Mean Reversion with Changing Coefficients
In a mean-reverting SDE, the drift pulls the state toward a target level, while Brownian noise continually perturbs it. If the strength of reversion, target, or volatility changes with time, the solution still has an explicit form: past values and past noise are weighted by an integrating kernel. That kernel also gives the conditional mean and variance directly.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to \((\mathcal{F}_t)\). In particular, Brownian increments after time \(s\) are independent of \(\mathcal{F}_s\). We use \(B_t\) for Brownian motion. Let \(\kappa_t,m_t,\sigma_t\) be deterministic measurable functions. On each finite time interval \([0,T]\), assume \(\int_0^T|\kappa_u|\,du<\infty\), \(\int_0^T|\kappa_um_u|\,du<\infty\), and \(\int_0^T\sigma_u^2\,du<\infty\). Assume \(X_0\) is \(\mathcal{F}_0\)-measurable and square-integrable.
These assumptions make the deterministic integrals finite and the stochastic integrals well-defined. For the latter, the integrand must be adapted and progressively measurable with \(\mathbb{E}\int_0^T H_u^2\,du<\infty\); the deterministic integrands used below satisfy this condition. The equation is understood in its integral form, with drift coefficient \(\kappa_t(m_t-X_t)\) and diffusion coefficient \(\sigma_t\). When \(\kappa_t\ge0\), the drift points toward \(m_t\) for states above or below that target, which is the mean-reverting case. Bounded \(\kappa_t\) gives a drift globally Lipschitz in \(x\), with Lipschitz constant \(|\kappa_t|\); if \(\kappa_t\) and \(\kappa_tm_t\) are bounded on \([0,T]\), the drift also has linear growth there. The explicit derivation below uses the stated integrability assumptions directly.
The Explicit Solution
The kernel attenuates the contribution of a state observed at time \(s\). When \(\kappa\) is positive, a longer time gap generally produces greater attenuation. The forcing term \(\kappa_um_u\) contributes a deterministic weighted integral, and the noise contributes a stochastic integral with the same kernel.
Proof. Fix a finite horizon and define \(A(t)=\int_0^t\kappa_u\,du\) and \(F(t)=e^{A(t)}\). The function \(F\) is absolutely continuous and satisfies \(F'(t)=\kappa_tF(t)\) for almost every \(t\). Apply Itô's formula to \(f(t,x)=F(t)x\): its derivatives are \(f_t=F'(t)x\), \(f_x=F(t)\), and \(f_{xx}=0\), so the \(\tfrac12 f_{xx}\,dt\) correction is zero. Using the product rule with the SDE gives \(d(F(t)X_t)=F(t)\kappa_tm_t\,dt+F(t)\sigma_t\,dB_t\). Integrating from \(s\) to \(t\) and multiplying by \(F(t)^{-1}\) yields the displayed formula, since \(F(u)/F(t)=G(u,t)\). The deterministic integrals are finite by the assumptions, and \(u\mapsto F(u)\sigma_u\) is square-integrable on each finite interval, so the stochastic integral exists. For uniqueness, the difference \(D_t\) of two solutions driven by the same Brownian motion satisfies \(dD_t=-\kappa_tD_t\,dt\). Multiplication by \(F(t)\) gives \(d(F(t)D_t)=0\). If the solutions have the same initial value, then \(D_0=0\), hence \(D_t=0\) for every \(t\). \(\square\)
Conditional Mean and Variance
Condition on the information at time \(s\). The initial term \(G(s,t)X_s\) is then known, and the forcing integral is deterministic. The stochastic term has conditional mean zero because it uses Brownian increments after \(s\). Its conditional variance follows from the Itô isometry and independence of those increments from \(\mathcal{F}_s\).
