# Mean finger volume: Is a GLM with log link function appropriate?

I have a model where the volume ($$V$$) of a finger is normally distributed, with mean $$\mu = \beta_0 L^{\beta_1}D^{\beta_2}$$ (where $$L=$$ length, $$D=$$ diameter and $$\beta_i \in \Bbb R$$ for $$i=0,1,2$$) and some variance $$\sigma^2$$.

I was thinking that if this is a GLM (although I am not sure if we have an exponential family here), a log link function would be appropriate since $$\ln (\mu)$$ is a linear function of the natural log of the predictors: $$\ln (\mu) = \ln(\beta_0 L^{\beta_1}D^{\beta_2}) = \ln(\beta_0) + \beta_1\ln(L) + \beta_2\ln(D)$$

However, a Poisson distribution is not ideal since $$V, L, D \in \Bbb R$$ are not counts.

So, I am not sure if I should proceed with this, or if another link function would be more appropriate. Any suggestions would be appreciated.

• @Tim I mean that $V$ itself is normally distributed. Sep 17, 2021 at 8:06
• What you suggest is a glm that can be fitted with glm(V ~ log(L) + log(D), family=gaussian(link="log")). But I would expect lm(log(V) ~ log(L) + log(D)) (assuming that $V$ conditional on $L$ and $D$ follows a lognormal distribution) to give a better fit to the data. This is also (for good reasons) how most allometric data such as this is analysed, see en.wikipedia.org/wiki/Allometry Sep 17, 2021 at 10:32
• @JarleTufto Just to clarify, are you saying that my model $V$ ~ $N(\beta_0 L^{\beta_1}D^{\beta_2}, \sigma ^2)$ is fitted with glm(V ~ log(L) + log(D), family=gaussian(link="log")), but you think it would be better fit by lm(log(V) ~ log(L) + log(D))? Sep 17, 2021 at 10:46
• @whuber Yes, I fully agree that these are different models (and not different ways of fitting the same model as perhaps implied by @Manuel). Sep 17, 2021 at 15:38
• Manuel, the question expressed in your latest comment about whether to include an intercept is answered at stats.stackexchange.com/questions/7948. The short answer is that you ought to include it. In fact, in what way would a nonzero intercept make no sense? In your model its exponential is a multiplicative factor in $V$ and so it would have to vary if you re-expressed $V$ with different units of measurement. That suggests including an intercept is necessary.
– whuber
Sep 17, 2021 at 15:58

I have a model where the volume ($$V$$) of a finger is normally distributed, with mean $$\mu = \beta_0 L^{\beta_1}D^{\beta_2}$$

You can rewrite this as

a model where the volume ($$V$$) of a finger is normally distributed, with mean $$\mu = \exp \left( \beta_0^\prime + \beta_1 L^\prime + \beta_2 D^\prime \right)$$ where $$L^\prime = \log(L)$$, $$D^\prime = \log(D)$$ , $$\beta_0^\prime = \log(\beta_0)$$

So the mean can be expressed as a linear function of the independent variables $$\beta_0^\prime + \beta_1 L^\prime + \beta_2 D^\prime$$ wrapped inside a non-linear function.

That classifies as a GLM. (provided that the deviation parameter $$\sigma$$ is independent from $$L$$ and $$D$$, ie. constant)

although I am not sure if we have an exponential family here

The normal distribution is in the exponential family.

Instead of the Poisson distribution you can use other distributions. The log link does not restrict this. Jarno's comment shows how you can do it in R glm(V ~ log(L) + log(D), family=gaussian(link="log"))