Girsanov's Theorem: Drift Changes and Risk-Neutral Pricing

What Girsanov's Theorem Actually Changes

Girsanov's theorem changes the drift seen under a probability measure. It does not turn a "Brownian motion with drift" into a martingale by relabelling it. Rather, it starts with a Brownian motion under $\mathbb P$, constructs an equivalent measure $\mathbb Q$, and identifies a shifted process that is Brownian under $\mathbb Q$.

That distinction keeps the signs straight. It also explains the theorem's role in finance: an asset may have physical drift $\mu$ under $\mathbb P$, yet drift at the short rate $r$ under a pricing measure $\mathbb Q$. The Radon-Nikodym derivative is the likelihood ratio between those measures; Girsanov tells us what it does to Brownian-driven dynamics.

I. V. Girsanov published the general result in 1960. The original paper and abstract remain useful for one central fact: equivalent diffusion measures can alter drift while retaining the same diffusion matrix.

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A Sign-Consistent Statement

Fix a horizon $T$. Let $W^{\mathbb P}$ be a Brownian motion on a filtered probability space $(\Omega,\mathcal F,(\mathcal F_t),\mathbb P)$. Let $\theta$ be progressively measurable, with $\int_0^T\theta_s^2\,ds<\infty$ almost surely, and suppose Novikov's sufficient condition holds:

$$\mathbb E^{\mathbb P}\!\left[ \exp\!\left(\frac12\int_0^T\theta_s^2\,ds\right) \right]<\infty.$$

Define the stochastic exponential

$$Z_t=\exp\!\left( -\int_0^t\theta_s\,dW_s^{\mathbb P} -\frac12\int_0^t\theta_s^2\,ds \right).$$

Novikov's condition makes $Z$ a true $\mathbb P$-martingale, not merely a non-negative local martingale. We may therefore define $\mathbb Q$ on $\mathcal F_T$ by

$$\left.\frac{d\mathbb Q}{d\mathbb P}\right|_{\mathcal F_T}=Z_T.$$

Then

$$W_t^{\mathbb Q}=W_t^{\mathbb P}+\int_0^t\theta_s\,ds$$

is a Brownian motion under $\mathbb Q$.

This convention uses a minus sign in the density and a plus sign in the Brownian shift. Some books define $\lambda=-\theta$, which reverses both signs. Either convention works; mixing the two does not.

The University of Chicago lecture notes derive the exponential density, state Novikov's criterion and prove the Brownian shift with one consistent sign convention.

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Why the Density Works

Apply Itô's formula to the exponential. Its differential is

$$dZ_t=-Z_t\theta_t\,dW_t^{\mathbb P}.$$

The $dt$ term from the exponent cancels the quadratic-variation correction. This proves only that $Z$ is a local martingale. Novikov supplies the missing integrability and gives $\mathbb E^{\mathbb P}[Z_T]=1$. Since $Z_T>0$ almost surely, $\mathbb P$ and $\mathbb Q$ are equivalent on $\mathcal F_T$.

For the shifted process,

$$[W^{\mathbb Q}]_t=[W^{\mathbb P}]_t=t,$$

because the added time integral has finite variation. To check the martingale part, use integration by parts:

$$d(Z_tW_t^{\mathbb Q}) =Z_t\,dW_t^{\mathbb Q} +W_t^{\mathbb Q}\,dZ_t +d[Z,W^{\mathbb Q}]_t.$$

Substituting $dW_t^{\mathbb Q}=dW_t^{\mathbb P}+\theta_tdt$ and $dZ_t=-Z_t\theta_t dW_t^{\mathbb P}$ gives

$$d(Z_tW_t^{\mathbb Q}) =Z_t(1-\theta_tW_t^{\mathbb Q})\,dW_t^{\mathbb P}.$$

The drift cancels. After localisation, Bayes' formula shows that $W^{\mathbb Q}$ is a continuous $\mathbb Q$-local martingale. Lévy's characterisation then makes it a $\mathbb Q$-Brownian motion. The localisation qualification matters: "zero drift" by itself is not a proof that an unbounded process is a true martingale.

