Timeline for Removing undefined (NaN) values during log-likelihood maximization
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2016 at 16:10 | vote | accept | Danielle | ||
| Jun 24, 2016 at 16:10 | history | edited | Danielle | CC BY-SA 3.0 |
added 65 characters in body
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| Jun 24, 2016 at 9:27 | answer | added | Jarle Tufto | timeline score: 2 | |
| Jun 24, 2016 at 1:24 | comment | added | Danielle | I am trying to understand if this is statistically the correct thing to do, or if it introduces an unfair "cheat" that should be avoided. | |
| Jun 24, 2016 at 1:21 | comment | added | Danielle | I am concerned that by just dropping these bins, without any further action, I am not taking an adequate statistical penalty for the loss of information. I also don't fully understand the term "statistical penalty"; I had the impression that not only am I unable to use that bin to provide the statistics needed to tighten my confidence intervals, I also need to analytically account for the fact that this bin cannot be used. For example, my implementation also explicitly sets some elements of the array to zero if we don't want them to count, such as in the case a detector is bad. | |
| Jun 23, 2016 at 20:42 | comment | added | whuber♦ | How can cases with $k=m=0$ possibly be part of the likelihood? What information do you get when the model says nothing will appear and nothing appears? (A Poisson model with a parameter of $0$ assigns all the probability to $0$, after all.) The only place you can run into trouble is where the model predicts nothing will appear and something does: that would be definitive evidence the model is wrong. | |
| Jun 23, 2016 at 20:38 | history | asked | Danielle | CC BY-SA 3.0 |