I have that $\text{var}(X) = \frac13v^2$ and I want to find the lower bound for $P(|X| \le v)$

I tried doing the following: $P(|X| \le v) = 1 - P(|X| \ge v)$ so using Chebyshev's inequality I have that

$1 - P(|X| \ge v) \le \frac13$

$\frac23 \le P(|X| \ge v)$,

but I don't know what can I say about $P(|X| \le v)$ since everything is now on terms of $P(|X| \ge v)$.


You are making a mistake in using Chebyshev's inequality. What you get $P(|X| \geq v) \leq 1/3$ (assuming that the mean is $0$) and not $1-P(|X| \geq v) \leq 1/3$

  • $\begingroup$ so if 𝑃(|𝑋| ≥ 𝑣) ≤ 1/3, then 𝑃(|𝑋| ≤ 𝑣) ≥ 2/3 ? @KaviRamaMurthy $\endgroup$ – Anna Mar 19 at 5:43
  • $\begingroup$ @Anna Yes, that is the lower bound. $\endgroup$ – Kavi Rama Murthy Mar 19 at 5:44
  • $\begingroup$ Thank you, I see now the mistake I was making $\endgroup$ – Anna Mar 19 at 5:45

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