# Is it possible to obtain upper bound for this probability using Chebyshev inequality?

$$X$$ 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 $$P[|X-\mu|< k\sigma]\geq 1- \frac1{k^2}$$

But,

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

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

• 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. – dnqxt May 7 at 1:23