Let $X_1, \dotsc,X_n$ be i.i.d. from $N(\mu,\sigma^2)$, then we know that sample mean $\bar X\equiv \frac{1}{n}\sum_{i=1}^nX_i$ and $S^2=\frac{1}{n-1}(X_i-\bar X)^2$ are independent. Obviously, they both depend on $\bar X$ and depend on all observations as well. So this independence result is really surprising. I'm wondering what's the intuition behind this result.
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2$\begingroup$ Related? $\endgroup$– DaveCommented Nov 20, 2022 at 5:42
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1$\begingroup$ First of all, having an overlapped part of two random variables does not rule out the possibility of the independence between them. As another example, $X, Y \text{ i.i.d. } \sim N(0, 1)$, then $X + Y$ and $X - Y$ are independent. $\endgroup$– ZhanxiongCommented Nov 20, 2022 at 16:36
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2$\begingroup$ For this question, a geometric interpretation can be found in Section 1.3 of the textbook The Coordinate-Free Approach to Linear Models by M. Wichura. At high-level, if view $X := (X_1, \ldots, X_n)$ as a vector in $\mathbb{R}^n$, then $\bar{X}e, X - \bar{X}e$ are orthogonal projections of $X$ onto $[e]$ and $[e]^\perp$ respectively, thus are "uncorrelated", which are in turn independent by the normality assumption. Then note $(n - 1)S^2 = (X - \bar{X}e)^T(X - \bar{X}e)$. $\endgroup$– ZhanxiongCommented Nov 20, 2022 at 16:42
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2$\begingroup$ You're unlikely to find any intuition which doesn't at some point rely on yet other properties of the normal distribution (plus what will be some mathematical argument to relate them), for which other properties you'll presumably want intuition (as with the comment immediately above this one,, which does just that - uses a mathematical argument in order to invoke yet another property of the normal which I assume you don't already have intuition for). The normal has a number of properties that are unique to it. $\endgroup$– Glen_bCommented Nov 20, 2022 at 17:23
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2$\begingroup$ This is only true for normal distributions, see stats.stackexchange.com/a/4359/11887 $\endgroup$– kjetil b halvorsen ♦Commented Mar 31, 2023 at 17:18
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