# Markovian Excursions

In the previous post, I talked about the excursions of a Brownian motion. Today I’m thinking about how to extend these ideas to more general Markov chains. First we want to rule out some situations. In particular, we aren’t hugely interested in discrete time Markov chains. The machinery is fairly well established for excursions, whether or not the chain is transient. Furthermore, if the state space is discrete, as for a Poisson process for example, the discussion is not hugely interesting either. Remember that the technical challenges in the constructions of local time arise because of the Blumenthal 0-1 law property that Brownian motion visits 0 infinitely often in the small window after the start time. We therefore want the process to be regular at the point of the state space under discussion. This is precisely the condition described above for BM about 0.

Why is it harder in general?

The informal notion of a local time should transfer to a more general Markov chain, but there are some problems. Firstly, to define something in terms of an integral is not general enough. The state space E needs some topological structure, but any meaningful definition must be in terms of functions from E into the reals. There were also all sorts of special properties of Brownian motion, including the canonical time-space rescaling that came in handy in that particular case. It turns out to be easiest to consider the excursion measure on a general Markov chain through its Laplace transform.

Definition and Probabilistic interpretations

The resolvent is the Laplace transform of the transition probability $P_t(x,A)$, viewed as an operator on functions $f:E\rightarrow \mathbb{R}$.

$R_\lambda f(x):=\mathbb{E}_x\left[\int_0^\infty e^{-\lambda t}f(X_t)dt\right]=\int_0^\infty e^{-\lambda t}P_tf(x)dt$.

We can interpret this in terms of the original process in a couple of ways which may be useful later. The motivation is that we would like to specify a Poisson process of excursions, for which we need to know the rate. We hope that the rate will in fact be constant, so it will in fact to suffice to work out the properties of the expected number of excursions (or whatever) up to various random times, in particular those given by exponential RVs.

So, we take $\zeta\sim\text{Exp}(\lambda)$ independent of everything else, and assume that we ‘kill the chain’ at time $\zeta$. Then, by shuffling expectations in and out of the integral and separating independent bits, we get:

$R_\lambda f(x)=\mathbb{E}_x\int_0^\zeta f(X_s)ds = \frac{1}{\lambda}\mathbb{E}_xf(X_\zeta)$.

As in the Brownian local time description

$R_\lambda 1_A(x)=\mathbb{E}(\text{time spent in }A\text{ before death at time }\zeta_\lambda)$.

Markovian property

We want to show that excursions are Markov, once we’ve sorted out what an ‘excursion’ actually means in this context. We do know how to deal with the Markovian property once we are already on an excursion. It is relatively straightforward to define an extension of the standard transition probability operator, to include a condition that the chain should not hit point a during the transition. That is

$_aP_t(x,A):= \mathbb{P}_x(X_t\in A\cap H_a>t)$.

This will suffice to define the behaviour once an excursion has started. The more complicated bit will be the entrance law $n_t(A)$, being the probability of arriving at A after time t of an excursion. To summarise, as with BM, all the technical difficulties with an excursion happen at the beginning, ie bouncing around the start point, but once it is ‘up-and-running’, its path is Markovian and controlled by $_aP_t$.

Marking

The link between the resolvent and the excursions, is provided as in the Brownian case, by supplying a PPP of marks at uniform rate $\lambda$ to real time. This induces a mark process on excursions, weighted by an (exponential) function of excursion length. We make no distinction between an excursion including one mark or many marks. Then the measure on marked excursion is, in a mild abuse of notation:

$n_\lambda=(1-e^{-\lambda\zeta(f)})\cdot n.$

We compare with the Laplace transform $n_\lambda(dx)=\int_0^\infty e^{-\lambda t}dtn_t(dx)$ using a probabilistic argument.

We can apply the measure to a function in the usual way: $\lambda n_\lambda(1_A)$ is the measure of those excursions for which the first mark occurs in $A$. So by taking $A=E$, we get

$\lambda n_\lambda(1)=\text{ Excursion measure }=\int_U n(df)(1-e^{-\lambda\zeta(f)}).$

We have therefore linked the exponential mark process on excursion measure with the Laplace transform of the entrance law. So in particular:

$\frac{\lambda n_\lambda(A)}{\lambda n_\lambda(1)}=\mathbb{P}(\text{first mark when in }A)=\int_0^\infty \lambda e^{-\lambda t}P_t(0,A)dt=\lambda R_\lambda 1_A(0)$.

The resolvent is relatively easy to calculate explicitly, and so we can find the Laplace transform $n_\lambda(A)$. From this it is generally possible to invert the transform to find the entrance law $n_t$.

References

A Guided Tour Through Excursions – L. C. G. Rogers.

This pair of posts is very much a paraphrase of chapters 3 and 4 of this excellent text. The original can be found here (possibly not open access?)