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Let $X_1,...,X_n$ be an iid (independent and identically distributed) sample with mean $ \mu $ and variance $\sigma^2$.

How to show $$ (n-1)S^2 = \sum_{i=1}^n (Xi-\overline X)^2 = \sum_{i=1}^n (Xi-\mu ) ^2 - n(\mu-\overline X) ^2 $$

I found that if I treat $\mu$ equals $\overline X$ , it would do. But I don't think that's right.

BadAtAlgebra
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gong.y
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  • Expand $X_i-\overline X=X_i-\mu+\mu-\overline X$. –  Apr 22 '19 at 10:05
  • @YvesDaoust I have to expand the first X_i-$\overline X $as X_i-$\mu $+$\mu $-$\overline X$ and the second X_i-$\overline X$ as X_i-$\mu$-$\mu$+$\overline X $to get there. – gong.y Apr 22 '19 at 10:29
  • What ? What do you mean by first and second ? –  Apr 22 '19 at 10:30
  • @YvesDaoust (X_i-$\overline X$)^2 has two X_i-$\overline X$ . – gong.y Apr 22 '19 at 10:32

1 Answers1

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Let $Y_i:=X_i-\mu$. We have $\overline Y=\overline X-\mu$.

Then

$$\sum_i(Y_i-\overline Y)^2=\sum_iY_i^2-2\overline Y\sum_i Y+n\overline Y^2=\sum_iY_i^2-n\overline Y^2.$$