# Tag Info

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You are not using OLS, so why do you think $R^2$ should increase when you add more variables? the premise of your question is flawed. You should show what command you used to generate the output. I can see that you probably used robustfit function in MATLAB. This function is not OLS. It uses an iterative algorithm to weight the observations, looking for what ...

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Even very weak nonzero effects will become statistically significant if you have enough data. A significant F test simply means that the proportion of variance explained is larger than would be expected if there was no effect at all, i.e., it is a statement about relative effect strength. It does not mean that the effect is large in absolute terms. You can ...

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Adding more variables will always increase the R squared. Additionally, there is no reason to remove variables which fail to reject the null as I explain here.

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The difficulty here is that a level of a categrical variable is not a variable --- it is an outcome of the variable. So this is a bit like asking if it is possible to get a measure of variance explained for a continuous variable for the outcome $X=3$. While it is an odd question, I suppose that the answer is yes, it is possible (though it would no longer be ...

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Yes, you can use $R^2$ as a goodness-of-fit measure in a VAR model. There are multiple dependent variables in a VAR model and there is an equation for each one. You can use the $R^2$ for each equation separately. If it is not part of standard output, you can obtain $R^2$ as the square of correlation between the true values and the fitted values. For a ...

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ORIGINAL ANSWER: All bets are off without knowing what MATLAB is doing here under the heading (robust fit). The results show quite different models: one forced through the origin (intercept zero) and the other omitting a predictor that the first model declares strongly significant. It's implausible -- despite the otherwise appealing figures of merit -- that ...

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One visual interpretation of adding more variables to your model, which implies adjusting for them and thus slicing your data, is that you may end up with only the points that are closer to the curve you're fitting. This explains why your $R^2$ is increasing when you add more variables. Indeed, your model fits better the data [that is left], however, this ...

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Keep in mind what the equation for $R^2$ is. $$R^2 = 1 - \dfrac{ \sum_{i=1}^n \big( y_i - \hat y_i \big)^2 }{ \sum_{i=1}^n\big( y_i - \bar y \big)^2 }$$ If you are measuring model performance in terms of the numerator (or something equivalent, like $MSE$ or $RMSE$), and you find that to be acceptable despite a low $R^2$, it means that the denominator is ...

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I'm aware that the question was asked a long time ago but in case anyone stumbles across it in the future - An Introduction to Statistical Learning (Chapter 3.2) has a good explanation to that question: The $R^2$ statistic has an interpretational advantage over the $RSE$, since unlike the $RSE$, it always lies between 0 and 1. However, it can still be ...

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