# Tag Info

0

I don't understand what is intended with the phrase "information content". That being said, you might investigate any one of several pseudo r-square measures. Efron's pseudo r-square relies on the difference between the y values predicted by the model and the observed y values. So, it's pretty easy to explain. Some other pseudo r-square values compare ...

6

No. $R^2$ in instrumental variables regression is not useful. Since one of the explanatory variables $x$ is correlated with the error $\epsilon$ we can't decompose the variance of the outcome $y$ into $\beta^2 Var(x) + Var(\epsilon)$, so the obtained $R^2$ has neither a natural interpretation, nor can it be used for computation of F statistics. ...

2

It is well-known that the relationship between $R^2$ and adjusted $R^2$ in a linear regression (and ultimately, a fixed-effects regression can also be seen as a linear regression, see e.g. Difference between fixed effects dummies and fixed effects estimator?) is (see e.g. Is $R^2_{adjusted}$ both unbiased and consistent under the alternative in simple ...

2

Suppose we have a linear regression model $$y=y_{12\ldots p-1}+\varepsilon_{12\ldots p-1}\,,$$ where $y_{12\ldots p-1}=\beta_0+\beta_1 x_1+\beta_2x_2+\cdots+\beta_{p-1}x_{p-1}$ is the part of $y$ explained by $(x_1,x_2,\ldots,x_{p-1})$ and $\varepsilon_{12\ldots p-1}$ is the unexplained part. Parameters $(\beta_0,\beta_1,\ldots,\beta_{p-1})$ are estimated ...

0

If you are targeting on getting better accuracy instead of interpretability, DO NOT do feature selection. Use all features and add regularization. In general, more features means more information, in an extreme case, the feature has nothing to do with the prediction target, the fit will get the coefficient to be 0 automatically. So, more feature will not ...

0

Yes, there is an advantage to $R^2$: It has a direct interpretation as the proportion of variance in the dependent variable that is accounted for by the model. Adjusted $R^2$ does not have this interpretation. Also, you write that adjusted $R^2$ "penalizes the model for useless variables". That is true but incomplete. First, almost no variable is totally ...

0

Yes, you can use the residuals of the fitted model (uGARCHfit class object) to calculate the $R^2$ and the $R^2_{adj.}$ manually. Note: you need raw residuals, not standardized ones. You do not use the estimated sigma from the fitted model to weight an lm object and extract the $R^2$ from there; the classical definition of $R^2$ does not involve any ...

3

It depends on what kind of outlier / the location of the outlier with respect to the regression line. Imagine a bivariate data cloud without the outlier. X and Y covary, so the cloud is oblong. Draw the regression line. If the outlier lies along that regression line, but far from the cloud so that the observation has high leverage, then the outlier ...

2

For the question: "How much of a dependent variable is explained by each of a bunch of independent variables?", the short answer is the so-called adjusted R-squared. It is a modified version of R-squared that has been adjusted for the number of predictors (degrees of freedom) in the model. The adjusted R-squared increases only if the new variable actually ...

6

The important thing is that the influence that each predictor has depends on which other predictors are in the model. So the effect of each predictor depends on the model you're using, and which other variables are in it. For example, you might have two highly collinear variables that both have a large $R^2$ when used in a bivariate linear model, but, ...

Top 50 recent answers are included