Timeline for How can we determine the generator of this Markov process (at least formally)?
Current License: CC BY-SA 4.0
13 events
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| S Jun 6, 2022 at 22:00 | history | bounty ended | CommunityBot | ||
| S Jun 6, 2022 at 22:00 | history | notice removed | CommunityBot | ||
| Jun 3, 2022 at 13:04 | history | edited | Nawaf Bou-Rabee | CC BY-SA 4.0 |
deleted 1 character in body
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| Jun 3, 2022 at 12:59 | comment | added | 0xbadf00d | @NawafBou-Rabee Thanks for mentioning. I've fixed the definition. However, meanwhile I think the question boils down to the following easier to describte question: math.stackexchange.com/q/4464715/47771. Maybe you can take a look. | |
| Jun 3, 2022 at 12:58 | history | edited | 0xbadf00d | CC BY-SA 4.0 |
added 50 characters in body
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| May 31, 2022 at 19:29 | history | edited | Nawaf Bou-Rabee | CC BY-SA 4.0 |
corrected a misprint in (2)
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| S May 29, 2022 at 20:22 | history | bounty started | 0xbadf00d | ||
| S May 29, 2022 at 20:22 | history | notice added | 0xbadf00d | Canonical answer required | |
| May 28, 2022 at 17:49 | answer | added | John Dawkins | timeline score: 1 | |
| May 28, 2022 at 9:27 | comment | added | 0xbadf00d | @NateEldredge Assuming $E$ is discrete, do you mean to directly calculate $\operatorname P_\mu[X_{s+t}=x\mid\mathcal F_s]=\sum_{k=1}^\infty\operatorname P_\mu\left[s+t\in\left[\sum_{i=1}^{k-1}\tau_i,\sum_{i=1}^k\tau_i\right),Y^{(k)}_{s+t-\sum_{i=1}^{k-1}\tau_i}=x\mid\mathcal F_s\right]$? Seems still to be quite cumbersome. | |
| May 28, 2022 at 9:10 | comment | added | 0xbadf00d | @NateEldredge Thank you very much for your helpful comment. Do you sugget to directly calculate $(Cf)(x)=\lim_{t\to0+}\frac{(T_tf)(x)-f(x)}t$, where $(T_tf)(x)=\operatorname E_x[f(X_t)]$, in these cases? | |
| May 28, 2022 at 0:59 | comment | added | Nate Eldredge | A good start would be to take $A=0$, so that $X$ is a continuous-time Markov chain, and convince yourself that the formula gives the correct generator. Letting $E$ be discrete may make it even more clear. | |
| May 27, 2022 at 14:27 | history | asked | 0xbadf00d | CC BY-SA 4.0 |