12
$\begingroup$

I had to explain in informal terms what a Feller process was, to undergraduate students who understand Markov property, Poisson processes and such. It was easy to define Levy process as generalisation of random processes with stationary and independent increments. The closest to a layman explanation about a Feller process I could get was that it is a subset of strong Markov process. Beyond that, the limiting mechanism in a Feller definition is too complicated for me to explain in simple language.

How does one do this? Even the counterexamples of non-Markovian Feller processes are too contrived to give simple intuition.

$\endgroup$

3 Answers 3

10
$\begingroup$

By definition, the transition probability of a Feller process depends continuously on the starting point (in topology of weak convergence of measures, but maybe you don't have to say this in your pedagogical context). So if you change the starting point just a little bit, the distribution at, say, time $1$ will also be deformed only a bit.

An example of a non-Feller situation is the (deterministic) process solving $\dot X(t)=sign(X(t))$. The solution at time $1$ depends on its initial condition in a discontinuous way at $0$.

$\endgroup$
1
$\begingroup$

I found George Lowther's blog on Feller processes quite instructive, with several helpful examples. The comments in particular give some simple constructions for Markov processes that are almost, but not quite, Feller processes:

Just take something standard like Brownian motion or reflecting Brownian motion and remove the point $\{0\}$ from the state space. Seems like a bit of a cheat, but I think all examples are like this. They are Feller processes in a larger state space with some points removed.

$\endgroup$
1
$\begingroup$

I would use the setting of discrete time and finite state space, where it has less technicalities and I am more comfortable myself.

Starting with an example, I would suggest a family of random variables $\ldots,X_{-1},X_0,X_1,\ldots$ with values in $\{0,1\}$ with the property that the conditional probability of $X_t$ given the past depends on the weighted density of $1$s seen in the past, where $k$ step into the past is weighted with $2^{-k}$. More specifically,

$$\mathbb{P}(X_t=1\,|\,X_{t-1},X_{t-2},\ldots)= \frac{1}{2}X_{t-1}+\frac{1}{4}X_{t-2}+\frac{1}{8}X_{t-3}+\cdots\;.$$

This is not a Markov process (of any order), because of the dependence on the states arbitrarily far in the past. But this dependence becomes more and more negligible as we go farther to the past, and this we call the Feller property.

I would then go on with an abstract process with finite state space $S$ and introduce a transition kernel $K(\cdot,\cdot)$ that given values for $(\cdots,X_{t-2},X_{t-1})$, returns a distribution for $(\cdots,X_{t-1},X_t)$. More specifically, let $\mathbf{X}_t:=(\cdots,X_{t-1},X_t)$ be the history of the process up to time $t$ and denote by $\mathcal{H}=\{(\cdots,x_{-1},x_0): x_i\in S\}\cong S^{-\mathbb{N}}$ the space of all possible histories. Then,

$$\mathbb{P}(\mathbf{X}_t\in E\,|\,\mathbf{X}_{t-1})=K(\mathbf{X}_{t-1},E)$$

almost surely, for every measurable set $E\subseteq\mathcal{H}$. Here, I consider $\mathcal{H}$ as a topological space with the product topology and use the Borel $\sigma$-algebra on it. The transition kernel $K$ is Feller if $K(\cdot,E)$ is continuous for every cylinder set $E=\{(\cdots,x_{-1},x_0)\in\mathcal{H}: x_i=w_i \text{ for $i=0,-1,\ldots,-n$}\}$.

Finally, I would point out how a transition kernel acts on probability measures on $\mathcal{H}$ (from the right) and on bounded measurable functions $f:\mathcal{H}\to\mathbb{R}$ (from the left), in analogy with the transition matrix of a Markov chain. I would demonstrate (or ask the students to argue) that the Feller property is equivalent to either of the following properties:

  • $\pi\mapsto\pi K$ is continuous on probability measures on $\mathcal{H}$ with weak topology;

  • $f\mapsto K f$ maps every continuous function $f:\mathcal{H}\to\mathbb{R}$ to a continuous function.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .