- #1

chiropter

- 25

- 0

Comparing the statistical errors to the residuals, the former involves taking a subset of observations from a population and comparing them to the population mean, while the latter involves comparing the observations to the sample mean calculated from the same subset of observations. The differences between a sample of observations and the population mean obviously does not have to sum to 0, while the differences between the observations and the sample mean does sum to 0.

Thus, if you are trying to estimate the 'average' of the difference between the population mean and the set of values in the population, using sample residuals (which are derived from a set of sample observations and a sample mean) to do so will result in the last 'residual' NOT being free to vary over the 'full range' a true statistical error would normally do, it will just have to be whatever the value is that will allow the sum to add up to 0.

So, if this is correct so far, it seems pretty straightforward to then claim that final sample residual won't contribute the required quantity to the sum of (squared) residuals that would be needed to estimate the 'average' of n (squared) residuals, and so instead, we redistribute that final quantity to all the other residuals and divide the sum by n-1 instead. So it's like you don't actually have n bits of variability from which you can estimate the errors (variance).

Ok, if that makes sense so far, my question is, why does n-1 take care of it? Why is it as if we have exactly one fewer residuals to estimate the population variance, wherein we take the remaining amount of 'variability' and distribute it equally to the other residuals before dividing by the number of things to get the average (my interpretation of dividing by n-1)? Why isn't it n-2 or 0.5n?

It bothers me that something as intuitive as calculating the average, or dividing, or summing can suddenly become so cryptic when we come to sample statistics. Hoping to resolve some of this confusion.