Phase 9 - Lesson 9.4
Girsanov's Theorem and Change of Measure
Rewriting the drift of a Brownian motion by re-weighting probabilities.
Builds on: 9.3 gave GBM under the real-world drift \(\mu\).
Leads to: 9.6 uses Girsanov to switch to the risk-neutral measure.
Leads to: 9.6 uses Girsanov to switch to the risk-neutral measure.
Learning Objectives
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- Explain what it means for two measures to be equivalent and define the Radon–Nikodym derivative.
- State Girsanov's theorem and identify the new Brownian motion under the changed measure.
- Compute the drift change induced by a given market price of risk.
- Explain intuitively why changing measure can remove drift but not volatility.
- Apply Girsanov to turn \(dS=\mu S\,dt+\sigma S\,dW\) into a driftless discounted process.
Key Vocabulary
- Equivalent measures
- Two probability measures \(\Prob,\mathbb{Q}\) that agree on which events have probability zero (same null sets).
- Radon–Nikodym derivative
- The density \(Z=d\mathbb{Q}/d\Prob\) re-weighting outcomes; \(\mathbb{Q}(A)=\E_\Prob[Z\,\ind_A]\).
- Market price of risk
- The quantity \(\theta=(\mu-r)/\sigma\); excess return per unit of volatility.
- Girsanov's theorem
- Result: subtracting a drift from \(W\) corresponds to an explicit change of measure with a known density.
- Novikov condition
- A sufficient integrability condition \(\E[e^{\frac12\int\theta^2 dt}]\lt \infty\) ensuring \(Z\) is a true martingale.
- Exponential martingale
- The density process \(Z_t=\exp(-\int_0^t\theta\,dW-\tfrac12\int_0^t\theta^2 du)\).
Intuition & Motivation
Intuition
A Brownian motion with drift, \(\tilde W_t=W_t+\theta t\), looks like it is trending up. But ‘trending’ is a statement about probabilities: up-moves are slightly more likely than down-moves. If we simply re-weight the outcomes - make the down-paths a bit more probable and the up-paths a bit less - the very same process becomes driftless. Girsanov's theorem says exactly how to re-weight (the Radon–Nikodym density) and confirms that volatility is untouched: you can bend the mean, not the wiggle.
Equivalent measures and densities
Two measures \(\Prob\) and \(\mathbb{Q}\) on the same space are equivalent if they have the same null sets: something impossible under one is impossible under the other. Then there is a positive random variable \(Z=d\mathbb{Q}/d\Prob\) with \(\E_\Prob[Z]=1\) such that \(\E_\mathbb{Q}[X]=\E_\Prob[ZX]\). Changing measure is re-weighting, not relabeling outcomes.
Theorem - Girsanov (one dimension)
Let \(W_t\) be a \(\Prob\)-Brownian motion and \(\theta_t\) adapted with \(\E_\Prob[e^{\frac12\int_0^T\theta_u^2 du}]\lt \infty\) (Novikov). Define \(Z_t=\exp\!\big(-\int_0^t\theta_u\,dW_u-\tfrac12\int_0^t\theta_u^2\,du\big)\) and \(d\mathbb{Q}=Z_T\,d\Prob\). Then
\[\tilde W_t=W_t+\int_0^t\theta_u\,du\quad\text{is a }\mathbb{Q}\text{-Brownian motion.}\] (9.6)
In differential form \(d\tilde W=dW+\theta\,dt\), so \(dW=d\tilde W-\theta\,dt\). Substituting into any SDE shifts its drift by \(-\sigma\theta\) while leaving the diffusion \(\sigma\) alone.
Worked Example - Removing the drift of a stock
1
Real-world dynamics: \(dS=\mu S\,dt+\sigma S\,dW\) under \(\Prob\). Choose \(\theta=(\mu-r)/\sigma\), the market price of risk.
2
Then \(dW=d\tilde W-\theta\,dt\), so \(dS=\mu S\,dt+\sigma S(d\tilde W-\theta\,dt)=(\mu-\sigma\theta)S\,dt+\sigma S\,d\tilde W.\)
3
But \(\mu-\sigma\theta=\mu-(\mu-r)=r\). Hence under \(\mathbb{Q}\): \(dS=r S\,dt+\sigma S\,d\tilde W.\)
4
The drift became the risk-free rate \(r\); volatility \(\sigma\) is unchanged. The discounted price \(e^{-rt}S_t\) is now a \(\mathbb{Q}\)-martingale.
