From what I've learnt, standard error is a concept related to sampling distribution. So why is the term 'residual standard error' used to refer to $\sqrt{\frac{RSS}{n-2}}$ and not $\sqrt{E(r_i - E(r_i))^2}$ in SLR context?

(I mentioned Simple Linear Regression only because I have not learnt other regression techniques yet)


The residual standard error in this context is the square root of the estimated variance of the error term. In a linear regression with an intercept term and a single explanatory variable we have:

$$\hat{\mathbb{V}}(\varepsilon) = \text{MSE} = \frac{\text{RSS}}{\text{df}_\text{Res}} = \frac{\text{RSS}}{n-2}.$$

So the quantity you are referring to is $\hat{\mathbb{S}}(\varepsilon) = \sqrt{\text{RSS}/(n-2)}$. The other quantity you refer to is the standard deviation of the $i$th residual, which is not a fixed quantity over $i=1,...,n$ since it depends on the leverage values of the data points.

  • $\begingroup$ Well I used r_i here as a random variable and not a specific single residual value from a sample. So my question was like, since it is named 'standard error' I thought it must be something related to the sampling distribution of randome variable r_i, and not RSS/n-2 which is a statistic calculated from a specific sample (of n observation of r_i) $\endgroup$ – user8931048 Aug 9 '18 at 9:34
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    $\begingroup$ Each of the random variables $R_1,...,R_n$ have a different variance, which is affected by the leverage of each data point. That means that there is no single generic variance for the residual. $\endgroup$ – Reinstate Monica Aug 9 '18 at 9:58

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