Markov inequality:
For non-negative random variable $X$ and real number $v$, then $Pr[X \geq v] \leq \frac{E(X)}{v}$
My question is: Let $0<a,b<1$, and Let $Y$ be a random variable ranging in the interval $[0,1]$ such that $E(Y)=a+b$. Give a lower bound on $Pr\left[Y\geq a+ \frac{b}{2} \right]$.
What I do know now is to construct (an) auxiliary random variable(s), but I don't know how to do that.
Best Answer
$$P(Y\geq a+b/2)=1-P(1-Y> 1-a-b/2)\geq 1-\frac{E[1-Y]}{1-a-b/2}=\frac{b/2}{1-a-b/2}$$