Timeline for Why is the link function a function of the mean and not the linear predictor?
Current License: CC BY-SA 3.0
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| Apr 7, 2017 at 18:31 | comment | added | user135912 | Fair enough. But even though it's pointless, maybe there's something elegant about being able to use a constant link function. Eg, if $\mu=h(\eta)$ is constant, then you still have response distribution for the data, but if $\eta=g(\mu)$ is constant, then there is no well-defined response distribution | |
| Apr 7, 2017 at 18:05 | comment | added | Tim | @51413 well, it needs to be a function that makes sense in this context. E.g. neither of the sources says that it cannot be a constant function, but obviously using constant function as a link function would be absolutely pointless... | |
| Apr 7, 2017 at 17:17 | comment | added | user135912 | ...which might have some use in the way that linear models are often not identifiable without constraints, (but I don't know if that would actually be useful because I don't know anything about GLMs). | |
| Apr 7, 2017 at 17:15 | comment | added | user135912 | Thanks for your detailed response. It does seem sort of trivial and maybe "intuitive" was the wrong word. But to me, it seems like if you define it as $g(\mu)=\eta$ then you should mention that $g$ has to be 1-1. For example, wikipedia's page on GLMs makes no mention of this, although they do imply it by using $g^{-1}$. It's a minor point, but I still feel it's slightly confusing to an unfamiliar person looking for a concise description of what is allowed by "generalized linear model." Moreover, it seems like by defining it as $\mu=h(\eta)$ you could allow for $h$ that are not 1-1... | |
| Apr 7, 2017 at 17:00 | vote | accept | CommunityBot | ||
| Apr 7, 2017 at 10:49 | history | edited | Tim | CC BY-SA 3.0 |
added 300 characters in body
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| Apr 7, 2017 at 10:41 | history | answered | Tim | CC BY-SA 3.0 |