Proof. Set \(H(u)=G(u,t)\sigma_u\) for \(s\le u\le t\). It is deterministic and \(\int_s^tH(u)^2\,du<\infty\). The Itô integral \(N_{s,t}=\int_s^tH(u)\,dB_u\) is square-integrable, with \(\mathbb{E}[N_{s,t}]=0\) and \(\mathbb{E}[N_{s,t}^2]=\int_s^tH(u)^2\,du\) by the Itô isometry. It is independent of \(\mathcal{F}_s\): deterministic step-function approximations are sums of Brownian increments after \(s\), independent of \(\mathcal{F}_s\), and their \(L^2\) limit retains that independence. Thus \(\mathbb{E}[N_{s,t}\mid\mathcal{F}_s]=0\) and \(\mathbb{E}[N_{s,t}^2\mid\mathcal{F}_s]=\int_s^tH(u)^2\,du\). Substituting the explicit solution gives the stated conditional mean and variance. The deterministic-integrand Itô integral is normal, as follows by the same step-function approximation from normal Brownian increments; its limit has variance \(\int_s^tH(u)^2\,du\). This proves the conditional-law statement, including the zero-variance case. \(\square\)
For clarity, for each fixed terminal time \(t\), the process \(r\mapsto\int_s^rG(u,t)\sigma_u\,dB_u\), \(s\le r\le t\), is a square-integrable martingale: its integrand is progressively measurable and square-integrable, and its conditional increments have mean zero. But the time-indexed noise term \(Z_t=\int_0^tG(u,t)\sigma_u\,dB_u\) generally is not a martingale, because its integrand changes when \(t\) changes. In the constant case \(\kappa=\sigma=1\), for example, \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=e^{-(t-s)}Z_s\), not generally \(Z_s\). This distinction prevents an incorrect martingale claim about the whole convolution.
Worked Calculations
Worked Example: A Changing Target
Take \(\kappa_u=1\), \(m_u=1+u\), and \(\sigma_u=2\). Find the conditional mean and variance at time \(t\), given \(\mathcal{F}_s\), where \(0\le s\le t\).
Here \(G(u,t)=e^{-(t-u)}\). The forcing integral is $$ \int_s^t e^{-(t-u)}(1+u)\,du =e^{-t}\int_s^t e^u(1+u)\,du =e^{-t}[ue^u]_s^t =t-se^{-(t-s)}. $$ Therefore the conditional mean is \(e^{-(t-s)}X_s+t-se^{-(t-s)} =t+(X_s-s)e^{-(t-s)}\).
The conditional variance is $$ 4\int_s^t e^{-2(t-u)}\,du =4\left[\tfrac12 e^{-2(t-u)}\right]_{u=s}^{u=t} =2\left(1-e^{-2(t-s)}\right). $$ For \(s=0\), \(X_0=0\), and \(t=1\), the mean is \(1+(0-0)e^{-1}=1\), and the variance is \(2(1-e^{-2})\). Check: the forcing integral at these endpoints is \(\int_0^1e^{-(1-u)}(1+u)\,du=1\), agreeing with the mean. The variance is nonnegative and is zero at \(t=s\), as required.
Worked Example: Constant Reversion and Volatility
Let \(\kappa_u=2\), \(m_u=1\), and \(\sigma_u=1\). Starting from \(X_s\), compute the conditional moments after an interval of length \(h=\ln 2\).
The kernel is \(G(s,s+h)=e^{-2h}=e^{-2\ln2}=\tfrac14\). The forcing term is \(\int_s^{s+h}2e^{-2(s+h-u)}\,du=1-e^{-2h}=\tfrac34\). Thus \(\mathbb{E}[X_{s+h}\mid\mathcal{F}_s]=\tfrac14X_s+\tfrac34\).
The conditional variance is $$ \int_s^{s+h}e^{-4(s+h-u)}\,du =\frac{1-e^{-4h}}{4} =\frac{1-e^{-4\ln2}}{4} =\frac{1-\tfrac1{16}}4 =\frac{15}{64}. $$ Check: if \(X_s=1\), the conditional mean is \(\tfrac14+\tfrac34=1\), the target level. As \(h\) increases, \(e^{-2h}\) decreases toward zero, so the initial state has less influence and the variance approaches \(1/4\).