Conditional expectations transform through Bayes' rule:

$$\mathbb E^{\mathbb Q}[X\mid\mathcal F_t] =\frac{\mathbb E^{\mathbb P}[Z_TX\mid\mathcal F_t]}{Z_t},$$

whenever the quantities are integrable. This identity is often more useful in code than the theorem's pathwise wording.

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The Drift Rule for a General Diffusion

The reusable calculation is not confined to geometric Brownian motion. Suppose an $n$-dimensional state process has $\mathbb P$-dynamics

$$dX_t=b^{\mathbb P}(t,X_t)\,dt +\Sigma(t,X_t)\,dW_t^{\mathbb P},$$

where $W$ has $m$ components and $\Sigma$ is $n\times m$. Under the convention above,

$$dW_t^{\mathbb P}=dW_t^{\mathbb Q}-\theta_t\,dt.$$

Substitution gives

$$dX_t= \bigl(b^{\mathbb P}(t,X_t)-\Sigma(t,X_t)\theta_t\bigr)dt +\Sigma(t,X_t)\,dW_t^{\mathbb Q}.$$

Hence

$$b^{\mathbb Q}=b^{\mathbb P}-\Sigma\theta.$$

This is the line to audit in any implementation. The volatility matrix is identical on both sides, the matrix dimensions match, and the sign agrees with the density. If a derivation produces $b^{\mathbb P}+\Sigma\theta$ while retaining $Z=\mathcal E(-\int\theta^{\mathsf T}dW^{\mathbb P})$, one convention has been reversed.

The same rule also clarifies what Girsanov cannot do. A desired drift shift must lie in the column space of $\Sigma$. A component of drift orthogonal to all Brownian volatility directions cannot be removed by choosing $\theta$.

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Black-Scholes Measure Change

Under the physical measure, suppose a non-dividend-paying stock follows

$$dS_t=\mu S_t\,dt+\sigma S_t\,dW_t^{\mathbb P}, \qquad \sigma>0.$$

With a constant short rate $r$, set

$$\theta=\frac{\mu-r}{\sigma}.$$

Since $dW_t^{\mathbb P}=dW_t^{\mathbb Q}-\theta\,dt$,

$$\begin{aligned} dS_t &=\mu S_t\,dt+\sigma S_t(dW_t^{\mathbb Q}-\theta\,dt)\\ &=(\mu-\sigma\theta)S_t\,dt+\sigma S_t\,dW_t^{\mathbb Q}\\ &=rS_t\,dt+\sigma S_t\,dW_t^{\mathbb Q}. \end{aligned}$$

Consequently $e^{-rt}S_t$ is a $\mathbb Q$-martingale. In the idealised one-stock, one-Brownian Black-Scholes market, the market is complete and this equivalent martingale measure is unique. Suitable claims can then be valued by the risk-neutral pricing formula. Girsanov alone does not prove completeness or make every payoff attainable.

The quantity $\theta=(\mu-r)/\sigma$ is the market price of Brownian risk under this sign convention. If time is measured in years, $\mu$ and $r$ have units year$^{-1}$, $\sigma$ has units year$^{-1/2}$, and $\theta$ has units year$^{-1/2}$. Thus $\theta W_T$ and $\theta^2T$ are dimensionless, as an exponential requires.

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Checked Numerical Example

Take $S_0=100$, $\mu=10\%$, $r=5\%$, $\sigma=20\%$ and $T=1$ year. Then

$$\theta=\frac{0.10-0.05}{0.20}=0.25\ \text{year}^{-1/2}$$

and

$$Z_T=\exp(-0.25W_T^{\mathbb P}-0.03125).$$

On the particular path $W_T^{\mathbb P}=0.5\sqrt{\text{year}}$,

$$Z_T=e^{-0.15625}=0.855345\ldots.$$

This is a likelihood ratio at that path, not a probability of the path. Positive Brownian realisations are down-weighted here because $\mu>r$.

For a one-year call with $K=100$, the $\mathbb Q$-dynamics give

$$S_T=100\exp\!\left((0.05-0.02)+0.20W_T^{\mathbb Q}\right).$$

The Black-Scholes inputs are $d_1=0.35$ and $d_2=0.15$, so

$$C_0=100\Phi(0.35)-100e^{-0.05}\Phi(0.15) =10.4506.$$

Both the density value and the option price follow directly from the displayed parameters. See the Black-Scholes derivation for the lognormal integral.

