Let $0 < \epsilon$ and $\delta < 1$, and let $Y$ be a random variable ranging in the interval $[0,1]$ such that $E(Y)=\delta + \epsilon$. Give a lower bound on $Pr[Y ≥ \delta + \epsilon/2].$
The standard application of Markov's Inequality gives the upper bound instead of lower. I tried to start with the basic proof of Markov's Inequality on $Y$ and used conditions on $Y$ to get a lower bound of the probability as $ \delta + \epsilon $. Can we make it stronger? Thanks.
Let $Z=1-Y$. Then $P\{Z >1-\delta - \epsilon /2\} \leq \frac {EZ} {1-\delta -\epsilon /2}$. This gives $P\{ Y< \delta +\epsilon /2\} \leq \frac {1-\delta -\epsilon} {1-\delta -\epsilon /2}$. Hence $P\{ Y\geq \delta +\epsilon /2\} \geq 1-\frac {1-\delta -\epsilon} {1-\delta -\epsilon /2} =\frac {\epsilon /2} {1-\delta -\epsilon /2} $. Note that $\delta +\epsilon =EY \leq 1$ by hypothesis so the denominator is positive.
$$\delta+\epsilon=\mathbb{E}[Y]=\int_{Y\geq \delta+\frac{1}{2}\epsilon}Yd\mathbb{P}+\int_{Y< \delta+\frac{1}{2}\epsilon}Yd\mathbb{P}$$ $$ \leq \mathbb{P}(Y\geq \delta+\frac{1}{2}\epsilon)+(1-\mathbb{P}(Y\geq \delta+\frac{1}{2}\epsilon))(\delta+\frac{1}{2}\epsilon)$$
where in the last step we used that $0 \leq Y \leq 1$.
You can then get a lower bound provided $1-\delta-\frac{1}{2}\epsilon>0$ which as @KaviRamaMurthy pointed out holds by assumption
