# Tag Info

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Heteroskedasticity means quite general that the residual variance is not homogenous, i.e. depends on groups or some other variable. Note that, according to this definition, if you find that var(res) ~ fitted, you clearly have heteroskedasticity However, if you don't find a relationship between var(res) ~ fitted, you could still have heteroskedasticity, for ...

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In my opinion "examining fitted vs residuals" is a graphical alternative to the Box-Cox test When (and why) should you take the log of a distribution (of numbers)? which can help to evaluate the hypothesis that the variability of residuals is or is not linearly related (sympathetic) to the level i.e.the fitted values of the series. IFF you are analyzing ...

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The wikipedia link you give are using imprecise/confounded language, so you should maybe find some better tutorial! It also confuses assumptions and mathematical conclusions from those assumptions (it isn't an assumption that $S^2$ follows a $\chi^2$-distribution with $n − 1$ degrees of freedom, that can be concluded from the normal assumption and ...

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I just found that the question has been posed before and answered here Quoting Florian Hartig's response: Hi, this is expected behavior - the default of the plot function is to do the quantile regression for n <= 2000, and for larger datasets a nonparametric smoother, because the quantile regressions can take a long time to calculate. You can ...

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What this plot is saying is that you only have two predicted values from your model. One way that this could happen is that you only have one independent variable and it has only two levels. So, for instance, if you are trying to model weight based on sex, this sort of thing could happen. It is also showing that the residuals are much bigger than the ...

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@Erik Ruzek's answer is correct for the larger question - you shouldn't correlate your factors in a bifactor model. For the smaller question - can latent variables have residuals, the answer is yes. But don't think of them as residuals, think of them as unexplained variance. If two latent variables have a common cause, and that common cause is not included ...

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Yes. The generalization is most clear if you think of linear regression not with an error term, but as a model for the conditional distribution of $Y \mid X$: $$Y \mid X \sim Normal(X \beta, \sigma)$$ When using the parametric bootstrap, we can think of our new $y_i$'s as samples from these conditional distributions, one for each $x_i$. This ...

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The formula that divides by $n-p$ is an estimator for unbiased mean squared error for regression, where you divide by the degrees of freedom. Imagine you would like to compare results returned by different machine learning algorithms, in many cases it would be hard to say what does exactly $p$ is, and for some models, like non-parametric models, or neural ...

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First look at simple linear regression: $$y_i = \mu + \beta x_i + \epsilon_i$$ for $i=1, \dotsc,n$ and the usual assumptions, specifically that the error terms $\epsilon_i$ are iid. Estimating the parameters with ordinary least squares, it is well known that the residuals $e_i= y_i - \hat{y}_i$ satisfy $\sum_{i=1}^n e_i =0$, see Why do residuals in ...

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So, I figured where the problem is with high probability. The real issue is my data... I created target variable $Y$ so that for each random subject the dependancy on covariates was 100%. Call: lm(formula = score_diff_delta ~ score_diff, data = ddt) Residuals: Min 1Q Median 3Q Max -3.202e-15 -2.600e-18 7.460e-17 7.460e-...

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There could multiple reasons why you observe this systematic behavior. A couple are: You have modeled appropriately the relationship between $Y$ and $X$. For example, it could be that $X$ is nonlinearly associated with $Y$. You could investigate that by relaxing the linearity assumption either using polynomials or splines. You have missing data in $Y$ which ...

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