Regression computation $\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - n \bar{X}^2$

Question

Prove that $\sum_{i=1}^{n}\left(X_i - \bar{X}\right)^2 = \sum_{i=1}^{n}\left(X_i^2\right) - n \bar{X}^2$

Here's what I have so far:

$$\begin{align} \sum_{i=1}^{n}\left(X_i - \bar{X}\right)^2 &= \sum_{i=1}^{n}\left(X_i^2 - 2X_i \bar{X} + \bar{X^2}\right) = \sum_{i=1}^{n}\left(X_i^2-2X_i\bar{X}\right) + n\bar{X}^2 \\ &= \sum_{i=1}^{n}\left(X_i^2 - 2 X_i\left(\frac{1}{n}\sum_{i=1}^{n} X_i \right)\right) + n \bar{X}^2 \end{align}$$

I'm lost after this step. Any help really appreciated

Answer 1

In the second line, you peroperly wrote $$A=\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} (X_i^2 - 2X_i \bar{X} + \bar{X^2}) = \sum_{i=1}^{n} (X_i^2-2X_i\bar{X}) + n\bar{X}^2$$ That is to say $$A=\sum_{i=1}^{n} X_i^2-2\bar{X}\sum_{i=1}^{n}X_i+n\bar{X}^2$$ But $$\sum_{i=1}^{n}X_i=n\bar{X}$$ So ... ???

Answer 2

Let's start from here

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

One more step

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

We know,

$$\sum_1^n - 2 X_i X = -2 \bar X \sum_1^n X_i = -2 \bar{X} (n \bar{X})= -2n {\bar X }^2$$

Plugging this back into the last expression above, we get

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

Answer 3

In the part of expression before the last equal sign, just note that $\sum_{i=1}^n X_i \bar{X} = \bar{X}\sum_{i=1}^n X_i=n\bar{X}^2$. The result is then immediate.