$\sigma_x^2 \leq \frac{R_x^2}{4}$

Question

Can anyone please help me to prove this - $\sigma_x^2 \leq \frac{R_x^2}{4}$ ? Here $\sigma_x$ is the variance of the given data set and $R_x$ is the range of the given data set.

Answer 1

Proof:

Let $\min_{1\leq i \leq n} x_i = a$ and $\max_{1\leq i \leq n} x_i = b$

Then $R_x = b - a$

$\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$

Now $\sum_{i=1}^n (x_i - c)^2$ is least when $c = \bar{x}$

Hence $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_{i=1}^n (x_i - \frac{a+b}{2})^2 = \sum_1 (x_i - \frac{a+b}{2})^2 + \sum_2 (x_i - \frac{a+b}{2})^2$

where $\sum_1$ includes the values of $x$ less than or equal to $\frac{a+b}{2}$ and $\sum_2$ includes values of $x$ greater than $\frac{a+b}{2}$

Or, $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_1 (a - \frac{a+b}{2})^2 + \sum_2 (b - \frac{a+b}{2})^2 = \sum_1 \frac{R_x^2}{4} + \sum_2 \frac{R_x^2}{4} = n \frac{R_x^2}{4}$

Or, $\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \leq \frac{R_x^2}{4}$

Answer 2

Also known as the Popoviciu inequality, a proof sketch is as follows : if $X$ is the random variable generating the data set (i.e. whose samples are being taken) with mean $\mu$ and variance $\sigma^2$, then for any $t$ we have : $$ \mbox{Var}(X) = E[(X-\mu)^2] = E[(X-t)^2] - (t - \mu)^2 $$

which can be checked by expansion. This implies that $\mbox{Var}(X) \leq E[(X-t)^2]$ for all $t$. Let $X$ have range $[m,M]$ and put $t = \frac{m+M}{2}$. Try to now bound the RHS of the inequality above by $\frac{(M-m)^2}{4}$ using elementary inequalities.

Answer 3

It is first important to realize that if this is true for the case $n = 2$, then it will be true for all cases. This is because for all datasets $X$ taking $X_1, X_n$ as a new dataset will only increase the LHS and keep the RHS constant. Now the problem is simple algebra.

$$Var(\{X_1, X_2\}) = \frac{(X_1 - \frac{X_1 + X_n}{2})^2 + (X_2 - \frac{X_1 + X_n}{2})^2}{2} = \frac{(X_1 - X_n)^2 + (X_n - X_1)^2}{8} = \frac{R_x^2 + R_x^2}{8} = \frac{R^2_x}{4}$$