Timeline for Probability that no three events happen in a pre-defined window
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 3, 2013 at 18:39 | vote | accept | Pankaj Gupta | ||
| Jul 3, 2013 at 18:39 | comment | added | Pankaj Gupta | Coincidentally, I was doing the same a few minutes ago. Thanks Gustav! | |
| Jul 3, 2013 at 14:43 | comment | added | Gustav | I now edited the answer to answer the correct question. | |
| Jul 3, 2013 at 14:43 | history | edited | Gustav | CC BY-SA 3.0 |
added 14 characters in body
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| Jul 3, 2013 at 14:32 | history | edited | Gustav | CC BY-SA 3.0 |
Now finally answering the original question.
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| Jun 28, 2013 at 9:45 | comment | added | Gustav | Ah, again I did not read your question carefully. So you want the probability that all gaps are larger than W. So you need to change all the < to >. That makes the integrals slightly more fiddly. To save time and avoid errors I do such straightforward but tedious integrals with a computer algebra system, see for example wolframalpha.com/share/… | |
| Jun 28, 2013 at 6:25 | comment | added | Pankaj Gupta | Sorry, it is not $1 - P(S_{k-2})$ that we are after, but 1 - Prob(some gap is less than W). Seems fixable. | |
| Jun 28, 2013 at 6:18 | comment | added | Pankaj Gupta | Gustav, while this is trivially fixable, also note that the question asks not that all gaps are less than W, but that no gap is less than W (i.e., $1 - P(S_{k-2})$) | |
| Jun 27, 2013 at 23:43 | comment | added | Pankaj Gupta | Thanks. But I get a much more complicated result of that triple integral. Do you mind double-checking your results and/or adding a few intermediate steps in that calculation? | |
| Jun 27, 2013 at 20:12 | comment | added | Gustav | Pankaj, I have added an extra line to the calculation in the above answer, showing how to express the joint probability as an integral over the probability densities of the exponential random variables $X_i$. The limits on the integrals are such that $x_j+x_{j+1}<W$ and $x_{j+1}+x_{j+2}<W$. | |
| Jun 27, 2013 at 20:10 | history | edited | Gustav | CC BY-SA 3.0 |
Added a calculational detail
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| Jun 27, 2013 at 17:40 | comment | added | Pankaj Gupta | Can you explain the line that calculates $P(G_j, G_{j-1})$ ? Also, in that line, there is a typo as $G_{j-1}$ should correspond to $X_j + X_{j+1}$ | |
| Jun 27, 2013 at 10:20 | review | First posts | |||
| Jun 27, 2013 at 10:24 | |||||
| Jun 27, 2013 at 10:03 | history | answered | Gustav | CC BY-SA 3.0 |