Range，Mean deviation and Standard deviation

statistics

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Definition:

$$RD = max(x_i)-min(x_i)\\ MD = {1 \over n} \sum_{i=1}^{n} {\left|x_i - \overline{x} \right|}\\ \sigma^2 = {1 \over n} \sum_{i=1}^{n} (x_i - \overline{x})^2$$

Result:

$$RD \geq \sigma \geq MD$$

Proof:
First, we show that $RD \geq \sigma$, or equivalently, $RD^2 \geq \sigma^2$.

$$RD^2 = \left[max(x_i) - min(x_i) \right]^2\\ = \left[max(x_i) - \overline{x} + \overline{x} - min(x_i) \right]^2 \\ = \left[max(x_i) - \overline{x} \right]^2 + \left[\overline{x} - min(x_i) \right]^2 + 2\left[max(x_i) - \overline{x} \right]\left[\overline{x} - min(x_i)\right]\\ \ge \left[max(x_i) - \overline{x} \right]^2 + \left[\overline{x} -min(x_i) \right]^2$$

Note that for every observed value $x_i$ which is greater than $\overline{x}$, $(x_i-\overline{x})^2 \le [max(x_i) - \overline{x}]^2$. And $x_i$ which less than $\overline{x}$ satisfies $(x_i-\overline{x})^2 \le [min(x_i) - \overline{x}]^2$. Supoose the number of $x_i$ which less than $\overline{x}$ is $s$ and define set of subscript $I = \{ i\in [1, n]: x_i \le \overline{x}\}$. Then

$$\sum_{i=1, i\in I}^s (x_i - \overline{x})^2 \le s [min(x_i) - \overline{x}]^2\\ \sum_{i=1, i\notin I}^{n-s} (x_i - \overline{x})^2 \le (n-s) \left[max(x_i) - \overline{x} \right]^2.$$

$$RD^2 \ge {1 \over s}\sum_{i=1, i\in I}^s (x_i - \overline{x})^2 + {1 \over {n-s}}\sum_{i=1, i\notin I}^{n-s} (x_i - \overline{x})^2\\ \ge {1 \over n} \left[ \sum_{i=1, i\in I}^s (x_i - \overline{x})^2 + \sum_{i=1, i\notin I}^{n-s} (x_i - \overline{x})^2 \right] = \sigma^2$$

To complete the proof, we need to show $\sigma \ge MD$.

$$MD = {1 \over n} \sum_{i=1}^{n} {\left|x_i - \overline{x} \right|} = {1 \over n} \sum_{i=1}^{n} \sqrt{(x_i - \overline{x})^2} = {1 \over n} \sum_{i=1}^{n} \sqrt{t_i}$$
$$\sigma = \sqrt{{1 \over n} \sum_{i=1}^{n} (x_i - \overline{x})^2} = \sqrt{{1 \over n} \sum_{i=1}^{n} t_i}$$
Let $f(t_i) = \sqrt{t_i}$ and obviously $f(\cdot)$ is cancave, in accord with Jensen's Inequality, we have
$$f(E(t)) \ge E(f(t))$$
and hence,
$$\sigma = \sqrt{{1 \over n} \sum_{i=1}^{n} t_i} \ge {1 \over n} \sum_{i=1}^{n} \sqrt{t_i} = MD.$$