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

Question

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.

Answer 1

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.

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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"))

See also What is the objective function to optimize in glm with gaussian and poisson family?