Doob's h-transform

Remarks

The kernels $P^h_t$ form a sub-Markov semigroup. Indeed, $P^h_t(x, E) = P_t h(x)/h(x) \le 1$ since $h$ is excessive, and $$\int P^h_s(x, dy) P^h_t(y, dz) = \frac{h(z)}{h(x)} \int P_s(x, dy) P_t(y, dz) = \frac{h(z)}{h(x)} P_{s+t}(x, dz).$$ If $P_t h = h$ for all $t$, then each $P^h_t$ is a Markov kernel and the h-process has infinite lifetime. Example: let $(P_t)$ be Brownian motion on $(0, \infty)$ killed at $0$, and $h(x) = x$. Then $P_t h(x) = E^x[B_t ; t < \tau_0] = x$ by optional stopping, so $h$ is invariant, and the h-process is the three-dimensional Bessel process, Brownian motion conditioned never to hit $0$. Non-example: for Brownian motion on $\R$ and $h(x) = 1 + x^2$, one has $P_t h(x) = 1 + x^2 + t > h(x)$, so $h$ is not excessive and $P^h_t(x, \R) > 1$. If $L$ is the infinitesimal generator of $(P_t)$, the generator of the h-process acts by $L^h f = h^{-1} L(h f)$ on suitable $f$. For $L = \frac12 \Delta$ and harmonic $h$, this is $L^h f = \frac12 \Delta f + \nabla(\log h) \cdot \nabla f$, a Brownian motion with drift $\nabla \log h$.