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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.

kjetil b halvorsen
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ExcitedSnail
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    Related? – Dave Nov 20 '22 at 05:42
  • @Dave Thanks, Dave! It's indeed related, but didn't give any intuition though. – ExcitedSnail Nov 20 '22 at 06:49
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    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. – Zhanxiong Nov 20 '22 at 16:36
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    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 $(n - 1)S^2 = (X - \bar{X}e)^T(X - \bar{X}e)$. – Zhanxiong Nov 20 '22 at 16:42
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    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. – Glen_b Nov 20 '22 at 17:23
  • @Zhanxiong Thank you very much! This is very helpful! – ExcitedSnail Nov 20 '22 at 23:49
  • @Glen_b I see. This is very helpful! – ExcitedSnail Nov 20 '22 at 23:50
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    @Glen_b Indeed, I guess what I asked for was not real intuition, but merely deeper understanding of why this is true. – ExcitedSnail Nov 21 '22 at 07:37
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    This is only true for normal distributions, see https://stats.stackexchange.com/a/4359/11887 – kjetil b halvorsen Mar 31 '23 at 17:18
  • @kjetilbhalvorsen You are right! – ExcitedSnail Apr 02 '23 at 14:11

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