# How to prove that Adjusted R^2 is less than R^2

The adjusted R^2 formula is :

$$\overline{R}^{2}=1-\left( \left( 1-R^{2}\right) \cdot > \dfrac{n-1}{n-k}\right)$$

In case of k > 1 , I continue like that;

$$\overline{R}^{2}=1-\left( \dfrac{n-1}{n-k}-\dfrac{n-1}{n-k}R^{2}\right)$$

then

$$\left( n-k\right) \cdot \overline{R}^{2}=n-k-\left( n-1\right) +\left( n-1\right) R^{2}$$

so $$\left( n-k\right) \cdot \overline{R}^{2}-\left( n-1\right) R^{2}=1-k$$

but after that I don't know how to proceed, is there someone who has an idea?

$$SSRes=\sum_{i=1}^n\left( y_i-\hat y_i \right)^2\\ SSTotal=\sum_{i=1}^n\left( y_i-\bar y \right)^2$$

$$R^2=1-\dfrac{ SSRes/(n-1) }{ SSTotal/(n-1) }$$

$$\bar R^2=1-\dfrac{ SSRes/(n-k) }{ SSTotal/(n-1) }$$

When $$k>1$$, the numerator of the second equation will be larger, meaning that the second equation subtracts a larger number from $$1$$ than does the first equation.

Thus, $$R^2\ge \bar R^2.$$ $$\square$$

A way to think about $$R^2$$ is that it compares the error variance to the total variance, and adjusted $$\bar R^2$$ does the same but with an estimate of the error variance that is slightly larger. Since that estimate of the error variance is slightly larger, adjusted $$\bar R^2$$ must be slightly less favorable (slightly smaller).

There is an alternative form of writting $$\bar{R}^2$$ in terms of $$R^2$$, with $$p$$ being the number of covariates in the model and $$n$$ the number of observations: $$\bar{R}^2 = \frac{(n-1)R^2 - p}{n-p-1}$$ Now, subtract $$p$$ and add $$p$$ inside $$(n-1)$$: $$\bar{R}^2 = \frac{(n-1)R^2 - p}{n-p-1} = \frac{(n-p-1+p)R^2 - p}{n-p-1} = \frac{(n-p-1)R^2 + pR^2 - p}{n-p-1} = \frac{(n-p-1)R^2 - p(1-R^2)}{n-p-1} = R^2 - \frac{p(1-R^2)}{n-p-1}$$
We can see that $$\bar{R}^2$$ is $$R^2$$ minus something nonnegative whenever $$n > p + 1$$, so $$\bar{R}^2$$ will be less or equal to $$R^2$$. For the "equal" part, $$\bar{R}^2 = 1$$ only when $$R^2 = 1$$ (knowing that $$0 \le R^2 \le 1$$).
Continuing from \begin{align*} \left( n-k\right) \cdot \overline{R}^{2}-\left( n-1\right) \cdot R^{2} & = 1-k \\ \left( n-k\right) \cdot \overline{R}^{2} + (k-1) \cdot 1 & = \left( n-1\right) \cdot R^{2}. \end{align*} So $$R^{2}$$ is a weighted average of $$\overline{R}^{2}$$ and $$1$$, and since $$\overline{R}^{2} \leq 1$$: \begin{align*} \left( n-k\right) \cdot \overline{R}^{2} + (k-1) \cdot \overline{R}^{2} & \leq \left( n-1\right) \cdot R^{2} \\ \left(n-1\right) \cdot \overline{R}^{2} & \leq \left( n-1\right) \cdot R^{2} \\ \overline{R}^{2} & \leq R^{2} \\ \end{align*}