Markov's Inequality and Probability distribution?

Question

I know that Markov's inequality provides an upper bound for probability of a random variable being greater than a certain value, but in what instances/distributions would the upper bound be the exact probability?

Answer 1

Markov's inequality says $EX \geq aP\{X\geq a\}$ for any non-negative random variablse $X$ and any $a \geq 0$. Equality holds iff $X$ is equal to the constant $a$ with probability $1$. To see this assume that $EX = aP\{X\geq a\}$. Then $aP\{X\geq a\} =EX= EXI_{X\geq a} +EXI_{X<a} \geq aP\{X\geq a\}+ EXI_{X<a}$ so $EXI_{X<a}=0$. This implies that $X \geq a$ almost surely. Now $E(X-a)=aP\{X\geq a\}-a=0$ and $X-a$ is non-negative, so $X=a$ almost surely.