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Measure Change Is Not Discounting

Two operations are often compressed into the phrase "go risk-neutral". Girsanov changes the probability law. Choosing a numeraire determines which prices must be martingales after normalisation.

With the bank account $B_t$ as numeraire, the pricing condition is that $S_t/B_t$ is a local martingale under $\mathbb Q^B$. With a zero-coupon bond $P(t,T)$ as numeraire, it is $S_t/P(t,T)$ that must be a local martingale under the $T$-forward measure. The drift adjustment between these measures follows from Girsanov, but the numeraire theorem identifies the density process.

This separation matters when rates are stochastic. The money-market price of a payoff $X_T$ is

$$V_t=B_t\mathbb E^{\mathbb Q^B}\!\left[ \frac{X_T}{B_T}\,\middle|\,\mathcal F_t \right],$$

not generally $e^{-r(T-t)}\mathbb E^{\mathbb Q}[X_T\mid\mathcal F_t]$ with a fixed $r$. A measure change removes the appropriate risk premia; it does not license pulling a random discount factor outside the expectation.

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Several Risk Factors

For an $m$-dimensional Brownian motion and vector process $\theta_t$,

$$Z_t=\exp\!\left( -\int_0^t\theta_s^{\mathsf T}dW_s^{\mathbb P} -\frac12\int_0^t\|\theta_s\|^2ds \right), \qquad W_t^{\mathbb Q}=W_t^{\mathbb P}+\int_0^t\theta_sds.$$

If traded assets have excess-drift vector $\mu-r\mathbf 1$ and volatility matrix $\Sigma$, a candidate market-price-of-risk vector solves

$$\Sigma\theta=\mu-r\mathbf 1.$$

There need not be one solution. No solution signals inconsistency with an equivalent Brownian pricing measure; several solutions are one route to market incompleteness. It is therefore wrong to say that every asset freely has its own unrelated market price of risk.

In rates, a change of numeraire changes the associated measure and Girsanov supplies the drift adjustment. The HJM framework imposes its no-arbitrage forward-rate drift this way. The LIBOR market model moves among forward measures when different rates are most convenient as martingales.

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Limits and Failure Modes

Novikov is sufficient, not necessary. A density may be a true martingale even when Novikov fails, and Kazamaki's criterion may still apply. Conversely, a positive local martingale can be strict, with expectation below one; then it cannot define the intended probability measure.

Equivalent measure change preserves pathwise quadratic variation. For a diffusion this means the diffusion coefficient is not changed by Girsanov, although its distribution under the new measure can change when the coefficient depends on the state. In semimartingale versions, the compensator of jumps may change too. The continuous covariance characteristic still does not.

Finally, constructing an equivalent measure is not the same as identifying the economically appropriate price in an incomplete market. Several martingale measures can satisfy no-arbitrage. Preferences, calibration or a hedging criterion must then select among them.

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Frequently Asked Questions

What does Girsanov's theorem say?

An exponential change of measure converts $W^{\mathbb P}+\int\theta\,dt$ into Brownian motion under $\mathbb Q$. Equivalently, it changes drift while preserving continuous quadratic variation.

Why is the sign easy to get wrong?

There are two common conventions. With $d\mathbb Q/d\mathbb P=\mathcal E(-\int\theta\,dW^{\mathbb P})_T$, the new Brownian motion is $W^{\mathbb Q}=W^{\mathbb P}+\int\theta\,dt$. Reverse one sign only and the Black-Scholes drift will not become $r$.

Is Novikov's condition necessary?

No. It is a convenient sufficient condition ensuring the stochastic exponential is a true martingale. Bounded $\theta$ on a finite horizon satisfies it; vague conditions such as "slow growth" do not guarantee it.

Can the market price of risk be stochastic?

Yes. The theorem allows progressively measurable $\theta_t$, provided the stochastic integral exists and the density is a true martingale. Novikov is one way, but not the only way, to establish that.

Does Girsanov change volatility?

It does not change the continuous quadratic variation or diffusion coefficient in the transformed SDE. A state-dependent volatility process may nevertheless have a different law after the drift change.

Does Girsanov cover jumps?

Semimartingale versions can change both continuous drift characteristics and jump compensators. They require a density with jump terms, not merely the Brownian exponential displayed here.