# Is it a problem to get a negative adjusted r-squared?

Background: I have the cross-sectional model:

$Y_{i} = a + b X_{1,i} + c X_{2,i} + d X_{3,i} + e X_{4,i} + \nu_i$.

The application is corporate finance. So each $Y_i$ is something like the change in return on assets over a 1 year period for firm $i$, and the regressors are typical corporate finance variables.

In corporate finance, very small values of $R^2$ are common, even sometimes $1\%$. My $R^2$ is around $1\%$ but my Adjusted $R^2$ is $-0.2\%$.

I have never seen papers report negative Adjusted $R^2$ but this could just be because they omit the publication of their adjusted $R^2$ when they see that it is negative.

### Question

Is there some problem when the adjusted $R^2$ is negative?

• It does mean that some or all of your independent variables have no relation with your dependent variable. Mar 3 at 23:58

The formula for adjusted R square allows it to be negative. It is intended to approximate the actual percentage variance explained. So if the actual R square is close to zero the adjusted R square can be slightly negative. Just think of it as an estimate of zero.

If $R^2_{adj} \leq 0$, then: \begin{align} 1 &\leq \dfrac{n-1} {n-p} .\dfrac{SSE} {SST} \\[8pt] \dfrac{SST}{SSE} &\leq \dfrac{n-1}{n-p} \\[8pt] \dfrac{SST-SSE}{SSE} &\leq \dfrac{n-1-n-p}{n-p} \\[8pt] R^2 &\leq \dfrac{p-1}{n-p} \end{align} and we know that if $R^2\leq1$, then: $$1 \leq \dfrac{p-1}{n-p}$$ $R^2_{adj}$ must be negative. Therefore, \begin{align} n-p &\leq p-1 \\[8pt] \dfrac{n+1}{2} &\leq p \end{align}

This means the number of variables must be more than $\dfrac{n+1}{2}$ to get a negative adjusted R-squared.

• Welcome to the site, @saeed_ans. I reformatted your answer in the hopes of making it easier to read. Please ensure it still says what you want. If it doesn't, or you otherwise don't like it, roll it back with my apologies. Dec 16, 2017 at 0:21
• This answer is not correct. If SSE = SST, you can very well have negative R^2 with just p = 2. Dec 16, 2017 at 0:44

The above derivation is incorrect. \begin{align*} R_{adj}^2 &< 0\\ 1-\frac{n-1}{n-p-1}(1-R^2) &<0\\ \frac{n-p-1}{n-1} &< 1-R^2\\ R^2 < 1-\frac{n-p-1}{n-1}\\ R^2 < \frac{p}{n-1}. \end{align*}

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