths are discontinuous
Brownian paths have infinite first variation on every interval
Brownian motion is not adapted
The integrand is not measurable
Q2 Medium
\(\int_0^T W_s\,dW_s\) equals:
\(\tfrac12 W_T^2\)
\(\tfrac12 W_T^2-\tfrac12 T\)
\(W_T^2-T\)
\(\tfrac12 W_T^2+\tfrac12 T\)
Q3 Medium
The Itô isometry states that \(\E[(\int_0^t\Delta\,dW)^2]\) equals:
\((\E\int_0^t\Delta\,ds)^2\)
\(\E\int_0^t\Delta_s^2\,ds\)
\(\int_0^t\Delta_s^2\,ds\)
\(t\,\E[\Delta_t^2]\)

Practical Exercise

Let \(X_t=\int_0^t s\,dW_s\). (a) Show \(X\) is a martingale with mean 0. (b) Compute \(\Var(X_t)\) using the Itô isometry. (c) What is the distribution of \(X_t\)?

▶ Show full solution

(a) The integrand \(\Delta_s=s\) is deterministic (hence adapted) and square-integrable, so by the theorem \(X_t\) is a martingale and \(\E[X_t]=0\).

(b) By the isometry (9.2), \(\Var(X_t)=\E[X_t^2]=\E\int_0^t s^2\,ds=\int_0^t s^2\,ds=t^3/3.\)

(c) An Itô integral of a deterministic integrand is a Gaussian process (a limit of sums of independent Gaussians). So \(X_t\sim\Normal(0,\,t^3/3)\).

After the reveal, answer for yourself: Where did you use that the integrand was deterministic rather than merely adapted?

Lesson Summary

Brownian paths have infinite first variation but finite quadratic variation \([W,W]_t=t\). This forces a new integral: sample the adapted integrand at the LEFT endpoint, take an \(L^2\) limit of simple processes, and obtain the Itô integral - a mean-zero martingale obeying the isometry \(\E[(\int\Delta dW)^2]=\E\int\Delta^2 ds\). The identity \(\int_0^T W\,dW=\tfrac12W_T^2-\tfrac12T\) shows how quadratic variation modifies ordinary calculus.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers
Q: Why is the Itô integral evaluated at the left endpoint?
A: So the integrand is non-anticipating (adapted): its value uses only past information. This makes the integral a martingale and gives the Itô (not Stratonovich) calculus used in finance.
Q: State the Itô isometry and one use.
A: \(\E[(\int_0^t\Delta\,dW)^2]=\E\int_0^t\Delta^2\,ds\). Use it to compute the variance of an Itô integral, e.g. \(\Var(\int_0^t s\,dW)=t^3/3\).

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