Introduction to Regression Analysis Proof

Question

Show that: $\sum_{i=1}^n(x_i-\bar x)^2 = $ $\sum_{i=1}^nx_i^2$ $-$$(\frac{\sum_{i=1}^n(x_i)^2}{n})$

I know $n$ is a positive number. I think I should start with $\bar x$ $=$ $\frac{\sum_{i=1}^n(x_i)}{n}$

Is this the right approach? Squaring a polynomial in a summation seems challenging, and I feel like this is a trivial problem.

Answer 1

There isn't much more too it: as you figured, the whole thing boils down to expanding the square. \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) = \sum_{i=1}^n x_i^2 - 2\sum_{i=1}^n x_i\bar{x}+\sum_{i=1}^n\bar{x}^2 \\ &= \sum_{i=1}^n x_i^2 - 2 \bar{x}\sum_{i=1}^n x_i+ n \bar{x}^2 = \sum_{i=1}^n x_i^2 - 2 n \bar{x}^2+ n \bar{x}^2 \tag{since $\sum_{i=1}^n x_i = n\bar{x}$}\\ &= \sum_{i=1}^n x_i^2 - n \bar{x}^2 = \sum_{i=1}^n x_i^2 - n\cdot\left(\frac{\sum_{i=1}^n x_i}{n}\right)^2 \\ &= \sum_{i=1}^n x_i^2 - \frac{\left(\sum_{i=1}^n x_i\right)^2}{n}\,. \end{align}