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Just take the squared error multiple times. If you have {...,(1,1),(1,2),(1,3),...} and you predict y=2 at x=1, the error sum of squares would be $...+(1-2)^2+(2-2)^2+(3-1)^2+...$ Then, just divide it by the number of observations to get MSE.


Pivoting $\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu=n-1),$ as suggested in comment gives a CI for $\sigma^2$ of the variance $\sigma^2$ of the normal population from which data were randomly sampled. A 95% Ci is $\left(\frac{(n-1)S^2}{U}, \,\frac{(n-1)S^2}{L}\right),$ where $L$ and $U$ cut probability $0.025$ from lower and upper tails, respectively, ...


The computation like it is shown above discards terms which have higher order than 1 when the function is expanded. Therefore, the proposed formula only applies to functions f which are ("well enough") representable as a linear function. This is the level of error propagation like you will typically find it. If you want to do error propagation ...


My question is, $x_i-\bar{x}$ is not always a small term!!! Why should we just discard it? Or, are there more formal references of this error propagation formula, because all I search online is some other kind of form. Indeed $x_i-\bar{x}$ is not always small but the average of it $\overline{x_i-\bar{x}}$ is zero. You could write is as $$\overline{x_i-\bar{...

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