Changing the Mean-Reversion Level
An Ornstein–Uhlenbeck (OU) process is a diffusion whose drift pulls the state toward a long-run mean. Changing the probability measure can change the strength of that pull without changing the paths being considered or the diffusion coefficient. The central calculation is to choose the Girsanov kernel with the correct sign, then check that the resulting stochastic exponential is a genuine density on the chosen time horizon.
Fix a finite horizon \(T>0\) and work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Assume \(\kappa>0\), \(\bar\kappa>0\), and \(\sigma>0\), and let \(m,x_0\) be constants. Under \(\mathbb{P}\), suppose \(X_0=x_0\) and
Here \(\kappa\) is the original mean-reversion level, \(m\) is the mean level, and \(\sigma\) is the diffusion coefficient. We seek a probability measure \(\mathbb{Q}\), equivalent to \(\mathbb{P}\) on \(\mathcal{F}_T\), under which the same process has drift \(\bar\kappa(m-X_t)\) and the same diffusion coefficient.
The sign follows from the convention in Girsanov’s theorem: if \(Z\) is a true density martingale and \(d\mathbb{Q}=Z_T\,d\mathbb{P}\) on \(\mathcal{F}_T\), then \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds\) is Brownian motion under \(\mathbb{Q}\). Thus \(dB_t=dB_t^{\mathbb{Q}}-\theta_t\,dt\). Substitution into the original equation gives drift \(\kappa(m-X_t)-\sigma\theta_t\), which is \(\bar\kappa(m-X_t)\) for the stated \(\theta_t\).
Worked Example: Checking the Kernel Sign
Take \(\kappa=2\), \(\bar\kappa=1\), and \(\sigma=1\). Then \(\Delta=2-1=1\), so the kernel is \(\theta_t=m-X_t\). Under \(\mathbb{Q}\),
The new mean-reversion level is therefore \(1\), as required. If instead one used the opposite sign, \(\theta_t=X_t-m=-(m-X_t)\), the drift would be \(2(m-X_t)-\theta_t=3(m-X_t)\). That choice changes the level to \(3\), not \(1\).
The Likelihood in Terms of the OU Path
The stochastic-exponential expression for \(Z_T\) is useful, but it still contains an Ito integral. For OU paths, that integral can be eliminated in favour of the endpoint and the time integral of the squared displacement. Set \(Y_t=X_t-m\) and \(y_0=x_0-m\). Under \(\mathbb{P}\), \(dY_t=-\kappa Y_t\,dt+\sigma\,dB_t\), while \(\theta_t=-\Delta Y_t/\sigma\).
Proof. Apply Ito’s formula to \(f(y)=y^2\). Its derivatives are \(f'(y)=2y\) and \(f''(y)=2\). Since \(d[Y]_t=\sigma^2dt\), the correction term is \(\tfrac12 f''(Y_t)\sigma^2dt=\sigma^2dt\), and hence
Integrating and rearranging gives \(\int_0^T Y_s\,dB_s=(Y_T^2-y_0^2+2\kappa\int_0^T Y_s^2\,ds-\sigma^2T)/(2\sigma)\). The stochastic-exponential definition and \(\theta=-\Delta Y/\sigma\) give \(\log Z_T=(\Delta/\sigma)\int_0^T Y_s\,dB_s-(\Delta^2/(2\sigma^2))\int_0^T Y_s^2\,ds\). Substituting the integral identity yields the endpoint coefficient \(\Delta/(2\sigma^2)\), the constant term \(-\Delta T/2\), and the coefficient of \(\int Y_s^2ds\) equal to \((2\kappa\Delta-\Delta^2)/(2\sigma^2)\). Since \(2\kappa\Delta-\Delta^2=\Delta(\kappa+\bar\kappa)=\kappa^2-\bar\kappa^2\), the stated expression follows. \(\square\)
The constant \(-\Delta T/2\) comes from the Ito correction. Omitting it, or changing either sign, does not give the logarithm of the stochastic exponential.
