# 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.

• Is there any particular reason to deal with $(M,M-W)$ instead of $(M,W)$ here? – Ian May 9 '15 at 21:58
• Not really. I used $(M,M-W)$ because state space is simpler and I found a book giving the semigroup of $(M,M-W)$.. Also, I suspect that we can say that $(M,M-W)$ has a deeper meaning, since that is what you might use to reconstruct $M$ and ultimately to define the local time.. In any case, I would not be able to give an answer to my question even in the case of $(M,W)$.. Having an answer for both cases would be even better of course! – Tom May 9 '15 at 22:15
• Can you put up the part about the semigroup for $(M,M-W)$? Also, I am not so sure about the Neumann condition either: the process does actually accumulate time on the boundary, in the sense that the time spent on the boundary has positive measure, but it has empty interior. I don't quite see how that would happen with a purely reflecting condition. – Ian May 9 '15 at 22:21
• I got it from the book Cinlar - Probability and Stochastics page 406. Back to the boundary conditions, I think I don't agree on the fact that the process accumulates time on the boundary. I think we can say that the set of times such that the process is on the $x$-axis has zero Lebesgue measure, almost surely (with respect to any probability $\mathbb{P}_{a,b}$). Having said that, let me stress that I am not sure about this, I am not sure at all about my conjecture and that in general I am not at all an expert in maths :) – Tom May 9 '15 at 23:10
• No, the process definitely accumulates time on the boundary, this is precisely the time where $W_t=M_t$, which is a type of local time. – Ian May 9 '15 at 23:11