Worked Example: Piecewise-Constant Reversion
On \([0,2]\), let \(m_u=\sigma_u=1\), and let \(\kappa_u=1\) for \(0\le u<1\) and \(\kappa_u=2\) for \(1\le u\le2\). With deterministic \(X_0=0\), compute the mean and variance at time \(2\).
The total integrated reversion is \(\int_0^2\kappa_u\,du=1+2=3\), so \(G(0,2)=e^{-3}\). Because the target is constantly \(1\), the forcing integral is \(\int_0^2G(u,2)\kappa_u\,du=1-G(0,2)=1-e^{-3}\); this identity follows by differentiating \(G(u,2)\) in \(u\), which gives \(\partial_uG(u,2)=\kappa_uG(u,2)\) almost everywhere. Hence the mean is \(1-e^{-3}\).
For \(0\le u<1\), \(G(u,2)=e^{-[(1-u)+2]}=e^{u-3}\). For \(1\le u\le2\), \(G(u,2)=e^{-2(2-u)}\). Therefore $$ \operatorname{Var}(X_2\mid\mathcal{F}_0) =\int_0^1e^{2u-6}\,du+\int_1^2e^{-4(2-u)}\,du =\frac{e^{-4}-e^{-6}}{2}+\frac{1-e^{-4}}{4}. $$ Check: both terms are nonnegative since \(e^{-4}>e^{-6}\) and \(1>e^{-4}\). The first term uses the earlier volatility interval but is attenuated by the reversion accumulated after time \(1\), as its kernel includes the factor \(e^{-2}\).
Reading the Formula Correctly
The conditional mean is not obtained by simply replacing \(X_t\) by the target \(m_t\). It combines the current state and the entire intervening target path, weighted by \(G\kappa\). Likewise, the variance depends on the volatility accumulated over the interval, weighted by the square of the kernel. These formulas separate the effects of drift and diffusion: \(\kappa,m\) determine the conditional mean, while \(\sigma\) determines the conditional variance.
A common error is to treat a time-varying target as if it were constant at its terminal value. For the first example, this would give the incorrect forcing term \((1+t)(1-e^{-(t-s)})\). At \(s=0,t=1\), that expression is \(2(1-e^{-1})\), whereas the correctly integrated forcing is \(1\). The target must remain inside the time integral. Another error is to call the entire noise convolution a martingale; only the integral with a fixed terminal-time integrand has the martingale property described above.
The kernel also makes useful limiting checks available. At \(t=s\), \(G(s,s)=1\), both integrals vanish, and the formula returns \(X_s\). If \(\sigma\) is identically zero, the conditional variance is zero. If \(\kappa\) is identically zero, then \(G=1\), the drift and forcing terms vanish, and the equation reduces to \(X_t=X_s+\int_s^t\sigma_u\,dB_u\). These checks test the signs, endpoints, and roles of the coefficients.
Check Your Understanding
Use the kernel formula and show the intermediate steps in each calculation.
- For \(\kappa_u=3\), \(m_u=2\), and \(\sigma_u=0\), find \(\mathbb{E}[X_t\mid\mathcal{F}_s]\) in terms of \(X_s\) and \(t-s\).
- For \(\kappa_u=1\), \(m_u=0\), and \(\sigma_u=1\), compute the conditional mean and variance over an interval of length \(\ln 2\).
- Explain why the forcing integral in the changing-target example is \(t-se^{-(t-s)}\), not \((1+t)(1-e^{-(t-s)})\).
- For a fixed terminal time \(t\), identify the integrand of the martingale indexed by \(r\le t\), and state the integrability condition it satisfies.
- Verify directly from the conditional variance formula that the variance is zero when \(t=s\) and when \(\sigma\) vanishes throughout \([s,t]\).