Phase 8 - Lesson 8.5

Gaussian Processes and the Foundations of Change of Measure

Brownian motion as a Gaussian process, the multivariate normal toolkit, and how Radon-Nikodym densities and Girsanov’s theorem re-weight probability to remove drift.

⏱ 55 min● Research🔗 Prereqs: 8.3, 7.4

↖ Phase 8 hub

Builds on: 8.3 gave BM’s Gaussian marginals; 7.4 gave transforms; 7.3 gave Radon-Nikodym.
Leads to: Girsanov and the equivalent martingale measure are the mechanism of risk-neutral pricing in Phases 9 and 14.

Learning Objectives

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Define a Gaussian process and show Brownian motion is one with mean 0 and covariance \(\min(s,t)\).
Use the multivariate normal to compute conditional distributions of Brownian values.
Define equivalent measures and the Radon-Nikodym derivative as a re-weighting of probability.
State Girsanov’s theorem and explain how it converts a drifting process into a martingale.

Key Vocabulary

Gaussian process: A process all of whose finite-dimensional marginals \((X_{t_1},\dots,X_{t_k})\) are jointly Gaussian; determined by mean and covariance functions.

Covariance function: \(c(s,t)=\Cov(X_s,X_t)\); for BM, \(c(s,t)=\min(s,t)\).

Equivalent measures: \(\Prob\sim\Q\): they agree on which events have probability zero (mutually absolutely continuous).

Radon-Nikodym derivative: The density \(Z=d\Q/d\Prob\ge0\) with \(\E_\Prob[Z]=1\) that re-weights \(\Prob\) into \(\Q\).

Girsanov’s theorem: Under a suitable exponential density, \(\tilde B_t=B_t+\theta t\) becomes a Brownian motion (a martingale) under the new measure.

Equivalent martingale measure: A measure \(\Q\sim\Prob\) under which discounted prices are martingales; its existence is no-arbitrage.

Intuition & Motivation

Intuition

Two big ideas close the phase. First, Brownian motion is a Gaussian process: pick any finite set of times and the values are jointly normal, so the entire process is pinned down by just a mean function and a covariance function \(\min(s,t)\). All the power of the multivariate normal - linear conditional expectations, Gaussian conditionals - is therefore available for Brownian calculations. Second, change of measure: probability is not sacred, it can be re-weighted. Two measures are equivalent if they agree on what is impossible; the Radon-Nikodym derivative is the density that converts one into the other. Girsanov’s theorem is the spectacular payoff: by re-weighting paths you can delete the drift of a Brownian motion, turning a process with a trend into a pure martingale. That is the exact mathematical move behind risk-neutral pricing - switch to a measure where the discounted stock is a fair game.

Brownian motion as a Gaussian process

Definition - Gaussian process

A process \((X_t)\) is Gaussian if for every finite \(t_1\lt \cdots\lt t_k\) the vector \((X_{t_1},\dots,X_{t_k})\) is multivariate normal. It is completely determined by its mean function \(m(t)=\E[X_t]\) and covariance function \(c(s,t)=\Cov(X_s,X_t)\).

Proposition - Brownian motion is Gaussian

Standard BM is the Gaussian process with \(m(t)=0\) and \(c(s,t)=\min(s,t)\). Any mean-zero, continuous Gaussian process with this covariance is a Brownian motion.

This gives a second, equivalent definition of BM and a powerful computational shortcut: to find the joint law of \((B_{t_1},\dots,B_{t_k})\) just assemble the covariance matrix \(\Sigma_{ij}=\min(t_i,t_j)\) and use standard multivariate-normal formulas.

Gaussian conditioning

For jointly normal \((X,Y)\), the conditional law is again normal with a linear mean and a variance reduced by the explained part:

\[\E[Y\mid X=x]=\mu_Y+\frac{\Cov(X,Y)}{\Var(X)}(x-\mu_X),\quad \Var(Y\mid X)=\Var(Y)-\frac{\Cov(X,Y)^2}{\Var(X)}.\] (8.8)

Applied to BM this recovers, e.g., the Brownian bridge: conditioning on both endpoints gives a Gaussian interpolation with a deterministic variance profile.

Change of measure

Definition - Equivalent measures and Radon-Nikodym derivative

Probability measures \(\Prob\) and \(\Q\) on the same space are equivalent (\(\Prob\sim\Q\)) if they have the same null sets. Then there is a density \(Z=d\Q/d\Prob\gt 0\) with \(\E_\Prob[Z]=1\), and for any integrable \(Y\),

\[\E_\Q[Y]=\E_\Prob[YZ],\qquad Z=\frac{d\Q}{d\Prob}.\] (8.9)

Re-weighting does not move probability mass onto previously impossible events (that is what equivalence guarantees); it only tilts the likelihoods of the possible ones. The tilt we want is one that removes drift.

Theorem - Girsanov (one-dimensional, constant drift)

Let \(B\) be a Brownian motion under \(\Prob\) and fix \(\theta\). Define the density process

\[Z_t=\exp\!\Big(-\theta B_t-\tfrac12\theta^2 t\Big),\qquad \E_\Prob[Z_t]=1.\]

Under the new measure \(\Q\) with \(d\Q/d\Prob=Z_T\), the process \(\tilde B_t=B_t+\theta t\) is a standard Brownian motion. More generally, a drift can be removed at the cost of the exponential-martingale density.

The density \(Z_t\) is itself a martingale (an exponential martingale), which is why \(\E_\Prob[Z_t]=1\) for all \(t\) and why \(\Q\) is a genuine probability measure. Girsanov says: to price, find the \(\theta\) that turns the discounted asset’s drift into zero, then compute expectations under \(\Q\).

