Can I apply the Girsanov theorem to an Ornstein-Uhlenbeck process?

Question

Let $W_t$ be a standard Brownian motion, and $X_t$ a measurable adapted process. Girsanov's theorem says that under certain conditions, the Brownian motion with drift $Y_t = W_t - \int_0^t X_s\,ds$ can be a Brownian motion under a certain equivalent probability measure.

I want to apply Girsanov's theorem with $X_t$ an Ornstein-Uhlenbeck process defined by $dX_t = dW_t - X_t dt$, $X_0 = 0$. In this case we would have $Y_t = X_t$, so I would learn that an Ornstein-Uhlenbeck process can be a Brownian motion under an equivalent measure.

The condition needed for Girsanov's theorem to hold is that $$Z_t = \exp\left(\int_0^t X_s\,dW_s - \frac{1}{2} \int_0^t X_s^2\,ds\right)$$ be a martingale.

Is this condition satisfied?

A sufficient condition, due to Novikov, is that $$E \exp\left(\frac{1}{2} \int_0^T X_s^2\,ds\right) < \infty.$$ I can't seem to see how to verify either of these conditions, though the Ornstein-Uhlenbeck process has so many nice properties that one would think something simple would work.

This question came up while studying the solution of the quantum harmonic oscillator via the Feynman-Kac formula. I am trying to understand the "ground state transformation" in terms of Girsanov's formula.

Thanks!

Answer 1

@Nate: I think your argument is fine. I didn't know the Ferinique's theorem. My argument on that part was this one.

Fix $T>0$ (as you did eventually $T+1$). Applying Jensen we get: $$\exp\left(\frac{1}{2}\int_S^{S+\epsilon} X_t^2\,dt\right)=\exp\left(\frac{1}{\epsilon}\int_S^{S+\epsilon} \frac{\epsilon}{2}X_t^2\,dt\right)\leq \frac{1}{\epsilon}\int_S^{S+\epsilon} \exp\left(\frac{\epsilon}{2}X_t^2\right)\,dt \qquad a.s..$$

Now taking the expectation and using Tonelli's theorem we have to study: $$\frac{1}{\epsilon}\int_S^{S+\epsilon} E\left[\exp\left(\frac{\epsilon}{2}X_t^2\right)\right]\,dt.$$

$X_t \sim N(\mu_t=X_0e^{-t}\,,\,\,\sigma_t^2=\frac{1-e^{-2t}}{2})$ so $$E\left[\exp\left(\frac{\epsilon}{2}X_t^2\right)\right]=E\left[\exp\left(\frac{\epsilon}{2}(\mu_t+\sigma_tZ)^2\right)\right]=$$ $$=e^{\frac{\epsilon}{2}\mu_t^2}\int_{\mathbb{R}} \frac{1}{\sqrt{2 \pi}}\, \exp\left(-\frac{x^2}{2}(1-\sigma_t^2 \epsilon)+\epsilon \mu_t \sigma_t x\right)dx. $$ Setting $\lambda_t=1-\epsilon \sigma_t^2=\frac{1}{2}[2-\epsilon(1-e^{-2t})]$, if $\lambda_t>0$ (for example with $\epsilon < 1$) the last integral is convergent and its value is:$$\exp\left(\frac {\epsilon}{2}\mu_t^2\right)\,\exp\left(\frac{\epsilon^2}{2 \lambda_t}\mu_t^2 \sigma_t^2\right)\frac{1}{\sqrt{\lambda_t}}.$$

Finally all the functions involved are continuous and since $\epsilon < 1$, $\lambda_t$ is away from 0 and so the moment generating function is integrable.

My first idea was to use instead of $\epsilon$ $T$, but moment generating function of the chi squared is not define to away from 0 but only in a neighbourhood.

Answer 2

I think I got it worked out.

Lemma. There exists $\epsilon > 0$ such that for any $S \in [0,T]$, we have $$E\left[\exp\left(\frac{1}{2}\int_S^{S+\epsilon} X_t^2\,dt\right)\right] < \infty.$$

Proof. Set $M = \sup_{[0, T+1]} |X_t|$. Then $$E\left[\exp\left(\frac{1}{2}\int_S^{S+\epsilon} X_t^2\,dt\right)\right] \le E\left[\exp\left(\frac{\epsilon}{2} M^2\right)\right].$$ By Fernique's theorem, there exists $\epsilon$ small enough that this is finite. ($X_t$, being a continuous Gaussian process, induces a Gaussian measure on the Banach space $C([0,T+1])$, and $M$ is the norm on this space.) Perhaps there is also a more direct way to get this.

Now we use Corollary 3.5.14 from Karatzas and Shreve, Brownian Motion and Stochastic Calculus, whose proof I'll paraphrase. Let $$Z_t = \exp\left(\int_0^t X_s\,dW_s - \frac{1}{2} \int_0^t X_s^2\,ds\right)$$ be the process we have to show is a martingale. $Z_t$ is a nonnegative local martingale (by Itô's formula) hence it is a supermartingale (using Fatou's lemma with a sequence of stopping times), so it is enough to show $E[Z_t ]= 1$ for all $t \in [0,T]$. We proceed by induction. Suppose we have shown $E[Z_t] = 1$ for $t \in [0,S]$ (the base case is $S=0$ which is trivial). Let $t \in [S, S+\epsilon]$. Set $$Z_t^S = \exp\left(\int_S^t X_s\,dW_s - \frac{1}{2} \int_S^t X_s^2\,ds\right).$$ By Novikov's condition, $Z_t^S$ is a martingale for $t \in [S, S+\epsilon]$. Now we have $$E[Z_t] = E[Z_S Z_t^S] = E[E[Z_S Z_t^S \mid \mathcal{F}_S]] = E[Z_S E[Z_t^S \mid \mathcal{F}_S]] = E[Z_S Z_S^S].$$ But $Z_S^S = 1$ by definition and $E[Z_S]=1$ by assumption. So we have $E[Z_t] = 1$, and this holds for any $t \in [0, S+\epsilon]$. Now we just repeat the induction $T/\epsilon$ times.

It appears this would work for any continuous Gaussian process $X_t$, which is nice to know.

Answer 3

$X^2$ follows a CIR type diffusion. Expressions such as occur in Novikov can be computed explicitly and seen to be finite at least for small t. If $Z_t$ is a martingale for small t I'm sure that using some combination of positivity and the markov property you can extend it. The result stated as prob 8.3.14 in Revuz & Yor, though without solution or substantial hints.