I am revising how to prove the variance of b1 is greater than that of b2. However I am having trouble with the maths. Could someone please tell me how I get from step 1 to step 2? Could you please explain all steps? My maths isn't very strong.
I just dont understand how we got from A to B with both b values.
On
Let's understand why the convexity argument holds for a vector $\boldsymbol x = (x_1, \ldots, x_n) \in \mathbb R^n$, as this is presently stated without any proof.
Let $\bar x = \frac{1}{n} \sum_{i=1}^n x_i$ be the arithmetic mean of the $x_i$s. Then $$\begin{align*} \sum_{i=1}^n x_i^2 &= \sum_{i=1}^n (x_i - \bar x + \bar x)^2 \\ &= \sum_{i=1}^n \left((x_i - \bar x)^2 + 2\bar x (x_i - \bar x) + \bar x^2 \right) \\ &= \sum_{i=1}^n (x_i - \bar x)^2 + 2 \bar x \sum_{i=1}^n (x_i - \bar x) + \sum_{i=1}^n \bar x^2 \\ &= \biggl( \sum_{i=1}^n (x_i - \bar x)^2\biggr) + 0 + n \bar x^2 \\ &\ge n \bar x^2, \end{align*}$$ where the middle term $$\sum_{i=1}^n (x_i - \bar x) = -n \bar x + \sum_{i=1}^n x_i = -n \bar x + \bar x = 0,$$ and $$\sum_{i=1}^n (x_i - \bar x)^2 \ge 0$$ with equality occurring if $x_i = \bar x$ for all $i$, because the sum of the squares of real numbers is never negative.
$$ \begin{align} % \sum x^{2} &\ge \frac{\left( \sum x \right)^{2}}{n} \\[5pt] % \frac{\sum x^{2}}{n} &\ge \frac{\left( \sum x \right)^{2}}{n^{2}} \\[5pt] % \overline{x^{2}} &\ge \bar{x}^{2} % \end{align} $$
The average of the squares is greater than or equal to the square of the average.