Continuous Markov Process and Change of Measure

Question

So I've been thinking about a problem for a little bit and I decided it's time to ask those that know more about the subject than I do. I've been working on some Stochastic Calculus (a new area of math for me), and I've been looking at change of measures, and how they affect different properties of stochastic processes. For instance, I know that the property of being a martingale is not invariant under a change of measure.

However, I've been thinking about whether or not the Markov property is invariant under a change of measure. More specifically, if I have a continuous time process X_{t} which is markov, and $Y_{t}$ is equivalent to $X_{t}$ via some change of measure, is $Y_{t}$ also necessarily Markov?

My initially feeling was that the answer would be no, and that I would want to apply the change of measure that changes the drift to something that depends on the whole path. For instance, I was thinking of something like:

$dX_{t} = dW_{t}$

$dY_{t} = \left(\int_{0}^{t} W_{s} ds \right) dt + dW_{t}$

However, when we say a process is Markov we mean that it is Markov with respect to its natural filtration; namely I've just showed that my process $Y_{t}$ will not be markov with respect to the filtration generated by $W_{t}$, but it looks like it might still be Markov under the natural filtration generated by $Y_{t}$. One way around this would be to modify this to be something like:

$dX_{t} = dW_{t}$

$dY_{t} = \left(\int_{0}^{t} Y_{s} ds \right) dt + dW_{t}$

but then I'd be worried that I applied Girsanov incorrectly. So at this point, I'm stumped. Does anyone have any suggestions for what I should be looking at or thinking about when considering this?

Answer 1

You have exactly the right idea. To make it precise, let $Y$ be a Brownian motion on some probability space $(\Omega,\mathcal{F},P)$, with respect to its natural filtration $\mathbb{F}$. Define a new measure $Q$ by

$\frac{dQ}{dP} = \exp\left(\int_0^T\int_0^tY_sdsdY_t - \frac{1}{2}\int_0^T\left|\int_0^tY_sds\right|^2dt\right)$.

(That this indeed has expectation 1 follows from Corollary 5.16 of Karatzas & Shreve.) By Girsanov's theorem, $W_t := Y_t - \int_0^t\int_0^sY_ududs$ defines a Brownian motion on $(\Omega,\mathcal{F},\mathbb{F},Q)$. This gives us a weak solution of the final SDE you wrote above; on $(\Omega,\mathcal{F},\mathbb{F},Q)$, we have

$dY_t = \left(\int_0^tY_sds\right)dt + dW_t$.

So, in summary, $Y$ is Markov on $(\Omega,\mathcal{F},\mathbb{F},P)$, but not on $(\Omega,\mathcal{F},\mathbb{F},Q)$, even though in each case the filtration $\mathbb{F}$ is the one generated by $Y$.