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Request a geometric proof that difference of sample-means has additive variance larger than the components?

You are not allowed to use the formula quoted.

In general, the variance of the sum of $\mathrm{n}$ variables is the sum of their covariances: $$ \operatorname{Var}\left(\sum_{i=1}^{n} X_{i}\right)=\sum_{i=1}^{n} \sum_{j=1}^{n} \operatorname{Cov}\left(X_{i}, X_{j}\right)=\sum_{i=1}^{n} \operatorname{Var}\left(X_{i}\right)+2 \sum_{1 \leq i<j \leq n} \operatorname{Cov}\left(X_{i}, X_{j}\right) $$ (Note: The second equality comes from the fact that $\operatorname{Cov}\left(X_{i}, X_{j}\right)=\operatorname{Var}\left(X_{j}\right)$.)

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    Could you rephrase your sentence. It is currently very difficult to read and it is difficult to make sense of it. – Sextus Empiricus Aug 05 '22 at 21:46
  • Prove Var(X-Y)=Var(X+Y) but you cannot use the formula quoted. –  Aug 05 '22 at 21:50
  • One solution is in my post at https://stats.stackexchange.com/a/267021/919. Another is at https://stats.stackexchange.com/a/122665/919. A third is at https://stats.stackexchange.com/a/535435/919. Take your pick. – whuber Aug 05 '22 at 21:57
  • Is there a geometric proof? –  Aug 05 '22 at 21:58
  • The polarization identity is geometry. Covariances are squared Euclidean distances and everything ultimately comes down to the Pythagorean Theorem. – whuber Aug 05 '22 at 22:57
  • "has additive variance larger than the components?" What does this mean? – Sextus Empiricus Aug 06 '22 at 07:23
  • Is this question about two independent variables? Say $X$ and $Y$ are independent (so covariance is zero) then why $$\text{Var}(X-Y) = \text{Var}(X) + \text{Var}(Y)$$ instead of $$\text{Var}(X-Y) = \text{Var}(X) - \text{Var}(Y)$$ The part 'larger than the components' is unclear to me. – Sextus Empiricus Aug 06 '22 at 07:31
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    If $X = Y$ (ie they are fully correlated), then $\text{Var}(X-Y)=0$ and it is smaller than the components. – Sextus Empiricus Aug 06 '22 at 07:42

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