Chapter 6 — Martingales and Change of Measure

Guiding question

A drifted Brownian motion $dX = \theta\,dt + dW$ is not a martingale — it has a trend. Can we change the probability measure so that $X$ becomes a martingale? If so, how does the change of measure interact with option pricing?

Girsanov's theorem says: yes, a drift can be removed by reweighting paths. The reweighting factor — the Radon–Nikodym derivative — is a stochastic exponential built from the drift itself. This is the mathematical engine behind risk-neutral pricing: the risk-neutral measure is obtained by Girsanov's theorem applied to the market price of risk.

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6.1 Equivalent measures and Radon–Nikodym derivatives

Two probability measures $\mathbb P$ and $\mathbb Q$ on $(\Omega,\mathcal F)$ are equivalent if they agree on which events have probability zero: $\mathbb P(A)=0\iff\mathbb Q(A)=0$.

For equivalent measures on a finite horizon $[0,T]$, the Radon–Nikodym theorem guarantees the existence of a density:

$$\frac{d\mathbb Q}{d\mathbb P}\bigg|_{\mathcal F_T} = Z_T > 0\quad\text{a.s.},$$

with $\mathbb E^{\mathbb P}[Z_T] = 1$. For any $\mathcal F_T$-measurable $X$:

$$\mathbb E^{\mathbb Q}[X] = \mathbb E^{\mathbb P}[Z_T\,X].$$

The conditional Radon–Nikodym derivative defines a $(\mathbb P,\mathcal F_t)$-martingale:

$$Z_t = \mathbb E^{\mathbb P}[Z_T\mid\mathcal F_t].$$

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6.2 The stochastic exponential

For an adapted process $\theta = (\theta_t)_{0\le t\le T}$ satisfying the Novikov condition

$$\mathbb E^{\mathbb P}\!\left[\exp\!\left(\tfrac12\int_0^T\theta_t^2\,dt\right)\right] < \infty,$$

define the stochastic exponential (Doléans–Dade exponential):

$$Z_t = \mathcal E_t(\theta) := \exp\!\left(\int_0^t\theta_s\,dW_s - \tfrac12\int_0^t\theta_s^2\,ds\right).$$

$Z_t$ is a martingale

Under the Novikov condition, $Z_t$ is a $(\mathbb P,\mathcal F_t)$-martingale with $Z_0 = 1$.

SDE for $Z_t$

Apply Itô's formula to $f(t,x) = e^x$ where $x = \int_0^t\theta_s\,dW_s - \frac12\int_0^t\theta_s^2\,ds$:

$$dZ_t = Z_t\,\theta_t\,dW_t.$$

The drift vanishes — $Z_t$ satisfies an SDE with zero drift coefficient. Under the Novikov condition, $Z_t$ is a true $L^2$-martingale (not merely a local martingale).

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6.3 Girsanov's theorem

Girsanov's theorem

Let $\theta = (\theta_t)$ be adapted and satisfy the Novikov condition. Define the measure $\mathbb Q$ by

$$\frac{d\mathbb Q}{d\mathbb P}\bigg|_{\mathcal F_T} = Z_T = \exp\!\left(\int_0^T\theta_t\,dW_t - \tfrac12\int_0^T\theta_t^2\,dt\right).$$

Then $\mathbb Q$ is a probability measure equivalent to $\mathbb P$, and the process

$$\widetilde W_t = W_t - \int_0^t\theta_s\,ds$$

is a standard Brownian motion under $\mathbb Q$.

Proof sketch

By Lévy's characterization theorem, it suffices to show that $\widetilde W_t$ is a $\mathbb Q$-martingale with quadratic variation $[\widetilde W]_t = t$.

Quadratic variation. $d\widetilde W_t = dW_t - \theta_t\,dt$, so $(d\widetilde W_t)^2 = (dW_t)^2 = dt$. The quadratic variation is unchanged by Girsanov.

Martingale. For $s\le t$, using Bayes' rule under $\mathbb Q$:

$$\mathbb E^{\mathbb Q}[\widetilde W_t\mid\mathcal F_s] = \frac{\mathbb E^{\mathbb P}[Z_t\widetilde W_t\mid\mathcal F_s]}{Z_s}.$$

Write $\widetilde W_t = \widetilde W_s + \int_s^t d\widetilde W_r$. An Itô computation shows $Z_t\widetilde W_t - Z_s\widetilde W_s = \int_s^t(\cdots)\,dW_r$ (pure stochastic integral), so $\mathbb E^{\mathbb P}[Z_t\widetilde W_t-Z_s\widetilde W_s\mid\mathcal F_s]=0$, giving $\mathbb E^{\mathbb Q}[\widetilde W_t\mid\mathcal F_s] = \widetilde W_s$.

Girsanov: drift θ removed by change of measure

θ = 1.00

Radon–Nikodym density dQ/dPexp(−θW_T − ½θ²T)

W̃_t = W_t + θt is Q-BMdW̃ = dW + θdt

Red/orange paths (drifted under P) and blue paths (pure BM under Q) use IDENTICAL Brownian increments — only the measure labels differ. Girsanov shows how multiplying probabilities by exp(−θW_T − ½θ²T) exactly cancels the drift, transforming P into Q.

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6.4 The key formula: drift and measure

Under $\mathbb P$, the process $X_t = \int_0^t\theta_s\,ds + W_t$ has drift $\theta$.

Under $\mathbb Q$ (with $\widetilde W = W - \int\theta\,ds$), we can write $X_t = \int_0^t\theta_s\,ds + W_t = \widetilde W_t + 2\int_0^t\theta_s\,ds$. Wait — let's redo this cleanly.

Under $\mathbb P$: $W_t$ is BM, so $X_t = \int_0^t\theta_s\,ds + W_t$ has drift $\theta_t$.

Under $\mathbb Q$: $\widetilde W_t = W_t - \int_0^t\theta_s\,ds$ is BM, so $W_t = \widetilde W_t + \int_0^t\theta_s\,ds$, meaning $X_t = \widetilde W_t$: the drift is gone. Girsanov removes the drift.

More generally, if $dX_t = b_t\,dt + dW_t$ under $\mathbb P$, then under $\mathbb Q$ (with $\theta_t = -b_t$):

$$dX_t = d\widetilde W_t,$$

so $X$ becomes pure BM under $\mathbb Q$.

Drift as the geometry of measure change

A drift $\theta_t$ tilts the probability measure in path space: paths going upward become relatively more likely under $\mathbb P$ than under the baseline BM. Girsanov's theorem reverses this tilt: $Z_T$ reweights each path by a factor that exactly compensates the drift.

The stochastic exponential $Z_T = e^{\int\theta\,dW - \frac12\int\theta^2\,dt}$ is large for paths that move against the drift (providing extra weight) and small for paths that move with the drift (reducing their weight).

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6.5 Martingale representation theorem

Martingale representation (Clark–Ocone)

Let $(M_t)_{0\le t\le T}$ be any square-integrable $(\mathcal F_t)$-martingale (where $\mathcal F_t = \sigma(W_s: s\le t)$). Then there exists a unique adapted process $H\in\mathcal L^2$ such that

$$M_t = M_0 + \int_0^t H_s\,dW_s.$$

This theorem says that every martingale in the Brownian filtration is an Itô integral. There are no other sources of randomness — Brownian motion is the only martingale "atom."

Completeness and perfect hedging

In a model driven by a single Brownian motion, the martingale representation theorem implies that every contingent claim can be perfectly replicated by a dynamic trading strategy in the underlying asset. The replication is exactly the stochastic integral $\int H_t\,dS_t$, where $H_t$ is the hedge ratio.

Markets with $d$ independent Brownian motions and $d$ tradable assets are complete: every claim is replicable. Incompleteness (stochastic volatility, jumps) introduces additional sources of randomness not spanned by the tradable assets.

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6.6 Finance: the risk-neutral measure

In the Black–Scholes model:

$$dS_t = \mu S_t\,dt + \sigma S_t\,dW_t^{\mathbb P},$$

the discounted stock price $\tilde S_t = e^{-rt}S_t$ satisfies

$$d\tilde S_t = (\mu-r)\tilde S_t\,dt + \sigma\tilde S_t\,dW_t^{\mathbb P}.$$

For $\tilde S_t$ to be a martingale (as required by no-arbitrage), the drift must vanish. Apply Girsanov with $\theta_t = -\frac{\mu-r}{\sigma}$ (the market price of risk):

$$\widetilde W_t = W_t^{\mathbb P} + \frac{\mu-r}{\sigma}t$$

is a $\mathbb Q$-Brownian motion. Under $\mathbb Q$:

$$dS_t = r S_t\,dt + \sigma S_t\,d\widetilde W_t.$$

The drift equals the risk-free rate: the stock earns $r$ under $\mathbb Q$.

