$X$ is random variable with $p(-1)=\frac18, \; p(0)= \frac68, \;p(1)=\frac18$ Can I calculate upper bound for $P[-1\leq X \leq 1]$ using Chebyshev's inequality?

Clearly the mean value, $\mu = 0$ and the standard deviation $\sigma = \frac12$

From Chebyshev's inequality $P[|X-\mu|< k\sigma]\geq 1- \frac1{k^2}$


$P[-1\leq X \leq 1] = P[|X-0|\leq 1 ] \geq 1 - \frac{1}{2^2} = 0.75 $ since $1=2\cdot\frac12$ but this is not an upper bound.

Also shouldn't $P[-1\leq X \leq 1]=1$?

  • 1
    $\begingroup$ It's not an upper bound since it's a lower bound. And, yes, 𝑃[−1≤𝑋≤1]=1 should be so as it is by the initial definition. $\endgroup$ – dnqxt May 7 at 1:23

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