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Mar 22, 2018 at 10:23 comment added Mateusz Kwaśnicki @sharpe: You're more than welcome, and thank you too: this question will now be my favourite example when I teach Feller processes.
Mar 22, 2018 at 10:21 comment added sharpe I understood. Thank you very much for teaching me carefully!
Mar 22, 2018 at 10:19 history edited Mateusz Kwaśnicki CC BY-SA 3.0
minor corrections
Mar 22, 2018 at 10:17 comment added Mateusz Kwaśnicki @sharpe: Oh, yes, you are right! (I am used to the convention that a BM is generated by $\Delta$, not $\tfrac{1}{2} \Delta$). I updated the definition of $u$ to compensate this $1/2$.
Mar 22, 2018 at 9:59 vote accept sharpe
Mar 22, 2018 at 9:57 vote accept sharpe
Mar 22, 2018 at 9:59
Mar 22, 2018 at 9:56 comment added sharpe Thank you for your reply. This is a minute thing, but $P^{(x,y)}(T_{K} \ge 1/2) \le 1/2$, right? According to my calculations, for large $x$, $$0 \le E^{(x,y)}[u(X_{T_K})]<u(x)-(1/2)E^{(x,y)}[T_K].$$ This implies $E^{(x,y)}[T_K] <1/4$ and $P^{(x,y)}(T_{K} \ge 1/2) \le 1/2$.
Mar 22, 2018 at 9:01 comment added Mateusz Kwaśnicki @sharpe: I expanded the last part of my answer (and fixed a typo: what I wanted to write is $T_K < \tfrac{1}{2}$, sorry).
Mar 22, 2018 at 8:59 history edited Mateusz Kwaśnicki CC BY-SA 3.0
expanded the last part of the answer
Mar 22, 2018 at 6:52 comment added sharpe Thank you for your reply. By using Skorohod type SDE for $X$ and Ito's forumula, I was able to obtain $P_{(x,y)}(T_K<1) \ge 3/4$. However, how can I use strong Markov property of $X$ to get lower estimate of $E_{(x,y)}[1_{K}(X_1)]$?
Mar 21, 2018 at 21:03 history answered Mateusz Kwaśnicki CC BY-SA 3.0