Worked Example: Evaluating the Log-Likelihood
Use \(\kappa=2\), \(\bar\kappa=1\), \(\sigma=1\), \(T=1\), \(y_0=0\), and suppose the path quantities are \(Y_1^2=1\) and \(\int_0^1Y_s^2\,ds=1\). These values illustrate evaluation of the pathwise formula; they are not an assertion that a particular path has positive probability.
Here \(\Delta=2-1=1\) and \(\kappa^2-\bar\kappa^2=4-1=3\). Substitution gives
Check: the stochastic-exponential logarithm is \(-\int_0^1( m-X_s)\,dB_s-\tfrac12\int_0^1(m-X_s)^2ds=\int_0^1Y_s\,dB_s-\tfrac12\int_0^1Y_s^2ds\). The Ito identity gives \(\int_0^1Y_s\,dB_s=(1-0+2\cdot2\cdot1-1)/2=2\), so this expression is \(2-\tfrac12=\tfrac32\), matching the endpoint formula.
When the Density Is a Probability Density
A positive local martingale is not automatically a true martingale, so the formal change of drift alone does not establish a probability measure. Novikov’s condition is a standard sufficient criterion: for a progressively measurable \(\theta\), if \(\mathbb{E}\exp(\tfrac12\int_0^T\theta_s^2ds)<\infty\), then \(Z_t=\mathcal{E}(-\int_0^\cdot\theta_s\,dB_s)_t\) is a true martingale on \([0,T]\), with \(\mathbb{E}[Z_T]=1\). This criterion and Girsanov’s theorem are standard results from the earlier material on exponential martingales and change of measure. Under these hypotheses, \(Z_t=\mathbb{E}[Z_T\mid\mathcal{F}_t]\), so the density has the required conditional-expectation property, and \(B^{\mathbb{Q}}\) is Brownian under \(\mathbb{Q}\).
For this OU kernel, \(\tfrac12\int_0^T\theta_s^2ds=\dfrac{\Delta^2}{2\sigma^2}\int_0^T Y_s^2ds\). The centered process \(Y_t-y_0e^{-\kappa t}\) is Gaussian. Its covariance kernel is \(C(s,t)=\dfrac{\sigma^2}{2\kappa}\big(e^{-\kappa|t-s|}-e^{-\kappa(s+t)}\big)\). The associated covariance operator on \(L^2([0,T])\) has norm at most \(\sigma^2/\kappa^2\): indeed, \(0\le C(s,t)\le\dfrac{\sigma^2}{2\kappa}e^{-\kappa|t-s|}\), and the integral of this upper bound in \(t\) over \([0,T]\) is at most \(\sigma^2/\kappa^2\), uniformly in \(s\). The same bound holds with \(s,t\) interchanged.
We use the standard Gaussian quadratic-integrability result: a Gaussian random element in a Hilbert space with covariance operator \(C\) has a finite exponential moment of its squared norm when the coefficient \(a\) satisfies \(2a\lVert C\rVert<1\). This sufficient result follows from the Gaussian quadratic-form formula; its proof is not needed here. A deterministic mean in the Hilbert space does not change this strict sufficient condition. Taking \(a=\Delta^2/(2\sigma^2)\), a convenient sufficient condition for Novikov is therefore \(|\Delta|<\kappa\). It is not a necessary condition, and failure of this bound does not by itself prove that \(Z\) fails to be a true martingale.
Worked Example: A Novikov Test for the OU Kernel
Again take \(\kappa=2\), \(\bar\kappa=1\), and \(\sigma=1\), but now consider any fixed finite \(T\) and any deterministic initial value \(x_0\). The parameter difference is \(\Delta=2-1=1\), and \(|\Delta|=1<2=\kappa\). The covariance-operator bound gives \(\lVert C\rVert\le1^2/2^2=1/4\). The exponential coefficient is \(a=\Delta^2/(2\sigma^2)=1/2\), so \(2a\lVert C\rVert\le2(1/2)(1/4)=1/4<1\). Thus the Gaussian quadratic-integrability result gives \(\mathbb{E}\exp(\tfrac12\int_0^T\theta_s^2ds)<\infty\), for every finite \(T\). Novikov’s condition applies, \(Z\) is a true martingale, and \(\mathbb{E}[Z_T]=1\).
