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Jul
12
awarded  Good Question
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Apr
7
accepted Is $|x-y|^n\leq 2^n(|x|^n+|y|^n)$?
Apr
7
comment Is $|x-y|^n\leq 2^n(|x|^n+|y|^n)$?
@lyj Oh yes of course, thank you.
Apr
7
asked Is $|x-y|^n\leq 2^n(|x|^n+|y|^n)$?
Apr
1
comment Approximating $C^2$ functions with compactly supported $C^2$ functions
@Igor, thank you for the comment. I just needed pointwise convergence, and cheap effective diet pills's suggestion seems to do the trick. Sorry I'm not sure how to follow through with your suggestion (not familiar with how to "blend spines together with the partition of the unity bump function").
Apr
1
comment Approximating $C^2$ functions with compactly supported $C^2$ functions
@cheapeffectivedietpills, thanks this works perfectly (for example, with $\phi(x)=\exp(1-1/(1-x^2))$ if $|x|<1$ and zero otherwise). If you copy paste your comment into an answer box, I'll accept and we can close the question.
Mar
30
asked Approximating $C^2$ functions with compactly supported $C^2$ functions
Mar
21
awarded  Popular Question
Mar
14
accepted Exercise from Rogers and Williams's Diffusions, Markov processes and martingales
Mar
14
accepted When are stable continuous time Markov chains Feller? Always?
Mar
11
asked The $K$-moment problem for measures with countable support?
Mar
5
comment When are stable continuous time Markov chains Feller? Always?
If it's fine with you, I'd like to keep the question open for a bit longer, see if anyone posts any conditions on $Q$ that ensure that the chain is Feller. By the way, just checking, $X_n$ should be indexed by the $[0,\infty)$ and not $1,2,3,\dots$ right?
Mar
5
comment When are stable continuous time Markov chains Feller? Always?
Nice counterexample, thanks -- intuitively, I always think of the first bullet point as a probabilistic bound on how fast the process can return to the "center" of the state space (where the mass of the $f$ is concentrated). Something like "for any $t\geq0$, if you start the process sufficiently far away, then by time $t\geq0$ it will still be sufficiently far away, at least on average". Your example breaks this by having a fixed probability of reaching to $0$ in less than some fixed amount of time irrelevant of where it initially started.
Mar
4
revised When are stable continuous time Markov chains Feller? Always?
added 8 characters in body
Mar
4
asked When are stable continuous time Markov chains Feller? Always?
Feb
26
comment Exercise from Rogers and Williams's Diffusions, Markov processes and martingales
@saz hi, thank you for the comments. I see how monotone class theorem then yields that $x\mapsto N(x,A)$ is measurable for any Borel set $A$ (included it in the answer). But how can we use it to show that $0\leq N(x,A)\leq 1$? The set functions $f$ such that $0\leq Nf\leq 1$ is not a vector space, so we can't apply the theorem. From the above it would be enough to show that the set of Borel sets $A$ such that $0\leq N(x,A)\leq 1$ is a sigma algebra (or a monotone class) but I'm having trouble showing its closed under complements (or intersections).
Feb
26
revised Exercise from Rogers and Williams's Diffusions, Markov processes and martingales
added 98 characters in body
Feb
25
answered Exercise from Rogers and Williams's Diffusions, Markov processes and martingales