1
$\begingroup$

Generalized linear models are formulated so that the link function $g$ of the mean $\mu$ of a random variable $Y$ is equal to the linear predictor $\eta=x'\beta$, i.e. \begin{align} g(\mu)=\eta. \end{align} My question is, why don't we call $h=g^{-1}$ the link function and write $\mu=h(\eta)$. That way, if $h$ is not one-to-one, there is still a distribution on the response variables but the $\beta$ are not identifiable. Isn't it problematic that for certain $g$, there is no well-defined model? It also is more intuitive to think of the mean as a function of the parameters, rather than the reverse. Is there a conceptual advantage to defining $g$ in this way? Is it mathematically necessary for some reason I'm not seeing?

$\endgroup$
3
  • 1
    $\begingroup$ Because someone has written it like this and we use it as so for historical reasons... moreover different people find different things to be "intuitive". $\endgroup$
    – Tim
    Apr 6, 2017 at 18:02
  • $\begingroup$ @Tim So it is just historical. Hmm, it just seems a little strange to me $\endgroup$
    – user135912
    Apr 6, 2017 at 18:14
  • $\begingroup$ stats.stackexchange.com/questions/41306/… $\endgroup$ Apr 7, 2017 at 11:00

1 Answer 1

2
$\begingroup$

First of all, the notation is not used consistently. For example, Nelder and Wedderburn (1972) write about linear predictor $Y$ and a linking function "$\theta = f(Y)$ connecting the parameter $\theta$ of the distribution of $z$ with the $Y$'s of the linear model" (p. 372). On another hand, McCullaugh and Nelder (1983) use the link function $\eta_i = g(\mu_i)$ (p. 27). So both notations were used in the classical literature on GLM's.

Basically it doesn't matter if you use $f = g^{-1}$ since what you need is a one-to-one mapping that works in both direction (through it's inverse). All the common link functions have this property. Nobody said that $g$ can be any function...

As about "intuitiveness", notice that if you waned to use linear regression for a count data, then you would usually transform the outcome using a log transformation and then fit a linear regression to it, so in some cases transforming the outcome is also "intuitive".

Moreover, I don't find anything more intuitive in considering transformed mean as a function of linear predictor vs mean as a function of transformed linear predictor, they are the same.

Nelder, J. and Wedderburn, R. (1972). Generalized Linear Models. Journal of the Royal Statistical Society. Series A (General). Blackwell Publishing. 135 (3): 370–384.

McCullagh, P. and Nelder, J. (1989). Generalized Linear Models, Second Edition. Boca Raton: Chapman and Hall/CRC.

$\endgroup$
4
  • $\begingroup$ 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... $\endgroup$
    – user135912
    Apr 7, 2017 at 17:15
  • $\begingroup$ ...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). $\endgroup$
    – user135912
    Apr 7, 2017 at 17:17
  • $\begingroup$ @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... $\endgroup$
    – Tim
    Apr 7, 2017 at 18:05
  • 1
    $\begingroup$ 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 $\endgroup$
    – user135912
    Apr 7, 2017 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.