Check: the strict criterion compares \(|\kappa-\bar\kappa|=|2-1|=1\) with \(\kappa=2\), not with the target level alone. The test verifies integrability; it does not say the random variable \(Z_T\) equals one path by path.
Dynamics and Mean Under the Changed Measure
When the stated sufficient condition holds, define \(\mathbb{Q}\) on \(\mathcal{F}_T\) by \(d\mathbb{Q}=Z_T\,d\mathbb{P}\). Positivity of the exponential and \(\mathbb{E}[Z_T]=1\) make this a probability measure; since \(Z_T>0\) almost surely, it is equivalent to \(\mathbb{P}\) on \(\mathcal{F}_T\). Girsanov’s theorem gives a \(\mathbb{Q}\)-Brownian motion \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_sds\), and substitution proves that \(X\) satisfies \(dX_t=\bar\kappa(m-X_t)dt+\sigma\,dB_t^{\mathbb{Q}}\).
This is a weak-solution statement under the new measure: the same coordinate process is viewed with a different probability law. Since the coefficients are globally Lipschitz and have linear growth, the explicit solution is also the unique strong solution for a given Brownian motion under either measure. Under \(\mathbb{Q}\), it is
The stochastic integral has mean zero: its integrand is deterministic and square-integrable on every finite interval, so the Ito integral is an integrable martingale. Consequently \(\mathbb{E}^{\mathbb{Q}}[X_t]=m+(x_0-m)e^{-\bar\kappa t}\). For the parameters in the first example with \(m=3\) and \(x_0=4\), this mean at \(t=1\) is \(3+(4-3)e^{-1}=3+e^{-1}\). Under the original dynamics it is \(3+(4-3)e^{-2}=3+e^{-2}\), consistent with the change in mean-reversion level.
Common Pitfalls
Do not reverse the kernel sign. With the convention \(dB=dB^{\mathbb{Q}}-\theta\,dt\), the new drift is the old drift minus \(\sigma\theta\). For \(\kappa=2,\bar\kappa=1,\sigma=1\), using \(\theta=X-m\) would produce \(2(m-X)-(X-m)=3(m-X)\), not \(m-X\).
Do not treat Novikov’s condition as necessary. The bound \(|\kappa-\bar\kappa|<\kappa\) is a convenient sufficient test obtained from a covariance estimate. If it is not met, the test is inconclusive; one must not infer that the density fails. Also distinguish the probability measures: expectations under the original model are \(\mathbb{E}\), while expectations after the change are \(\mathbb{E}^{\mathbb{Q}}\).
Check Your Understanding
In each question, assume \(\kappa,\bar\kappa,\sigma>0\) and that the density satisfies the stated integrability condition.
- For \(\kappa=3\), \(\bar\kappa=2\), and diffusion coefficient \(\sigma\), write the Girsanov kernel and verify the new drift by substitution.
- Starting from \(dY_t=-\kappa Y_tdt+\sigma dB_t\), derive the Ito identity for \(\int_0^T Y_s\,dB_s\), including the correction term.
- For \(\kappa=4\) and \(\bar\kappa=1\), does the sufficient bound \(|\kappa-\bar\kappa|<\kappa\) verify Novikov’s condition?
- Explain why \(\mathbb{E}[Z_T]=1\) is needed before \(Z_T\) can define a probability measure by \(d\mathbb{Q}=Z_Td\mathbb{P}\).
- Give the mean of \(X_t\) under the changed measure in terms of \(x_0,m,\bar\kappa\), and \(t\).