2
$\begingroup$

"To every Q-matrix corresponds a unique Markov process." I'm trying to understand Klenke's proof of the uniqueness part of this proposition.


Klenke's proof

Following is an adapted version of Klenke's proof of the uniqueness part. The original can be found here ("Now assume that...").

Let $(p_t)_{t\geq0}, (r_t)_{t\geq0}$ be Markov semigroups. Let $q$ be a bounded infinitesimal generator such that $$\lim_{s\downarrow0}\frac{1}{s}(p_s-I)=q=\lim_{s\downarrow0}\frac{1}{s}(r_s-I)$$

It can be shown that $\frac{d}{dt}p_t=qp_t$ and $\frac{d}{dt}r_t=qr_t$ for all $0\leq t$, and so $$p_t-r_t=\intop_0^t q(p_s-r_s)\space d s$$

Define $u_s:=p_s-r_s$. Then $\|u_s\|_2\leq2$ and $\|q\|_2\leq2\lambda$ (where $\lambda:=\sup |q_{i,i}|<\infty$ by assumption). Hence $$\sup_{s\leq t}\|u_s\|_2\leq\sup_{s\leq t}\intop_0^s\|qu_z\|_2dz\leq\|q\|_2\sup_{s\leq t}\intop_0^s\|u_z\|_2dz\leq2\lambda t\sup_{s\leq t}\|u_s\|_2$$

For $t<\frac{1}{2}\lambda$ this implies $u_t=0$. Recursively, we get $u_t=0$ for all $t\geq0$; hence $p_t=r_t$. $\square$


My questions

Here's what i don't understand about the proof.

  1. How can it be shown that $\frac{d}{dt}p_t=qp_t$? I was able to verify the right-hand derivative (Theorem 11), but not the left-hand derivative.

  2. What's $\|\cdot\|_2$? It looks like the Euclidean norm on the vectorized argument, but the Euclidean norm could evaluate to $\infty$ for some matrices.

  3. Why is $$\sup_{s\leq t}\|u_s\|_2\leq\sup_{s\leq t}\intop_0^s\|qu_z\|_2dz$$

  4. Why is $$\sup_{s\leq t}\intop_0^s\|qu_z\|_2dz\leq\|q\|_2\sup_{s\leq t}\intop_0^s\|u_z\|_2dz$$

  5. What does it mean "recursively"? It doesn't look like a proof by induction. What's the variable of recursion?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.