advanced33 min studyLesson 679 in Mathematics

Girsanov theorem and drift changes

Choose a valid density martingale and track the Brownian and drift signs under the new measure.

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What you’ll understand

  • State a finite-horizon Brownian Girsanov theorem with an explicit density convention.
  • Use Novikov's condition as a sufficient true-martingale criterion.
  • Compute the transformed drift without changing the pathwise quadratic variation.

Changing measure reweights possible paths without rewriting the realized path values. Under a suitable density, Brownian motion with an added finite-variation correction becomes Brownian for the new measure, allowing a diffusion drift to change. The exponential sign in the density determines the sign in that correction. We fix a finite horizon, a usual Brownian filtration, and one convention throughout, then verify the signs by substituting the corrected driver back into the original SDE.

A local density is not enough

Let θ\theta be predictable with a finite squared time integral almost surely on the horizon. The exponential with negative stochastic integral and negative half of its squared integral is a positive local martingale. To define a probability measure by its terminal value, it must have expectation one. A strict local martingale can have smaller expectation, so nonnegativity and a formally canceled drift do not by themselves supply a probability density. This is the central integrability requirement of the theorem.

Novikov's condition, finiteness of the expected exponential of half the accumulated squared shift, is a convenient sufficient criterion for the exponential to be a true martingale. It is not necessary in full generality. A bounded shift on a finite horizon satisfies it immediately. The assertion is horizon-specific: true-martingale densities for every finite time need not produce a uniformly integrable density on the infinite horizon or an equivalent measure on the full infinite-time path sigma algebra.

Zt=exp⁡ ⁣(−∫0tθs dWs−12∫0tθs2ds),Eexp⁡ ⁣(12∫0Tθs2ds)<∞Z_t=\exp\!\left(-\int_0^t\theta_s\,dW_s-\tfrac12\int_0^t\theta_s^2ds\right),\qquad \mathbb E\exp\!\left(\tfrac12\int_0^T\theta_s^2ds\right)<\infty
The displayed sufficient condition makes the positive local density a mean-one martingale on the finite horizon.↔ Scroll wide equations horizontally.

The corrected driver fixes the drift sign

Define the new measure on the terminal sigma algebra by density ZTZ_T relative to the original measure. Because the exponential is strictly positive almost surely, the two finite-horizon measures are equivalent. Under the new measure, the corrected process WtQ=Wt+∫0tθsdsW_t^Q=W_t+\int_0^t\theta_sds is Brownian for the filtration on that horizon. With the opposite sign in the stochastic exponential, the Brownian correction would have the opposite sign as well.

Substitute dW=dWQ−θdtdW=dW^Q-\theta dt into the original SDE. Its new drift is b−σθb-\sigma\theta, while the diffusion coefficient is unchanged. In several dimensions the same calculation uses a vector shift multiplied by the noise matrix. A drift change outside that matrix's range cannot be produced by such a shift. Pathwise quadratic variation is unchanged by adding finite variation or by passing between equivalent measures, even though the distribution and expected trajectory can change.

dQdP∣FT=ZT,WtQ=Wt+∫0tθsds,dXt=(bt−σtθt)dt+σt dWtQ\frac{dQ}{dP}\bigg|_{\mathcal F_T}=Z_T,\qquad W_t^Q=W_t+\int_0^t\theta_sds,\qquad dX_t=(b_t-\sigma_t\theta_t)dt+\sigma_t\,dW_t^Q
The plus correction to the new Brownian motion accompanies the negative sign in the chosen density exponential.↔ Scroll wide equations horizontally.

Reweighting has probabilistic limits

Equivalence preserves null events on the chosen horizon, not numerical probabilities or expected values. A process with one mean under the original measure can have another mean under the new one. Likewise, a drifted coordinate process need not become standard Brownian unless its new drift is zero and its covariance and initial value have the standard form. A volatility of two still has variance growth four after drift removal, so rescaling remains necessary.

When a desired shift is unbounded, stopping can verify local versions, but a global measure change still requires a mean-one terminal density or another justified criterion. One must not silently remove the stopping times from a density argument just because an SDE solution itself is nonexplosive. Integrability of the likelihood and nonexplosion of the state are distinct properties. Keeping them separate is particularly important when measure changes are used as a weak-existence method rather than merely to simplify a solved equation.

Remove a positive constant drift

Under P let Xt=1+4t+2WtX_t=1+4t+2W_t and take horizon T=3/4T=3/4. Choose a constant shift removing the drift under Q, verify Novikov, and find the endpoint law under both measures.

  1. Choose θ=2\theta=2, so b−σθ=4−2⋅2=0b-\sigma\theta=4-2\cdot2=0. The density is ZT=e−2WT−2TZ_T=e^{-2W_T-2T} and the corrected Brownian motion is WtQ=Wt+2tW_t^Q=W_t+2t.

  2. Novikov's exponential is deterministic, equal to e2T=e3/2e^{2T}=e^{3/2}, hence finite. Gaussian exponential moments also give EPZT=e−2Te2T=1\mathbb E_PZ_T=e^{-2T}e^{2T}=1. The proposed terminal density therefore defines an equivalent probability measure.

  3. Under Q the same coordinate process is Xt=1+2WtQX_t=1+2W_t^Q. At the horizon its law is N(1,3)N(1,3), whereas under P it is N(4,3)N(4,3). The mean changed but the quadratic-variation value 4T=34T=3 did not.

The result

The corrected sign removes the drift while retaining volatility two. The coordinate process under Q is a scaled and shifted Brownian motion, not a standard Brownian motion.

Explore the model

Interactive lab

Interactive lab · probability

A finite analogue of a likelihood tilt

Pause/Reset, then set both probability to 0.25 and trials to 8. Predict mean two and variance 1.5. Reset and set both probability to 0.75 and trials to 8; predict mean six and variance 1.5. For each count k, the ratio of the new bar to the old bar is 32k−83^{2k-8}.

A finite analogue of a likelihood tiltThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forExact finite likelihood reweighting illustrates how a measure change alters means. It is not the Brownian Girsanov density or a sampled importance-sampling experiment.

Check your understanding

With the negative stochastic exponential used here, what is the correct new drift?

Try it yourself

Under P let Xt=−t+3WtX_t=-t+3W_t. On any fixed finite horizon, choose the constant shift and density that make its Q drift zero. Write the corrected Brownian motion and check the transformed equation.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Solve −1−3θ=0-1-3\theta=0 using the fixed negative-exponential convention rather than changing conventions midway.

Compare with the solution

The shift is θ=−1/3\theta=-1/3. Thus ZT=exp⁡(WT/3−T/18)Z_T=\exp(W_T/3-T/18) and WtQ=Wt−t/3W_t^Q=W_t-t/3. Constant-shift Novikov holds on every fixed finite horizon. Substituting Wt=WtQ+t/3W_t=W_t^Q+t/3 gives Xt=−t+3WtQ+t=3WtQX_t=-t+3W_t^Q+t=3W_t^Q. The transformed drift vanishes while quadratic variation remains 9t9t. The positive sign of the stochastic term in this particular density comes from the negative value of the chosen shift.

Take it with you

  • A probability density needs a true mean-one martingale, not merely a local one.
  • Novikov is sufficient and finite-horizon, rather than universally necessary.
  • The density sign and corrected Brownian sign determine the drift together.

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