Interpretation of a Generator of a Feller Process

Question

I am working through a short lecture on Feller Processes. It was stated there (and made plausible on a heuristic level) that the generator of Brownian Motion $\mathcal{L}$ on $\mathcal{C}(\mathbb{R})$ is given by $\mathcal{L}f = \frac{f''}{2}$. A proof of the exact connection will probably happen later in the course.

I verified that this indeed is a generator. Now, the lecture notes ask of me to compute and interpret the operator $\mathcal{L}_\lambda = \mathcal{L} \circ (I - \lambda \mathcal{L})^{-1}$.

It has to be $\mathcal{L}_\lambda f = \frac{K_\lambda \star f - f}{\lambda}$ where $K_\lambda (t) = \frac{1}{\sqrt{2 \lambda}} \exp({-\sqrt{\frac{2}{\lambda}}|x|})$.

But how can I interpret this? Is there a stochastic process associated with this? And if so, how can I see this?

After all, I am not very familiar with the link between generators and Feller processes yet...

Answer 1

(In reply to your question — this wouldn't fit in a comment.)

Yes, $\mathcal L_\lambda$ (the Yoshida approximation of $\mathcal L$) is the generator of a Markov process $X_\lambda(t)$ that approximates the Brownian motion. (Each $\mathcal L_\lambda$ is a bounded operator.) It can be described as follows: If $X_\lambda(0)=x$ then $X_\lambda$ holds in state $x$ for an exponentially distributed time (mean $\lambda$), and then jumps to a new state whose distribution is given by the probability measure $A\mapsto \alpha U^\alpha(x,A)=\alpha U^\alpha(1_A)(x)$. (As before, $\alpha = 1/\lambda$.) $X_\lambda$ then holds in the new state (call it $x_1$) for an independent exponential time and then jumps to another state whose distribution is $A\mapsto \alpha U^\alpha(x_1,A)$, and so forth.

Another way to describe this would be to take $B(t)$ to be standard Brownian motion with $B(0)=x$, and $0=T_0<T_1<T_2<\cdots$ the arrival times of a rate-$\alpha$ Poisson process independent of $B$. Then $X_\lambda(t) =B(T_k)$ for $T_{k}\le t<T_{k+1}$, $k=0,1,2,\ldots$.

Answer 2

(I find it more convenient to work with $\alpha :=1/\lambda$.)

Heuristically, the transition operators of the Feller process are related to $\mathcal L$ by $$ P_t =e^{t\mathcal L}\qquad\qquad (1) $$ The resolvent operators $U^\alpha$ are then given by $$ U^\alpha =\int_0^\infty e^{-t\alpha}P_t \phantom{b}dt = (\alpha I-\mathcal L)^{-1},\qquad\alpha>0. $$ Now from (1), $$ {d\over dt}P_t =\mathcal L P_t, $$ so Laplace transforming (integrating by parts on the left side) $$ \alpha U^\alpha - I =\mathcal L U^\alpha=\mathcal L(\alpha I-\mathcal L)^{-1}. $$ Expressed in terms of $\lambda$ this is basically the relationship you have found — as you will have no trouble checking, the operator $U^\alpha$ is convolution with the kernel $u^\alpha(z)={1\over \sqrt{2\alpha}}e^{-|z|\sqrt{2\alpha}}$. (In your formula for $K_\lambda(x)$, the $x$ should be $|x|$.)