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…
Best Answer
(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 $|x|$.)