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How would you do a goodness of fit test for Lasso regression?

Im guessing that the $R^2$ value, as for linear regression, wont work anymore. Why is that?

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  • $\begingroup$ Why do you guess that? $\endgroup$
    – Dave
    Commented Sep 11, 2020 at 14:39
  • $\begingroup$ just a feeling, i dont have anymore justificaiton than that (which is worrying) @Dave $\endgroup$
    – Trajan
    Commented Sep 11, 2020 at 14:40
  • $\begingroup$ And what do you mean by "goodness of fit"? Would mean squared error suffice for your purposes? $\endgroup$
    – Dave
    Commented Sep 11, 2020 at 14:42
  • $\begingroup$ possibly, the first sentence i saw as an interview question, so i guess so @Dave $\endgroup$
    – Trajan
    Commented Sep 11, 2020 at 14:43

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The LASSO regression is still a linear regression.

In vanilla linear regression, we posit that the phenomenon follows the following:

$$y = X\beta + \epsilon$$

And we estimate $\beta$ via $\hat{\beta}_{ols} = (X^TX)^{-1}X^Ty$.

In ridge (linear) regression, we posit that the phenomenon follows the following:

$$y = X\beta + \epsilon$$

And we estimate $\beta$ via $\hat{\beta}_{ridge,\lambda} = (X^TX + \lambda I)^{-1} X^Ty$.

In LASSO (linear) regression, we posit that the phenomenon follows the following:

$$y = X\beta + \epsilon$$

And we estimate $\beta$ via $\hat{\beta}_{LASSO,\lambda} = \text{no closed-form expression}$.

Well we can argue forever about which estimator we want to use. Even for variance, it is not universal that we should be dividing by $n-1$ instead of $n$. However, we use all three estimators (OLS, ridge, and LASSO) to guess the $\beta$ in $y = X\beta + \epsilon$.

$R^2$ might not be ideal for LASSO regression, but it might not be ideal for OLS regression, either!

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  • $\begingroup$ i thought for OLS regression, it is least squares which is the estimator? $\endgroup$
    – Trajan
    Commented Sep 11, 2020 at 16:05
  • $\begingroup$ Squared error is the type of error that we aim to minimize, and it is the case that the $\hat{\beta}_{ols}$ that I gave minimizes $\sum(y_i - \hat{y}_i)^2$, hence the term "least squares" for "least squared error". $\endgroup$
    – Dave
    Commented Sep 11, 2020 at 16:09

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