everyone! I have a question when I studied matrices of norm and distance. Chebyshev inequality for standard deviation.
If $k$ is the number of entries of $x$ that satisfy $$ |x_i - \operatorname{avg}(x)| \ge a, $$ then $$ k/n \le (\operatorname{std}(x)/a)^2. $$ This inequality is only interesting for $$ a \gt \operatorname{std}(x) $$
Could you tell me how can we get this inequality. Thanks for your help.
That is the non-probabilistic (statistical) version of the Chebyshev inequality.
Let $S$ be the "tail set": $i\in S \iff |x_i - \overline{x}|\ge a$, and let $|S|=k$. Then $$\sum_{i=1}^n(x_i - \overline{x})^2= \sum_{i\in S} (x_i - \overline{x})^2+\sum_{i\notin S} (x_i - \overline{x})^2=\\ \ge \sum_{i\notin S} (x_i - \overline{x})^2 \ge k \,a^2 $$
Then, defining the sample variance $s_n^2=\sum_{i=1}^n(x_i - \overline{x})^2/ n $ we get:
$$ \frac{k}{n} \le \frac{s_n^2}{a^2}$$
The analogous probabilistic inequality is
$$ Pr(|x-u|\ge a) \le \frac{\sigma^2}{a^2}$$