Quant Academy Math · Probability · Quant Finance
Phase 9 - Lesson 9.2

Itô's Lemma

The chain rule of stochastic calculus, driven by the rule (dW)²=dt.

⏱ 55 min● Advanced🔗 Prereqs: 9.1
↖ Phase 9 hub
Builds on: 9.1 built the Itô integral and quadratic variation.
Leads to: 9.3 uses Itô to solve SDEs; 9.5 turns it into a PDE.

Learning Objectives

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Key Vocabulary

Itô process
A process \(dX=a\,dt+b\,dW\) with adapted drift \(a\) and diffusion \(b\).
Itô's lemma
The stochastic chain rule: it adds a \(\tfrac12 f_{xx}(dX)^2\) term to the ordinary chain rule.
Multiplication table
The formal rules \((dt)^2=dt\,dW=0\), \((dW)^2=dt\) used to expand differentials.
Itô correction
The second-order term \(\tfrac12 f_{xx}b^2\,dt\) absent from deterministic calculus.
Drift
The \(dt\) coefficient of an Itô process; the local mean rate of change.
Diffusion (volatility)
The \(dW\) coefficient; the local standard deviation rate.

Intuition & Motivation

Intuition
Ordinary Taylor expansion of \(f(X)\) keeps first-order terms and drops \((dX)^2\) as negligible. For an Itô process that is wrong: \((dX)^2\) contains \((b\,dW)^2=b^2(dW)^2=b^2\,dt\), which is first order in \(dt\). So we must keep the second-order term. Itô's lemma is exactly the chain rule plus this one extra piece.

The multiplication table

Because Brownian motion has quadratic variation \(t\), squared increments behave like \(dt\). Formally, when expanding differentials to first order in \(dt\):

\[(dt)^2=0,\qquad dt\,dW=0,\qquad (dW)^2=dt.\] (9.3)
Theorem - Itô's lemma (one dimension)
Let \(dX_t=a_t\,dt+b_t\,dW_t\) and let \(f(t,x)\) be \(C^{1,2}\) (once differentiable in \(t\), twice in \(x\)). Then
\[df(t,X_t)=\Big(f_t+a_t f_x+\tfrac12 b_t^2 f_{xx}\Big)dt+b_t f_x\,dW_t.\] (9.4)

Derivation sketch: Taylor-expand \(df=f_t\,dt+f_x\,dX+\tfrac12 f_{xx}(dX)^2+\cdots\), substitute \(dX=a\,dt+b\,dW\) and \((dX)^2=b^2\,dt\) from (9.3), and drop higher-order terms. The \(\tfrac12 b^2 f_{xx}\) term is the Itô correction.

Worked Example - d(log S) for geometric Brownian motion
1
Model a stock as \(dS_t=\mu S_t\,dt+\sigma S_t\,dW_t\), so \(a=\mu S,\ b=\sigma S\).
2
Take \(f(S)=\log S\): \(f_x=1/S,\ f_{xx}=-1/S^2,\ f_t=0\).
3
Apply (9.4): \(d(\log S)=\big(\mu S\cdot\tfrac1S+\tfrac12\sigma^2S^2\cdot(-\tfrac1{S^2})\big)dt+\sigma S\cdot\tfrac1S\,dW.\)
4
Simplify: \(d(\log S)=\big(\mu-\tfrac12\sigma^2\big)dt+\sigma\,dW.\) The drift of \(\log S\) is \(\mu-\tfrac12\sigma^2\), not \(\mu\) - the Itô correction in action. (This integrates directly; see 9.3.)
Worked Example - d(S²) and a martingale check
1
Let \(f(x)=x^2\) with \(dX=dW\) (so \(a=0,b=1\)): \(f_x=2x,\ f_{xx}=2\).
2
\(d(W^2)=(0+\tfrac12\cdot1\cdot2)dt+1\cdot2W\,dW=dt+2W\,dW.\)
3
Integrate: \(W_T^2=T+2\int_0^T W\,dW\) - matching 9.1's identity exactly.
4
Consequence: \(W_t^2-t\) has zero drift, so \(d(W_t^2-t)=2W\,dW\) is a pure Itô integral and hence a martingale.

