1
$\begingroup$

I've been reading the book "Generalized Linear Models with Examples in R" by Dunn and Smyth.

In chapter 5, they claim:

Variance-stabilizing transformations h(y) used with linear regression models are roughly equivalent to fitting a glm with variance function V (μ) = 1/h (μ)2 and link function g(μ) = h(μ)

So, is it the same using the GLMs with its respective link functions than transforming Y in linear regression? Can you please explain this in simple and practical terms to a person without a very good foundation in stats theory like me?

$\endgroup$

1 Answer 1

3
$\begingroup$

The details are mostly explained a couple of pages earlier, at the start of sec 5.8, and that in turn refers back to chapter 3 (in particular, see sec 3.9, especially 3.9.2).

Consider $\text{Var}(y)=\phi V(\mu)$ and the transformation $y^*=h(\mu)$. What effect does that have on the variance, approximately?

Under suitable conditions, we can obtain an approximation of the moments of transformed random variables via Taylor expansion.

Using a one-term approximation as given at that Wikipedia link:

$$\text{Var}(y^*)=\text{Var}\!\left(h(y)\right)\approx \left(h'(\mu _{y})\right)^{2}\sigma _{y}^{2}=\left(h'(\mu _{y})\right)^{2}.\phi V(\mu _{y})$$

How do we stabilize variance? That is, how do we choose $h$ to make $\text{Var}(y^*)$ approximately constant? By making $\left(h'(\mu _{y})\right)^{2}\propto V(\mu _{y})^{-1}$ $\ldots\:{(1)}$, so that the two functions of $\mu$ in the expression $\left(h'(\mu _{y})\right)^{2}\cdot\phi\cdot V(\mu _{y})$ cancel, so that this approximation to the variance is then a constant function of $\mu$.

So now back to Sec 5.8. What GLM specification for $y$ approximately corresponds (in terms of mean and variance) to taking the transformation $y^*=h(y)$ and then fitting a constant-variance linear regression?

If $E(y^*)=E[h(y)]$ is linear in $\eta = β_0 + β_1x_1 + ··· + β_px_p=X\beta$ (e.g. see sec 1.6 between eqn 1.5 and 1.6 and the first equation in 4.6.1), then the mean as a function of the $x$'s needs to be such that $g(\mu)=β_0 + β_1x_1 + ··· + β_px_p$ (see sec 1.6 p 13)

But now consider just the first term in the Taylor expansion about the mean; $E[h(y)] \approx h(\mu)$ (see the section on the first moment in the Wikipedia link); so we will need that the link function $g(\mu)\approx h(\mu)$.

In the transformed regression we also had $\text{Var}(y^*)=c$, so if $h$ was a variance-stabilizing transformation, then $\text{Var}(y)\propto 1/h'(μ)^2$, just as in eqn $(1)$ above.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.