Timeline for Independence of Brownian motion at hitting time from that hitting time
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 9, 2023 at 17:09 | history | edited | Martin Sleziak |
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| Dec 24, 2011 at 13:14 | vote | accept | madison54 | ||
| Dec 23, 2011 at 13:56 | comment | added | The Bridge | Hi I have downvoted the question because I found it was really sounding like homework, which is fine to as ask but rather in the Mathstackexchange forum, anyway from a "bad" question we sometime get really good answers like those of J. Dawkins and D. Zare. Many such questions periodically appears in the forum and are alomst immediately closed. If those two rich answers didn't appeared I think it would have been closed rapidly. Best regards | |
| Dec 22, 2011 at 5:59 | answer | added | Douglas Zare | timeline score: 5 | |
| Dec 22, 2011 at 2:27 | comment | added | Yemon Choi | @Ori: I haven't downvoted, but perhaps the reason for the downvote was that a baldly asked question, devoid of context (what has the OP already tried, why did they come to this problem) can seem like homework. It is, to be fair, a natural question; but I think the question would be better received if the OP acted along the lines suggested by David Roberts | |
| Dec 22, 2011 at 0:23 | comment | added | Ori Gurel-Gurevich | This question may be easy, but I don't think the downvote is justified. | |
| Dec 22, 2011 at 0:20 | comment | added | Ori Gurel-Gurevich | @The Bridge: I didn't understand your argument. Where does it fail for starting point other than 0? | |
| Dec 22, 2011 at 0:17 | answer | added | John Dawkins | timeline score: 4 | |
| Dec 21, 2011 at 23:50 | comment | added | The Bridge | $B_T$ is random variable taking value 1 and -1 with probability 1/2, it is independent from $T$, you can conditionate by replacing T, by T' the first time B gets to 1, then B_T is a constant so it is independent of T'. Then taking T'' first time to -1 tell the same as T=min(T',T'') you get the result. Regards, but not an MO question more suited for mathstackexchange. | |
| Dec 21, 2011 at 23:28 | comment | added | David Roberts♦ | What is your reason for asking this question? In what context did it arise? Also, reading mathoverflow.net/howtoask and implementing the suggestions from there may help attract answers... | |
| Dec 21, 2011 at 23:07 | history | asked | madison54 | CC BY-SA 3.0 |