Prove that $\frac1n\sum\limits^n_{i=1}(X_i-\overline{X})^2=\overline{X^2}-\overline{X}^{\ 2}$

Question

Prove that $\frac{1}{n}\sum\limits^n_{i=1}(X_i-\overline{X})^2=\overline{X^2}-\overline{X}^{\ 2}$ where $\overline{X}=\frac{1}{n}\sum\limits^n_{i=1}X_i$ and $\overline{X^2}=\frac{1}{n}\sum\limits^n_{i=1}X_i^2$

From the left, I can see that $$\frac{1}{n}\sum^n_{i=1}(X_i-\overline{X})^2=\frac{(X_1-\overline{X})^2+\cdots+(X_n-\overline{X})^2}{n} = \frac{(X_1-\frac{X_1+\cdots+X_n}{n})^2+\cdots+(X_n-\frac{X_1+\cdots+X_n}{n})^2}{n} = \cdots$$

I don't see how we can get to the right. Any suggestions?

Answer 1

Observe \begin{align} \frac{1}{n}\sum^n_{i=1}(X_i-\bar X)^2 =&\ \frac{1}{n}\sum^n_{i=1} (X_i^2-2\bar XX_i +\bar X^2)\\ =&\ \frac{1}{n}\sum^n_{i=1}X_i^2 -2\bar X \frac{1}{n}\left(\sum^n_{i=1}X_i\right)+\bar X^2 \\ =&\ \overline{X^2}-2\bar X^2+\bar X^2 = \overline{X^2}-\bar X^2. \end{align}

Answer 2

\begin{align} & \sum_{i=1}^n (X_i - \bar X)^2 = \sum_{i=1}^n(X_i^2 - 2X_i \bar X + \bar X^2) \\[10pt] = {} & \left( \sum_{i=1}^n X_i^2 \right) - \left( \sum_{i=1}^n (2X_i \bar X\,) \right) + \left( \sum_{i=1}^n \bar X \right) \tag 1 \end{align} Now observe that as $i$ goes from $1$ to $n$, then $X_i$ changes but $2\bar X$ does not. Therefore we can write $$ \sum_{i=1}^n (2X_i \bar X\,) = 2\bar X \sum_{i=1}^n X_i. $$ and that is equal to $$ 2\bar X\Big( n\bar X \Big) = 2 n \bar X^2. $$ In the last sum in line $(1)$, the quantity $\bar X$ does not change at all as $i$ goes from $1$ to $n$, so it is $$ \sum_{i=1}^n \bar X^2 = \underbrace{\bar X^2 + \cdots + \bar X^2}_{n \text{ terms}} = n \bar X^2. $$ So the sum in line $(1)$ becomes $$ \left( \sum_{i=1}^n X_i^2 \right) - 2n \bar X^2 + n \bar X^2. $$ Then do the obvious simplification and then divide both sides by $n$.