I am having trouble computing the infinitesimal generator of a rather simple process.

Let us consider a distribution $M$ taking values in the whole half line $\mathbb{R}_+$, and a process $(X_t)_t$ defined by the following dynamic :

  • If $X_t = \alpha > 0$, then $X_{t + h} = \alpha - h$ for $0 < h < \alpha$.
  • When the process reaches 0, (say at time $\tau$), we set $X_{\tau} = Y_i$, where $(Y_i)_i$ is an i.i.d. sequence of random variables following the distribution $M$, independent from everything else.

Hence, this is a toy example of a jump process in continuous time, but the jumps are somewhat deterministic (at a given time $t$, I know when the next jump will occur). I have no problem computin the infinitesimal generator of the process when $t > 0$, however I do not understand how to obtain the generator around the point 0.

For instance if I chose to make the jump happen exactly at time 0, it seems to me that I will end up with something similar to $$ \lim \limits_{t \to 0} \frac{1}{t} (\mathbb{E}[Y_1]-t) $$ which is certainly not finite.

I found several points about this case in the discrete setting, but this point 0 issue is not relevant in such situations.

Edit: In this discret setting, the jump occurs in a way that we never reach 0, the chain jumping from 1 to the value given by the next jump. This is the behavior which I find problematic in the continuous-time, $\mathbb{R}$ setting, since it means describing the transition probability at some limit point $0^+$.

  • 2
    $\begingroup$ After further research, it turns out that the generator for such problem is not obvious at all, and falls into the framework of the so-called Piecewise Deterministic Markov Process. Several articles, starting with the book of Davis (Markov Models and Optimisation) take an in-depth look at the problem. I am leaving the question there in case someone one else recognizes the same problem, so that one will know where to find the answer. $\endgroup$ – Gâteau-Gallois Jul 27 '18 at 13:24

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