Predict: what changes under Q?
Predict first: When you change measure to remove the drift of \(dS=\mu S dt+\sigma S dW\), does the volatility \(\sigma\) increase, decrease, or stay the same?
Why volatility is invariant
Quadratic variation \([S,S]_t=\int_0^t\sigma^2 S_u^2\,du\) is computed pathwise from the trajectory. Equivalent measures agree on which paths occur (same null sets), so they agree on quadratic variation - hence on \(\sigma\). Only the likelihood of paths (the drift) can change.
Common Mistakes to Avoid
- Thinking a change of measure alters the paths. It only re-weights their probabilities; every path still occurs.
- Trying to change volatility by changing measure - impossible; only drift moves.
- Sign errors: \(\tilde W=W+\int\theta\,dt\) so \(dW=d\tilde W-\theta\,dt\). Dropping the minus flips the drift.
- Forgetting the \(-\tfrac12\int\theta^2\) term in \(Z\); without it \(Z\) is not a mean-1 martingale.
Quant Practitioner Tips
- The market price of risk \(\theta=(\mu-r)/\sigma\) is the exact drift adjustment that turns \(\mu\) into \(r\).
- Girsanov is the bridge from the physical measure \(\Prob\) (used for risk, backtesting) to the risk-neutral \(\mathbb{Q}\) (used for pricing).
- Check any candidate density: \(\E_\Prob[Z_T]=1\) must hold - otherwise \(\mathbb{Q}\) is not a probability measure.
Knowledge Check
Q1 Medium
Under an equivalent change of measure, which of a Brownian-driven SDE changes?
Q2 Medium
To turn \(dS=\mu S\,dt+\sigma S\,dW\) into \(dS=rS\,dt+\sigma S\,d\tilde W\), the market price of risk is:
Q3 Easy
The Radon–Nikodym derivative \(Z=d\mathbb{Q}/d\Prob\) must satisfy:
Practical Exercise
A stock has \(\mu=0.10,\ r=0.02,\ \sigma=0.25\). (a) Compute the market price of risk \(\theta\). (b) Write the \(\mathbb{Q}\)-dynamics of \(S\). (c) State the density process \(Z_t\).
▶ Show full solution
(a) \(\theta=(\mu-r)/\sigma=(0.10-0.02)/0.25=0.32\).
(b) Under \(\mathbb{Q}\), \(dS=rS\,dt+\sigma S\,d\tilde W=0.02\,S\,dt+0.25\,S\,d\tilde W\), with \(\tilde W_t=W_t+0.32\,t\).
(c) \(Z_t=\exp(-0.32\,W_t-\tfrac12(0.32)^2 t)=\exp(-0.32\,W_t-0.0512\,t)\). One checks \(\E_\Prob[Z_t]=1\).
After the reveal, answer for yourself: If \(\mu=r\), what is \(\theta\) and what does the change of measure do?
Lesson Summary
Girsanov's theorem re-weights probabilities via the density \(Z_t=e^{-\int\theta dW-\frac12\int\theta^2 du}\) so that \(\tilde W=W+\int\theta\,dt\) is Brownian under \(\mathbb{Q}\). This shifts drift by \(-\sigma\theta\) while leaving volatility fixed. Choosing \(\theta=(\mu-r)/\sigma\) turns a stock's drift into \(r\), the key step toward risk-neutral pricing.
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: What does Girsanov's theorem change, and what does it preserve?
A: It changes the drift of a Brownian-driven process (by re-weighting path probabilities) but preserves the volatility, since quadratic variation is invariant under equivalent measures.
A: It changes the drift of a Brownian-driven process (by re-weighting path probabilities) but preserves the volatility, since quadratic variation is invariant under equivalent measures.
Q: Give the density and the new Brownian motion in Girsanov.
A: \(Z_t=\exp(-\int_0^t\theta\,dW-\tfrac12\int_0^t\theta^2 du)\), and \(\tilde W_t=W_t+\int_0^t\theta\,du\) is \(\mathbb{Q}\)-Brownian.
A: \(Z_t=\exp(-\int_0^t\theta\,dW-\tfrac12\int_0^t\theta^2 du)\), and \(\tilde W_t=W_t+\int_0^t\theta\,du\) is \(\mathbb{Q}\)-Brownian.
Completion Checklist
- I can explain the core ideas in my own words
- I worked the derivations/examples by hand
- I completed the interactive workbench(es)
- I passed the knowledge check
Confidence / mastery rating
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