Worked Example - Removing drift from an arithmetic price

Model a (driftful) price shock as \(Y_t=\mu t+B_t\) under the real-world measure \(\Prob\); it has upward trend \(\mu\), so \(Y\) is not a martingale.

Choose \(\theta=\mu\) and the Girsanov density \(Z_t=\exp(-\mu B_t-\tfrac12\mu^2 t)\).

Under \(\Q\) with \(d\Q/d\Prob=Z_T\), Girsanov says \(\tilde B_t=B_t+\mu t\) is a \(\Q\)-Brownian motion.

But \(Y_t=\mu t+B_t=\tilde B_t\): under \(\Q\), \(Y\) is exactly a Brownian motion - a martingale. The drift has been absorbed into the measure.

This is the whole engine of risk-neutral valuation: the real drift \(\mu\) is irrelevant to arbitrage-free prices; only the volatility (which change of measure cannot alter) matters.

Interactive: drift removal is a re-weighting, not a shift of outcomes

Common Mistakes to Avoid

Thinking change of measure alters the volatility - Girsanov changes only the drift; \([B]_t=t\) is measure-invariant.
Using a non-equivalent measure that assigns zero to a positive-probability event; then no Radon-Nikodym density exists.
Forgetting the \(-\tfrac12\theta^2 t\) correction in the exponential density; without it \(\E_\Prob[Z_t]\ne1\) and \(\Q\) is not a probability.
Assuming the real-world drift \(\mu\) enters an arbitrage-free price; under \(\Q\) it is replaced by the risk-free rate.

Quant Practitioner Tips

Any Brownian computation reduces to linear algebra: build \(\Sigma_{ij}=\min(t_i,t_j)\) and apply multivariate-normal formulas.
The risk-neutral measure is the equivalent martingale measure; its existence is precisely the first fundamental theorem of asset pricing.
In Monte Carlo, the Radon-Nikodym derivative is the importance-sampling weight - the same math, used to reduce variance.

Knowledge Check

Q1 Medium

A Gaussian process is completely determined by:

Its sample paths

Its mean function and covariance function

Its maximum over time

Its first four moments

Q2 Medium

Two equivalent measures \(\Prob\sim\Q\) must agree on:

The probability of every event

Which events have probability zero

The mean of every random variable

The variance of every random variable

Q3 Hard

Girsanov’s theorem is used in pricing to:

Change the volatility of the asset

Remove the drift so discounted prices become martingales

Make paths differentiable

Eliminate randomness

Practical Exercise

Let \((B_t)\) be standard BM. (a) Write the covariance matrix of \((B_1,B_2,B_3)\). (b) Using Gaussian conditioning, find \(\E[B_2\mid B_1=a,B_3=b]\).

▶ Show full solution

(a) With \(\Sigma_{ij}=\min(t_i,t_j)\): \(\Sigma=\begin{pmatrix}1&1&1\\1&2&2\\1&2&3\end{pmatrix}\).

(b) By the Markov/independent-increment structure, conditioning on \(B_1=a\) and \(B_3=b\), the value \(B_2\) is a Brownian bridge between times 1 and 3. The conditional mean interpolates linearly in time: \(\E[B_2\mid B_1=a,B_3=b]=a+\dfrac{2-1}{3-1}(b-a)=a+\tfrac12(b-a)=\tfrac{a+b}{2}.\)

The conditional variance is \(\dfrac{(2-1)(3-2)}{3-1}=\tfrac12\), independent of \(a,b\) - the hallmark of Gaussian conditioning: the mean is linear, the variance is reduced and deterministic.

After the reveal, answer for yourself: Why does the drift of the underlying not appear anywhere in the Brownian-bridge conditional mean?

Lesson Summary

Brownian motion is the Gaussian process with mean 0 and covariance \(\min(s,t)\), so every Brownian computation reduces to multivariate-normal algebra with linear, variance-reducing conditionals. Change of measure re-weights an equivalent probability by the Radon-Nikodym density \(d\Q/d\Prob\); Girsanov’s exponential-martingale density removes Brownian drift, turning a trending process into a martingale. This is the mathematical foundation of risk-neutral pricing, where only volatility - not real-world drift - survives.

Formula Sheet Additions

BM as Gaussian process

\[m(t)=0,\ c(s,t)=\min(s,t)\]

Two functions determine the whole process.

Girsanov density

\[Z_t=\exp(-\theta B_t-\tfrac12\theta^2 t),\ \tilde B_t=B_t+\theta t\]

Exponential martingale that removes drift.

Error Log Checklist

Did I keep volatility fixed under the change of measure?
Did I check the measures are equivalent (same null sets) before using a density?
Did I include the \(-\tfrac12\theta^2 t\) term so \(\E_\Prob[Z_t]=1\)?

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 Brownian motion a Gaussian process, and what determines it?
A: All finite-dimensional marginals are jointly normal; it is determined by mean 0 and covariance \(\min(s,t)\).

Q: What does it mean for two measures to be equivalent, and what object relates them?
A: They share the same null sets (mutual absolute continuity); the Radon-Nikodym derivative \(Z=d\Q/d\Prob\) re-weights one into the other.

Q: What does Girsanov’s theorem accomplish and what does it leave unchanged?
A: It removes (or changes) the drift of a Brownian motion via an exponential-martingale density, while leaving the volatility / quadratic variation unchanged.

Flashcards

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Gaussian process

Determined by mean function and covariance function; BM has \(m=0,\ c=\min(s,t)\).

Girsanov

Re-weight by \(Z_t=e^{-\theta B_t-\theta^2 t/2}\) to make \(B_t+\theta t\) a martingale under \(\Q\).

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

Personal notes

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