Raul Miller wrote:
> For example: in our simple example, $((X_i - \bar X)/(n-1))^2$ can be
> zero in some cases (when both sampled values are the same), which
> clearly is not the same as \sigma^2(population variance). So either
> I do not understand what you mean by = in
> variance of X_i=\sigma^2 (population variance)
> or I do not understand what you mean by variance of X_i
> or... something else.
X_i has the population distribution, so the variance of X_i is \sigma^2.
The parenthetical (population variance) is an explanatory note, not an
argument for \sigma^2. There is confusion in this notation because of the
limitations of plain text: I would normally write e.g. \sigma^2_{\bar X}
for the variance of the sample mena.
Best wishes,
John
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