An R-squared is the percentage of variance explained by a model. Let's say your data has a variance of 100: that is the sum of squared errors versus the mean and divided by $N-1$ (the degrees of ...

I can think of a topical example. If we look at cities overall, we see more coronavirus infections and deaths in denser cities. So clearly, density yields interactions yields infections yields deaths, ...

Heavy Tails or "Peakedness"? Kurtosis is usually thought of as denoting heavy tails; however, many decades ago, statistics students were taught that higher kurtosis implied more "...

Range, IQR, and really any quantile interval lengths; standard deviation/variance, kurtosis (not excess kurtosis), and all other even-power moments; and, likelihood and other chi-squared statistics (e....

You are doing nothing wrong. Ridge regression, the LASSO, and other penalized-coefficient regressions yield biased estimations. The idea is that maybe accepting a little bias will greatly reduce the ...

I would recommend A Distribution-Free Theory of Nonparametric Regression by Györfi, Kohler, Krzyżak and Walk. It looks like what you are looking for: proofs use sigma fields as needed, there is plenty ...

One approach would be to use a chi-square test. To do this, you would estimate the overall distribution of sales in various genres from all of the data. For example, say your data revealed this ...

The answer depends on contrasts: how you code these factors and what is your baseline. A lot of software operates from a perspective of experimental analysis and so takes the first factor level as the ...

This looks like you have a problem with multicollinearity: trait1 and trait3 are correlated. You can imagine creating such a scenario like so: Find a covariate (we'll call it trait.unseen) that is ...

The simple answer is because we have known distributions for those quantities. We generally believe differences of means and proportions should be asymptotically normal (and probably close to normal ...

There is no clear answer. If there is a theoretical reason to expect an interaction (or testing for one is important), then try out an interaction. For example: I hope people researching allergy ...

Not necessarily. Search around on feature selection and model selection. Model selection is not a solved problem and it is unlikely to be solved since it is NP-hard. In my own experience, I have seen ...

I suspect you will get lots of opinions here since a good introductory text is a subjective ask. That said, you might want to look at something like Weisberg's Applied Linear Regression. That is used ...

I think there are a few ways you could look at your data. On the one hand, you have a dataset that is almost always right-censored. You could try to model the maximum MWh output if the plant were not ...

Random effects models require having all of the data in memory for most estimation methods apart from ANOVA-based estimators. Historically, ANOVA-based estimators were used when calculations were done ...

A "Hard" Problem There does not exist a "best way" to select the variables in your model. If you have $p$ possible covariates, you get $2^p$ possible models without any ...

Financial prices are, in general, not stationary. However, for a number of theoretical and empirical reasons we do think that log-returns (differences of $\log($prices$)$) are closer to stationary. ...

This problem is quickly and easily solvable if you split up the regions into four pieces: $x_0\geq a \cap x_n\leq b$: Solve $\arg\min ||\mathbb{A}x-b||^2 + \text{other linear least squares terms}$ $... View answer Accepted answer 2 votes No, those are just eigenvectors. You need to save the output of prcomp into a variable and then look at the sdev component of that variable. Squaring the sdev component gets you the eigenvalues. View answer Accepted answer 2 votes There are indeed data-informed ways to transform the independent variables; however, they seem to have become a bit more obscure. You can indeed investigate using a Box-Cox transformation for your ... View answer 2 votes This can happen when your data are not independent but instead have some dependence structure. For example, consider homes from across the country with some being in large expensive cities while some ... View answer 2 votes The comments on blind source separation and independent components analysis are good. However, from what you have said, there might be an easier way. You said you have a sample of the pure noise ... View answer 2 votes Feature Selection Feature selection aka model selection is difficult. By that I mean it is an unsolved problem and there is evidence that it is an NP-hard problem. The title of Maymin (2011) hints at ... View answer Accepted answer 2 votes In general, you would use a paired$t$-test when there is variation among observations which is shared (and matchable) between the two samples. So, in your example #1, yes: use a paired$t$-test since ... View answer 2 votes You just need to do a frequency decomposition of your data. The fft command in$R\$ should get you what you want. The more difficult question is how to address variations in that frequency to determine ...

The term you are looking for is a mixture distribution. Mixture distributions get used in situations when: the observations involve a response to a choice (latent variable models), the data ...

I think there are some ways to better express your data that might be helpful and avoid the issues of nonstationarity as well as some other issues you have not mentioned. You have measured stocks and ...

A lack of seasonality at the monthly level need not imply a lack of seasonality at the weekly level. Layoffs may tend to happen at the beginning of the month after missing the prior month's targets. ...