Timeline for Process equivalent to conditional probability
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 30, 2010 at 21:56 | answer | added | Did | timeline score: 2 | |
| Jun 21, 2010 at 10:05 | answer | added | Grzenio | timeline score: -2 | |
| Jun 21, 2010 at 10:04 | vote | accept | Grzenio | ||
| Jun 13, 2010 at 18:22 | answer | added | Jeff Schenker | timeline score: 6 | |
| Jun 10, 2010 at 7:38 | comment | added | Grzenio | Hi @Mark, $t$ and $X$ are positive. It would be nice if $Z_t$ was positive as well, but its not really a requirement. | |
| Jun 9, 2010 at 17:43 | comment | added | Mark Meckes | Do you want to assume that $t$, $X$, and $Z_t$ are positive? | |
| Jun 9, 2010 at 17:19 | comment | added | vad | Never mind. I understand why my suggestion does not make sense. You have three random variables here Z, X and T. I was interpreting T somewhat differently ... | |
| Jun 9, 2010 at 17:13 | history | edited | Mark Meckes | CC BY-SA 2.5 |
Fixed typo I introduced.
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| Jun 9, 2010 at 16:07 | comment | added | vad | P[Z_t > T] cannot equal P[X>t] as the pdf of Z is greater than 0 for X_t > t and 0 otherwise. You can do a simulation to check the definition. By the way, if you want a closed form for Z_t you should provide more information for X_t. | |
| Jun 9, 2010 at 16:03 | history | edited | Grzenio | CC BY-SA 2.5 |
clarified
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| Jun 9, 2010 at 15:46 | answer | added | Mark Meckes | timeline score: 1 | |
| Jun 9, 2010 at 15:17 | comment | added | Grzenio | Hi @Anon, I might be wrong but I thing that with the definition you provided P[z_t > T] = P[X>T] and not P[X>T | X>t] (for T>t) | |
| Jun 9, 2010 at 15:10 | comment | added | vad | I am not sure I understand the difficulty. Define Z_t = X_t I(X_t > t) and you are done. Is there something I am missing here? | |
| Jun 9, 2010 at 14:52 | history | asked | Grzenio | CC BY-SA 2.5 |