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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?

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1 Answer 1

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

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

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

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).

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