<p>That's a great answer from Dave. I have another answer.</p>
<p>There is an alternative form of writting <span class="math-container">$\bar{R}^2$</span> in terms of <span class="math-container">$R^2$</span>, with <span class="math-container">$p$</span> being the number of covariates in the model and <span class="math-container">$n$</span> the number of observations:
<span class="math-container">$$
\bar{R}^2 = \frac{(n-1)R^2 - p}{n-p-1}
$$</span>
Now, subtract <span class="math-container">$p$</span> and add <span class="math-container">$p$</span> inside <span class="math-container">$(n-1)$</span>:
<span class="math-container">$$
\bar{R}^2 = \frac{(n-1)R^2 - p}{n-p-1} = \frac{(n-p-1+p)R^2 - p}{n-p-1} = \frac{(n-p-1)R^2 + pR^2 - p}{n-p-1} = \frac{(n-p-1)R^2 - p(1-R^2)}{n-p-1} = R^2 - \frac{p(1-R^2)}{n-p-1}
$$</span></p>
<p>We can see that <span class="math-container">$\bar{R}^2$</span> is <span class="math-container">$R^2$</span> minus something nonnegative whenever <span class="math-container">$n &gt; p + 1$</span>, so <span class="math-container">$\bar{R}^2$</span> will be less or equal to <span class="math-container">$R^2$</span>.
For the &quot;equal&quot; part, <span class="math-container">$\bar{R}^2 = 1$</span> only when <span class="math-container">$R^2 = 1$</span> (knowing that <span class="math-container">$0 \le R^2 \le 1$</span>).</p>

CC BY-SA 4.0
2024-04-01T14:42:23.030Z
2024-04-01T14:42:23.030Z
igorkf
<p>Continuing from
<span class="math-container">\begin{align*}
\left( n-k\right) \cdot \overline{R}^{2}-\left( n-1\right) \cdot R^{2} &amp; = 1-k \\
\left( n-k\right) \cdot \overline{R}^{2} + (k-1) \cdot 1 &amp; = \left( n-1\right) \cdot R^{2}.
\end{align*}</span>
So <span class="math-container">$R^{2}$</span> is a weighted average of <span class="math-container">$\overline{R}^{2}$</span> and <span class="math-container">$1$</span>, and since <span class="math-container">$\overline{R}^{2} \leq 1$</span>:
<span class="math-container">\begin{align*}
\left( n-k\right) \cdot \overline{R}^{2} + (k-1) \cdot \overline{R}^{2} &amp; \leq \left( n-1\right) \cdot R^{2} \\
\left(n-1\right) \cdot \overline{R}^{2} &amp; \leq \left( n-1\right) \cdot R^{2} \\
\overline{R}^{2} &amp; \leq R^{2} \\
\end{align*}</span></p>

CC BY-SA 4.0
2024-04-01T16:17:18.777Z
2024-04-01T16:17:18.777Z
Giskard
<p><span class="math-container">$$
SSRes=\sum_{i=1}^n\left(
y_i-\hat y_i
\right)^2\\
SSTotal=\sum_{i=1}^n\left(
y_i-\bar y
\right)^2
$$</span></p>
<p><span class="math-container">$$
R^2=1-\dfrac{
SSRes/(n-1)
}{
SSTotal/(n-1)
}
$$</span></p>
<p><span class="math-container">$$
\bar R^2=1-\dfrac{
SSRes/(n-k)
}{
SSTotal/(n-1)
}
$$</span></p>
<p>When <span class="math-container">$k&gt;1$</span>, the numerator of the second equation will be larger, meaning that the second equation subtracts a larger number from <span class="math-container">$1$</span> than does the first equation.</p>
<p>Thus, <span class="math-container">$R^2\ge \bar R^2.$</span> <span class="math-container">$\square$</span></p>
<p>A way to think about <span class="math-container">$R^2$</span> is that it compares the error variance to the total variance, and adjusted <span class="math-container">$\bar R^2$</span> 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 <span class="math-container">$\bar R^2$</span> must be slightly less favorable (slightly smaller).</p>

CC BY-SA 4.0
2022-11-25T14:12:42.520Z
2022-11-25T23:45:47.710Z
2022-11-25T23:45:47.710Z
Dave
<p>The adjusted R^2 formula is :</p>
<blockquote>
<p><span class="math-container">$$ \overline{R}^{2}=1-\left( \left( 1-R^{2}\right) \cdot   
&gt; \dfrac{n-1}{n-k}\right) $$</span></p>
</blockquote>
<p>In case of k &gt; 1 , I continue like that;</p>
<blockquote>
<p><span class="math-container">$$ \overline{R}^{2}=1-\left( 
 \dfrac{n-1}{n-k}-\dfrac{n-1}{n-k}R^{2}\right) $$</span></p>
</blockquote>
<p>then</p>
<p><span class="math-container">$$
\left( n-k\right) \cdot \overline{R}^{2}=n-k-\left( n-1\right) +\left( n-1\right) R^{2}
$$</span></p>
<p>so
<span class="math-container">$$
\left( n-k\right) \cdot \overline{R}^{2}-\left( n-1\right) R^{2}=1-k
$$</span></p>
<p>but after that I don't know how to proceed, is there someone who has an idea?</p>

<econometrics><regression><statistics>
How to prove that Adjusted R^2 is less than R^2
Tatanik501
# How to prove that Adjusted R^2 is less than R^2

(Asked by: username_0 on 2022-11-25)

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?

**Tags:** econometrics, regression, statistics

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## 3 answers

### Answer by username_3 on 2022-11-25. Score: 8 | Accepted answer

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

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

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

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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### Answer by username_1 on 2024-04-01. Score: 2

That's a great answer from Dave. I have another answer.

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 $0 \le R^2 \le 1$).


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### Answer by username_2 on 2024-04-01. Score: 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\*}


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