# Generator of Wiener process and its running maximum

This was originally posted on Math StackExchange a long time ago, but got no answer (even after a bounty). See https://math.stackexchange.com/questions/1274775/generator-of-wiener-process-and-its-running-maximum

If we let $W$ be a standard linear Wiener process issued from zero and $M$ its running maximum $$M_t := \sup \{ W_u: u \leq t \},$$ then we could show that $(X,Y):=(M,M-W)$ is a Markov process on $\mathbb{R}_+ \times \mathbb{R}_+$ with transition kernel $$P_t(a,b; dx,dy) = \mathbb{I}_{ \{x \geq a\} } \left[ \delta_a(dx) dy k_t(b,y) + dx dy 2 h_t(x-a+b+y) \right],$$ where $k$ is the transition density of a Wiener killed upon reaching the origin, i.e. $$k_t(x,y) = \frac{1}{\sqrt{2 \pi t}} \exp \left( -\frac{(x-y)^2}{2t} \right) - \frac{1}{\sqrt{2 \pi t}} \exp \left( -\frac{(x+y)^2}{2t} \right)$$ and $t \mapsto h_t(x)$ is the density of the first hitting time of $x$ by $W$, i.e. $$h_t(x) = \frac{x}{\sqrt{2 \pi t^3}} \exp \left( -\frac{x^2}{2t} \right).$$

Assuming $(X,Y)$ has the Feller property (which I think is true), my question is: what is the infinitesimal generator, say $G$, of $(X,Y)$? I am pretty sure that, since $(X,Y)$ moves as a Wiener only along the $y$-direction when it is away from the $x$-axis, $G$ is a degenerate differential operator like $$G = \frac12 \partial_{yy},$$ so my question is really about the boundary conditions, in particular I think the real issue is on the $x$-axis, since in any case the process never moves to the left. Since when $(X,Y)$ touches the $x$-axis (coming from above) it then goes either to the right or it's pushed back up (and also by thinking of an approximating random walk), I am led to conjecture that the correct boundary condition might be $$\partial_{u} v(x,0) := \lim_{h \to 0} \frac{v(x+h,h)}{h} = 0, \quad \forall x,$$ but I am far from being sure. I tried to derive it more formally with no success, so a real proof and/or a reference (I am sure this can be found somewhere but I didn't manage) would be really appreciated.

Many thanks in advance.

My guess would be indeed $\tfrac{1}{2} \partial_{yy}$, with $f$ in the domain if and only if:

(a) $f$ is in $C_0(\mathbb{R}_+^2)$;

(b) $f(x, \cdot)$ is $C_0^2(\mathbb{R}_+)$ for each $x$;

(c) $\partial_{yy} f$ is in $C_0(\mathbb{R}_+^2)$;

(d) $f(\cdot, 0)$ is $C_0^2(\mathbb{R}_+)$ and $\partial_{x} f(x, 0) = -\partial_{y} f(x, 0)$ for all $x$.

I do not know any reference for this. It should be relatively simple to show that any $f$ as above belongs to the domain (by proving convergence of $(P_t f - f) / t$ to $\tfrac{1}{2} \partial_{yy} f$, or, simpler, by using Dynkin's characteristic operator). The converse is always much more problematic.

• Thanks! I am not able to check if your answer is correct. I ll wait a little to see if this gets any other answer. – Tom Mar 8 at 15:27