Possible simplified variance

Question

I noticed during calculating variance:

$$S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline{X})^2$$

that the sum $\sum_{i=1}^n(X_i-\overline{X})^2$ is equal to $\sum_{i=1}^n(X_i-\overline{X})*X_i$.

However I wasn't able to prove why this is true, or if there are cases in which it won't be equal.

Answer 1

$$\sum_{i=1}^{n}\left(X_{i}-\overline{X}\right)^{2}=\sum_{i=1}^{n}X_{i}^{2}-2\sum_{i=1}^{n}X_{i}\overline{X}+\sum_{i=1}^{n}\overline{X}^{2}=\sum_{i=1}^{n}X_{i}^{2}-n\overline{X}^{2}$$

and also:

$$\sum_{i=1}^{n}\left(X_{i}-\overline{X}\right)X_{i}=\sum_{i=1}^{n}X_{i}^{2}-\overline{X}\sum_{i=1}^{n}X_{i}=\sum_{i=1}^{n}X_{i}^{2}-n\overline{X}^2$$

Answer 2

$$\sum_{i=1}^n(X_i-\overline{X})^2=\sum_{i=1}^n(X^2_i-2X_i\overline{X}+\overline{X}^2) $$

$$=\sum_{i=1}^n X^2_i-2\overline{X}\underbrace{\sum_{i=1}^n X_i}_{n\overline{X}} +n\overline{X}^2 $$

$$=\sum_{i=1}^n X^2_i-2n\overline{X}^2 +n\overline{X}^2 $$

$$=\sum_{i=1}^n X^2_i-n\overline{X}^2 $$

$$=\sum_{i=1}^n (X^2_i-\overline{X}^2) $$

On the other hand

$$\sum_{i=1}^n(X_i-\overline{X})*X_i ==\sum_{i=1}^n X^2_i-\overline{X}\cdot \underbrace{\sum_{i=1}^n X_i}_{n\overline{X}}$$

$$=\sum_{i=1}^n X^2_i-n\overline{X}^2$$