Risk-neutral pricing formula

Under the risk-neutral measure $\mathbb Q$, the price of any replicable claim with payoff $g(S_T)$ is

$$V_t = e^{-r(T-t)}\mathbb E^{\mathbb Q}[g(S_T)\mid\mathcal F_t].$$

The measure $\mathbb Q$ is called risk-neutral because under it all assets earn the risk-free rate — investors are indifferent to risk. The real-world drift $\mu$ disappears from the pricing formula. This is why you cannot infer $\mu$ from option prices.

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6.7 Worked example: pricing a binary call

A binary call pays \1$if$S_T > K$, zero otherwise. Under$\mathbb Q$:

$$V_0 = e^{-rT}\mathbb Q(S_T > K).$$

Under $\mathbb Q$, $S_T = S_0\exp((r-\sigma^2/2)T + \sigma\widetilde W_T)$. Then

$$\mathbb Q(S_T > K) = \mathbb Q\!\left(\widetilde W_T > \frac{\log(K/S_0)-(r-\sigma^2/2)T}{\sigma}\right) = N(d_2),$$

where $d_2 = \frac{\log(S_0/K)+(r-\sigma^2/2)T}{\sigma\sqrt{T}}$.

Therefore $V_0 = e^{-rT}N(d_2)$. The $N(d_2)$ component of Black–Scholes is exactly the risk-neutral probability that the option expires in the money.

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6.8 Problems

Problem 1: Girsanov density

Verify that $\mathbb E^{\mathbb P}[Z_T] = 1$ for $Z_T = e^{\theta W_T - \theta^2 T/2}$ (constant $\theta$). Compute $\mathbb E^{\mathbb P}[Z_T^2]$.

Problem 2: Measure change for OU

Under $\mathbb P$, let $dX = -\kappa X\,dt + dW$ (OU toward zero). Find the measure $\mathbb Q$ under which $X$ is a pure Brownian motion. Write down the Girsanov density $dQ/dP$.

Problem 3: Risk-neutral GBM

Under the risk-neutral measure $\mathbb Q$, verify that $\mathbb E^{\mathbb Q}[S_T] = S_0 e^{rT}$, consistent with risk-free growth.

Problem 4: Two-sided Girsanov

Suppose under $\mathbb P$, $X_t = \theta t + W_t$. Show that under $\mathbb Q$ with $dQ/dP = e^{-\theta W_T - \theta^2 T/2}$, $X_t$ is a $\mathbb Q$-Brownian motion.

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6.9 Selected solutions

Solution 1

$\mathbb E[Z_T] = \mathbb E[e^{\theta W_T - \theta^2 T/2}]$. Since $W_T\sim N(0,T)$:

$$\mathbb E[e^{\theta W_T}] = e^{\theta^2 T/2},$$

so $\mathbb E[Z_T] = e^{\theta^2 T/2}\cdot e^{-\theta^2 T/2} = 1$. For $\mathbb E[Z_T^2] = \mathbb E[e^{2\theta W_T - \theta^2 T}] = e^{2\theta^2 T}\cdot e^{-\theta^2 T} = e^{\theta^2 T}$.

Solution 2

The OU drift is $b_t = -\kappa X_t$. To remove this drift via Girsanov, set $\theta_t = \kappa X_t$ (the market price of risk equals the negative of the drift). Under $\mathbb Q$ with Radon–Nikodym derivative

$$\frac{d\mathbb Q}{d\mathbb P} = \exp\!\left(\kappa\int_0^T X_t\,dW_t - \tfrac12\kappa^2\int_0^T X_t^2\,dt\right),$$

the process $\widetilde W_t = W_t - \kappa\int_0^t X_s\,ds$ is a $\mathbb Q$-Brownian motion, and $dX = d\widetilde W$, making $X$ pure BM.

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6.10 What this chapter gives you

Concept	Content
Equivalent measures	same null sets, related by $Z_T > 0$ density
Stochastic exponential	$Z_t = e^{\int\theta\,dW - \frac12\int\theta^2\,dt}$, a martingale
Girsanov's theorem	removes drift: $\widetilde W = W - \int\theta\,ds$ is $\mathbb Q$-BM
Market price of risk	$\theta = (\mu-r)/\sigma$ converts $\mathbb P$ to $\mathbb Q$
Martingale representation	every BM-martingale is an Itô integral
Risk-neutral pricing	$V_t = e^{-r(T-t)}\mathbb E^{\mathbb Q}[g(S_T)\mid\mathcal F_t]$

The next chapter applies all of this to derive Black–Scholes from first principles.