How to get speed measure $m(dx)$, scale function $s$, and killing measure $k(dx)$ of a diffusion from the infinitesimal generator? [closed]

Question

This question comes from P13 and P17 of the book Andrei N.Borodin and Paavo Salminen.

Page P13 defines the speed measure $m(dx)$, the scale function $s$, and the killing measure $k(dx)$.

Case 9 on P17:

"We consider here the special case in which the basic characteristics are absolutely continuous with respect to the Lebesgue measure and have smooth derivatives. In other words,

$m(dx)=m(x)dy,k(dx)=k(x)dx,s(x)=\int^x s'(y)dy$,

where $m,s'$ are continuous and positive, and $k$ is continuous and non-negative. Morever, if $s''$ is continous, then the second order infinitesimal generator $$\mathcal{G}f(x)=\frac{1}{2}a(x)^2f''(x)+b(x)f'(x)-c(x)f(x),$$

The functions $a, b, c$ are the infinitesimal parameters of $X$. "

My question is, based on the statement above, how to get the following results:

$m(x)=2a^{-2}(x)e^{B(x)}, s'(x)=e^{-B(x)},k(x)=2a^{-2}(x)c(x)e^{B(x)}, B(x)=\int^x 2a^{-2}(y) b(y)dy$?

Answer 1

Intuitively, you are equating coefficients in the two different representations of the generator: $$ {1\over m(x)}\left[\left({f'(x)\over s(x)}\right)'-k(x)f(x)\right] = {1\over 2}a(x)^2f''(x)+b(x)f'(x)-c(x)f(x). $$ The left side expands out to $$ {1\over m(x)}\left[{f''(x)\over s(x)}-{s'(x)f'(x)\over s(x)^2}-k(x)f(x)\right]. $$ Therefore $$ {a^2\over 2} = {1\over ms},\quad b=-{s'\over ms^2},\quad c={k\over m}. $$ Dividing the first two of these: $$ (\log s)'={s'\over s} =-{2b\over a^2}, $$ so $$ s(x) = C_1\exp\left(-2\int^xb(t)/a(t)^2 dt\right)=C_1e^{-B(x)}. $$ Choosing the constant of integration $C_1$ to be $1$, the rest follows easily.