Predict before you compute

Predict first: For \(dS=\mu S\,dt+\sigma S\,dW\), is the drift of \(\log S\) equal to \(\mu\) or \(\mu-\tfrac12\sigma^2\)?
Common Mistakes to Avoid
  • Forgetting the \(\tfrac12 b^2 f_{xx}\) term and using the ordinary chain rule. This is the single most common error in stochastic calculus.
  • Using \((dW)^2\approx0\). It is exactly \(dt\).
  • Confusing the drift of \(S\) (\(\mu S\)) with the drift of \(\log S\) (\(\mu-\tfrac12\sigma^2\)).
  • Applying \(f_x b\) but forgetting \(f_t\) when \(f\) depends explicitly on time (e.g. discounting).
Quant Practitioner Tips
  • Mechanical recipe: write \(df=f_t\,dt+f_x\,dX+\tfrac12 f_{xx}(dX)^2\), then reduce with the table (9.3).
  • Zero drift after applying Itô is the fastest way to prove a process is a (local) martingale.
  • For products, \(d(XY)=X\,dY+Y\,dX+dX\,dY\) (Itô product rule); the cross term \(dX\,dY\) is the correction.

Knowledge Check

Q1 Easy
In the Itô multiplication table, \((dW)^2\) equals:
\(0\)
\(dt\)
\(dW\)
\((dt)^2\)
Q2 Medium
For \(dS=\mu S\,dt+\sigma S\,dW\), the SDE for \(Y=\log S\) is:
\(dY=\mu\,dt+\sigma\,dW\)
\(dY=(\mu-\tfrac12\sigma^2)dt+\sigma\,dW\)
\(dY=(\mu+\tfrac12\sigma^2)dt+\sigma\,dW\)
\(dY=\mu S\,dt+\sigma S\,dW\)
Q3 Medium
Why is \(W_t^2-t\) a martingale but \(W_t^2\) is not?
Both are martingales
Subtracting \(t\) removes the \(dt\) drift, leaving a pure Itô integral
Because \(W_t^2\) is negative
Because \(t\) is random

Practical Exercise

Use Itô's lemma to find the SDE for \(Y_t=e^{W_t}\). Then compute \(\E[e^{W_t}]\) from the drift and confirm it equals \(e^{t/2}\).

▶ Show full solution

Let \(f(x)=e^x\), \(dX=dW\) (\(a=0,b=1\)). Then \(f_x=f_{xx}=e^x\), so \(dY=(\tfrac12 e^{W})dt+e^{W}\,dW=\tfrac12 Y\,dt+Y\,dW.\)

Taking expectations kills the martingale \(dW\) term: \(\tfrac{d}{dt}\E[Y_t]=\tfrac12\E[Y_t]\), with \(\E[Y_0]=1\). Solving the ODE gives \(\E[e^{W_t}]=e^{t/2}\), the moment generating function of \(W_t\sim\Normal(0,t)\) at argument 1.

After the reveal, answer for yourself: Notice how Itô turned a probability question (\(\E[e^{W_t}]\)) into an ODE for the mean.

Lesson Summary

Itô's lemma is the chain rule of stochastic calculus: \(df=(f_t+af_x+\tfrac12 b^2 f_{xx})dt+bf_x\,dW\). The extra \(\tfrac12 b^2 f_{xx}\) correction arises because \((dW)^2=dt\). It yields \(d(\log S)=(\mu-\tfrac12\sigma^2)dt+\sigma\,dW\) for GBM, and zero drift after Itô is a quick martingale test.

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: State Itô's lemma and name the extra term.
A: \(df=(f_t+af_x+\tfrac12 b^2 f_{xx})\,dt+bf_x\,dW\). The extra term \(\tfrac12 b^2 f_{xx}\,dt\) is the Itô correction, from \((dW)^2=dt\).
Q: Why does \(\log S\) drift at \(\mu-\tfrac12\sigma^2\) rather than \(\mu\)?
A: Because \(f(S)=\log S\) is concave (\(f_{xx}\lt 0\)), and the Itô correction \(\tfrac12 f_{xx}(\sigma S)^2=-\tfrac12\sigma^2\) subtracts from the drift.

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