# How arbitrary is adjusted r-squared as a measure of fit?

I came across the adjusted $$R^2$$ for multivariate linear models, $$R^{2}_{adjusted} = 1 - \frac{SSE / (n-p-1)}{SSTO / (n-1)}$$, and I was curious what kinds of properties this satisfies. (Googling was not very helpful). I can tell that if we add additional predictors that don't help to explain the observations, then this quantity will strictly decrease, but are there other interesting properties this satisfies, or is just another somewhat arbitrary way of measuring fit?

Thanks.

• It's definitely only one way. If you're in 2D, or maybe even in 3D, it's not a bad idea to plot your residuals. If the residuals look random, you likely have captured most of the features of interest, particularly if $R^2$ is high. Mar 10, 2020 at 21:00
• @AdrianKeister Why only in 2D or 3D?
– Dave
Mar 16, 2020 at 12:15
• @Dave: It's difficult to plot residuals as a function of, say, a 4D independent variable. At that point, you're dealing with 5 dimensions, and our graphing abilities are pretty much non-existent in those higher dimensions. Naturally, you can project down into lower dimensions, but you'll always lose information when you do that. Mar 16, 2020 at 15:00
• @AdrianKeister The residuals exist in the space of the response variable ($\hat{y}_i-y_i$), so I am not sure what you mean.
– Dave
Mar 16, 2020 at 15:03
• @Dave: True, but that's the $y$ axis when you plot. What will be your $x$ axis? Mar 16, 2020 at 15:37

It makes a lot of sense.

$$R^2$$ measures the ratio of residual variance (numerator) to the total variance (denominator), calculating each variance as if the observations were the whole population (literally applying the discrete $$\mathbb E\left[\left(X-\mathbb E\left[X\right]\right)^2\right]$$ formula to the residuals (numerator) and the pooled distribution of all $$Y$$ values (denominator)).

$$R^2=1-\dfrac{ SSE/n }{ SSTotal/n }\\ SSE\text{: Sum of squared residuals (“errors”)}\\ SSTotal\text{: Total sum of squared deviations of } y\text{ values from }\bar y$$

However, these are biased estimates of the respective variances!

By dividing by $$n-p-1$$ in the numerator and $$n-1$$ in the denominator, we now have a ratio of unbiased estimates of the respective variances.

• (+1, I mean that's the correct answer I don't get it why 0 votes. Maybe a reference to the "residuals degrees of freedom" is missing if we are somewhat picky but still...) Nov 12, 2022 at 20:43