Timeline for Estimating the probability that a software change fixed a problem
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2010 at 12:22 | comment | added | PeterR | For any mean and variance a Gaussian has a non-zero probability of a negative value, and also a non-zero value of a non-integer value. Both these are not appropriate in this situation (what would it mean for the -2.7th trial to fail?). | |
| Nov 4, 2010 at 21:19 | comment | added | SiegeX | @PeterR: Does this assumption of independence in failures not apply to a normal/Gaussian distribution? | |
| Nov 3, 2010 at 19:06 | comment | added | PeterR | Assuming a geometric distribution is assuming that the failures are independent. In other words, on any trial the device fails with probability p, independent of the other trials (like tossing a coin). This is a reasonable starting point, but if the failure is due to something like a memory leak, then the longer the device runs, the more likely it is to fail, and the assumption of independent failures would be invalid. | |
| Nov 3, 2010 at 19:05 | comment | added | whuber♦ | @SiegeX: no, we're not playing fishing games here. When the probability of failure is constant over time the distribution of failure times will be exponential. Alternative theories of failure probabilities lead to other predictions of the failure time distribution. You have to be prepared to be wrong about your assumptions! (That's one reason to check them using techniques like probability plots.) | |
| Nov 3, 2010 at 18:21 | comment | added | SiegeX | @csgillespie: Thank you for taking the time to write this answer. I have added an Update section to my original answer. One more question I have for you is if there is an underlying reason why we believe failures rates to be exponentially distributed or are we just back fitting the empirical data we see from known failures to a distribution that fits? | |
| Nov 3, 2010 at 16:59 | comment | added | csgillespie | Each test does have it's own failure probability - it's 1. We've got the data to prove it ;) Anyway I'm assuming that each test is a replicate from the Geometric distribution. | |
| Nov 3, 2010 at 16:56 | comment | added | James | Then shouldn't each test have its own failure probability? | |
| Nov 3, 2010 at 16:14 | comment | added | csgillespie | No I don't think so. I took the question to mean that he ran device until he reached a failure. So it failed on 100 iterations, 22 iterations, ... | |
| Nov 3, 2010 at 16:09 | comment | added | James | Surely p^ is 7/100 (or however many trails were used in the original test)? | |
| Nov 3, 2010 at 15:16 | history | answered | csgillespie | CC BY-SA 2.5 |