5
$\begingroup$

I will be referring here to Nakagawa and Schielzeth (2013). As those authors state, $R^2$ for OLS regression could be defined as follows:

$$R^2 = \frac{\sum^n_{i=1}(\bar{y} - \hat{y_i})^2}{\sum^n_{i=1}(y_i - \bar{y})^2} = \frac{var(\hat{y_i})}{var(\hat{y_i})}$$

and so, if we define mixed model like this:

$$y_{ij} = \beta_0 + \sum^p_{h=1}\beta_hx_{hij} + \alpha_{j} + \epsilon_{ij}$$

then $R^2$ could be defined as follows:

$R^2$ for fixed effects ($R^2_m$):

$$R^2_m = \frac{\sigma^2_f}{\sigma^2_f + \sigma^2_\alpha + \sigma^2_\epsilon}$$

where:

$$\sigma^2_f = var\left( \sum^p_{h=1}\beta_hx_{hij}\right)$$

$R^2$ for random effects ($R^2_c$):

$$R^2_m = \frac{\sigma^2_f + \sigma^2_\alpha}{\sigma^2_f + \sigma^2_\alpha + \sigma^2_\epsilon}$$

However, this seems to be the same as ICC:

$$ICC = \frac{\sigma^2_\alpha}{\sigma^2_\alpha + \sigma^2_\epsilon}$$

and so:

$$R^2_c - R^2_m \approx ICC$$

This puzzled me a bit. Could you correct me if I am wrong? Why ICC is the same as $R^2$ for mixed models? If those two measures are equivalent, why are they not used like this? (or if they are, could you provide a reference?) Could you explain why they are equivalent?

Reference: Nakagawa, S. & Schielzeth, H. (2013). A general and simple method for obtaining $R^2$ from generalized linear mixed-effects models. Methods in Ecology and Evolution, 4, 133–142.

$\endgroup$

1 Answer 1

3
$\begingroup$

Nakagawa and Schielzeth (2013) call intra-class correlation ($ICC$) "closely connected to $R^2$" and refer to their 2010 paper. From posting this question I already seen few mentions of $ICC$ treated as "variance explained" since it in fact is a part of total variance explained by all the random effects and residuals. So it seems $ICC$ is used as an analog to $R^2$ while one have to remember that both statistics have their flaws (e.g. here and here).


Nakagawa, S. & Schielzeth, H. (2010). Repeatability for Gaussian and non-Gaussian data: a practical guide for biologists. Biological Reviews, 85, 935–956.

$\endgroup$

Your Answer

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

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