Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 30 / 3

The deterministic clock is $a(t)=e^{2\lambda t}$, so $a(0)=1$. For $s\le t$, the martingale property of Brownian motion givesIntegrability follows from the Gaussian second moment. Thus $X$ is a martingale in the time-changed filtration.

Its increments have variance $a(t)-a(s)$ and are independent of ${\mathcal{G}}_s$. Equivalently, $X_t^2-a(t)$ is a martingale. Since a quadratic variation starts at zero, its compensator must be normalized at $t=0$:The initial random variable $X_0=B_1$ does not contribute to quadratic variation. The expression $e^{2\lambda t}$ alone would violate $[X{]}_0=0$.

Define the stochastic integralIt is adapted to ${\mathcal{G}}_t$, starts at zero and has continuous paths. Integrals over disjoint clock intervals are centered Gaussian and independent, and an increment from $s$ to $t$ is independent of ${\mathcal{G}}_s$. The Itô isometry gives its varianceThese are precisely the defining properties of a ${\mathcal{G}}_t$-Brownian motion; in particular its future increments are independent of ${\mathcal{G}}_0={\mathcal{F}}_1$.

Deterministic substitution in the stochastic integral, first for step integrands and then by the Itô isometry, gives $d{X}_s=\sqrt{2\lambda }{e}^{\lambda s}d{W}_s$. HenceThe initial term is essential: a stochastic integral from time zero alone cannot equal $X$, since $X_0=B_1$. This is the unscaled part of the exponential Brownian time change to a stationary Ornstein-Uhlenbeck process.

Use the Itô product rule with the deterministic factor $e^{-\lambda t}$. Its quadratic covariation with $X$ is zero, givingMultiplication by the integrating factor $e^{\lambda t}$ solves the Ornstein-Uhlenbeck stochastic differential equation:The initial value $B_1$ is standard normal and independent of the future driving Brownian motion. Thus $Y$ is a centered Gaussian process with variance one. For $s\le t$, its independent future noise gives $\mathbb{E}(Y_s{Y}_t)=e^{-\lambda (t-s)}$. Its Gaussian finite-dimensional distributions consequently depend only on time differences, so it is a stationary process. This verifies the exponential Brownian time change to a stationary Ornstein-Uhlenbeck process directly.

The printed pathwise conclusion is false without specifying the initial value. We first prove exactly what the hypotheses do imply, by stationary drift saturation.

Stationarity gives $\mathbb{E}{Z}_s^2=1$. The Itô formula yieldsThe stochastic integral is a true square-integrable martingale on every finite horizon, since $\mathbb{E}{\int }_0^t{Z}_s^{2}ds=t$. Also $\mathbb{E}|Z_s{g}_s|\le \lambda$ by the Cauchy-Schwarz inequality. Taking expectations therefore givesEvery integrand is nonnegative, so $\mathbb{E}[Z_s{g}_s]=-\lambda$ for almost every $s$. For those times,It follows that $g_s=-\lambda Z_s$ for almost every time and outcome. Equality of these integrable drift terms gives the exact solutionThus $Z$ solves the same equation, but stationarity has not identified its initial random variable with $B_1$.

For an explicit counterexample, takeThis is adapted to ${\mathcal{G}}_t$. Its independent normal initial value and Brownian noise make it a centered stationary Gaussian process with covariance $e^{-\lambda |t-s|}$, exactly as for $Y$. Hence $\mathbb{E}{Z}_0^2=1$ and $\mathbb{E}{g}_t^2={\lambda }^2$, satisfying every printed hypothesis. Nevertheless $Z_t-Y_t=-2e^{-\lambda t}{B}_1$, which is nonzero almost surely for every finite $t$.

The corrected result is indistinguishability if one adds $Z_0=Y_0$ almost surely. Without that addition, there is equality in law of the stationary Ornstein-Uhlenbeck processes, not necessarily equality of their coupled paths. To see that the initial law is forced, independence of Brownian increments from ${\mathcal{G}}_0$ and stationarity give, for its characteristic function $\phi$,Letting $t\to \infty$ proves $\phi (u)=e^{-u^2/2}$. This is the distinction captured by same-noise stationary Ornstein-Uhlenbeck processes.