Given $n$ distinct values $(x_i)_{i=1}^{n}$ with mean $\mu$ and standard deviation $s$, for all $i$, we have $|x_i−\mu| ≤ s \sqrt{n − 1}$. How does this inequality compare with Chebyshev inequality as $n$ increases?
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1What do you think? What have you tried, and where did you get stuck? – Yuval Filmus Sep 03 '20 at 11:37
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Chebyshev's inequality is a bound on the probability that a random variable deviates from its mean. I don't see anything like that here. – Yuval Filmus Sep 03 '20 at 11:38
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Chebyshev's inequality states that $$ \Pr[|x_i - \mu| > c\sigma] < \frac{1}{c^2}. $$ Choosing $c = \sqrt{n}$, we deduce that with positive probability, $$ |x_i - \mu| \leq \sqrt{n} \cdot \sigma \text{ for all } i. $$
Yuval Filmus
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