Is adjusted R squared score still appropriate when number of regressors is larger than the sample size? So I have a really small sample size of 50, and I have 80 regressors.
The $R^2$ score is about 0.1, and according to the following equation on Wikipedia about how to compute adjusted $\bar{R}^2$,
$$
\bar{R}^2 = R^2 - (1-R^2)\frac{p}{n-p-1} \\
R^2 = 0.1 \\
p = 80 \\
n = 50
$$
Then the adjusted $\bar{R}^2$ shoots over to 2.42. But wikipedia says $\bar{R}^2$ should always be less than or equal to $R^2$, so what am I doing wrong here? or is it just the model is wrong since so many regressors?
Edit
Both $R^2$ and $\bar{R}^2$ were computed from lasso regression instead ordinary least squares.
 A: The adjusted $R^2$ value is specifically for linear regression where it's easy to know the effect of adding many predictors. If you were doing linear regression with more predictors than samples a linear regression would give $R^2=1$ so you must not be using a linear regression model. That means you can't adjust the $R^2$ figure regardless of how large your sample size is.
If you tried to adjust the $R^2$ value with your figures you'll notice that you get a value greater than $1$ and this is statistically meaningless.
But your question is still relevant if you had used linear regression with more predictors than samples and got $R^2=1$. You'll notice that when $p=n-1$ the adjusted $R^2$ is undefined, and in fact the adjustment isn't valid when $p\geq n-1$
A: To state notation, let $y$ be the $n$-vector of responses, let $X$ be the ($n\times p$) design matrix and let $\beta$ be the $p$-vector of unknown regression coefficients, with $n$ being the sample size. The well known least squares estimate of $\beta$ is $\hat\beta = (X^TX)^{-1} X^Ty$.
The coefficient of determination is $R^2 = 1-\frac{SS_{res}}{SS_{tot}}$, where $SS_{tot}$ is the total sum of squares and $SS_{res}$ is the residuals sum of squares. The adjusted $R^2$ is as you wrote. 
Coming to you question, when $n<p$, $\hat\beta$ is not anymore uniquely defined because the inverse of $X^TX$ is not defined. Hence, as far as $n<p$, no matter what algorithm you use to find $\hat\beta$, the latter will always be undefined and arbitrary. Essentially, in this case, the objective function of $\beta$ is a flat surface. Consequently, $R^2$ is also arbitrary and therefore meaningless. For this reason, adjusted $R^2$ will be meaningless as well. That's why you obtain such a strange value for the adjusted $R^2$.
A: The adjusted R-square value is always less than R-square when n>p that means number of observation is greater than the number of parameters. 
