= Equality in Chebyshev's inequality
{title2=$\mathbb P(X=\mu-a)=\mathbb P(X=\mu+a)=\sigma^2/(2a^2)$}
For $a\geq\sigma>0$, assign <probability> $\sigma^2/(2a^2)$ to each of $\mu-a$ and $\mu+a$, and the remaining <probability> $1-\sigma^2/a^2$ to $\mu$. This <probability distribution> has <expected value> $\mu$, <variance> $\sigma^2$, and <tail probability> $\mathbb P(|X-\mu|\geq a)=\sigma^2/a^2$. Thus <Chebyshev's inequality> is sharp when only these two <moments> are supplied. For example, $\mu=0$, $\sigma=1$, $a=2$ gives masses $1/8$, $3/4$, $1/8$. Replacing $\geq$ with $>$ would exclude the two extreme points in this example.
Back to article page
