2
$\begingroup$

the $R^2$ is the amount of variation of the dependend variable $Y$ explained by linear regression. It can take values between 0 and 1.

If I use the adjusted $R^2$ instead, what is the interpretation of this value?

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
0
$\begingroup$

I think there're many existing posts covering it. The wikipedia has a nice section on it, so I think you should take a look.

When you fit a regression model, you will get a R2 statistic. We all want a model with high R2, but we don't want to overfit our model. Our job is to create a model with variables that can explain the variability significantly. We don't want to pour all variables into the data-set.

Unfortunately, we can always add more variables into the model to achieve a higher R2, regardless of whether those variables are correlated with the dependent variable.

Adjusted R2 is an attempt to take care of it. It can decrease if a new variable isn't useful to explain the variability.

EDITED

Adjusted R2 has no direct and simple interpretation, it is a metric to compare two different models. You can think of it like AIC or BIC. It is a measure of fit.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I know how to calculate the adjusted $R^2$ and I know its usefulness, but what does it say? What is its interpretation? Adjusted $R^2$ can take negative values, $R^2$ cannot. $\endgroup$
    – spore234
    Jun 2, 2015 at 8:56
  • $\begingroup$ Check my EDITED answer. $\endgroup$
    – SmallChess
    Jun 2, 2015 at 9:25
  • 1
    $\begingroup$ but what if $n >> p$, then $R^2 \sim$ adj. $R^2$ ? $\endgroup$
    – spore234
    Jun 2, 2015 at 9:31
  • $\begingroup$ Yes, see the comments here: stats.stackexchange.com/questions/147688/… $\endgroup$
    – Christoph Hanck
    Jun 2, 2015 at 9:43
0
$\begingroup$

The interpretation of $1-R^2_{Adj}$ is the same as that of any other regularized loss function - penalizing a complicated model.

In particular this is similar to $L_0$ norm which also penalizes number of parameters but is not as popular as $ L_1$ and $L_2$ regularization because of their nice properties. See also their logistic counterparts, Elastic Net and Lasso regression.

Improve this answer
$\endgroup$

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