Timeline for How to deal with under-dispersion in negative binomial GLMM?
Current License: CC BY-SA 4.0
8 events
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| Apr 12 at 8:19 | comment | added | LT17 | Thank you very much, that is clearer now :) | |
| Apr 11 at 17:50 | comment | added | jbowman | It does in some circumstances, but... a rate can be > 1, so an offset doesn't act as an upper bound. A typical use would be if you are collecting Poisson data over intervals of different lengths $t$; the length of the interval would be used as an offset, thus allowing you to more easily estimate the rate parameter $\lambda$ (otherwise, you would have $\lambda t$ everywhere.) Instead of the link $\log (\lambda t) = X'\beta$, you'd have $\log (\lambda t) = \log(t) + X'\beta$, which you can see reduces to $\log(\lambda) = X'\beta$, almost certainly what you actually want to estimate. | |
| Apr 11 at 15:18 | comment | added | LT17 | 1. Okay, but even if it's a covariate with a fixed coefficient, I understood that its functions in this case would be to allow me to treat counts as rates (looking at, e.g., stats.stackexchange.com/questions/66791/… and stats.stackexchange.com/questions/11182/…). 2. Yes, I tried it in that way. The model is more under-dispersed (p-value of 0.001 instead of 0.016), the curve of the QQ-plot looks more pronounced, but the residuals look slightly better. | |
| Apr 11 at 14:16 | comment | added | jbowman | 1. No, that's not what an offset is. An offset is really a covariate that's included in the model with a coefficient fixed at $1$, not estimated. 2. You'd treat $p$ as the count of successes and $POAR-p$ as the count of failures. | |
| Apr 11 at 9:36 | comment | added | LT17 | Thank you for the suggestion. Yes, p is always ≤ POAR. I'm not sure a binomial distribution would be adequate, as I don't have success/failures. If I do have to use a proportion (thus I am indirectly giving the upper boundary), am I not doing something similar as a negative binomial with offset? My understanding so far was that the offset was giving an upper bound to my response variable. | |
| Apr 11 at 1:15 | comment | added | jbowman | Since $p$ is a count that is $\leq POAR$, if I've understood you correctly, it would seem that a Binomial distribution would be better than a Negative Binomial, which does not have an upper bound on the variate. A Binomial variate is underdispersed relative to a Poisson, so that might fix your apparent underdispersion problem as well. | |
| Apr 10 at 16:29 | history | edited | kjetil b halvorsen♦ | CC BY-SA 4.0 |
deleted 46 characters in body; edited tags
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| Apr 10 at 16:11 | history | asked | LT17 | CC BY-SA 4.0 |