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**azdang** Let h: **R** --> [0,$\displaystyle \alpha$] be a nonnegative bounded function. Show that for $\displaystyle 0\leq a<\alpha$,

$\displaystyle P(h(X)\geq a)\geq \frac{E(h(X))-a}{\alpha - a}$.

The only idea I have is that we want $\displaystyle a\leq h(X)\leq \alpha$. So, $\displaystyle P(h(X)\geq a)$ becomes $\displaystyle P(a\leq h(X)\leq \alpha)$.

I know that $\displaystyle P(h(X) \geq a)\leq \frac{E(h(X))}{a}$.

Then, shouldn't it be that $\displaystyle P(a\leq h(X))\leq \frac{E(h(X))}{a}$ and $\displaystyle P(h(X)\leq \alpha)\geq 1 - \frac{E(h(X))}{\alpha}$ In fact, you have $\displaystyle \color{red} P(h(X)\leq \alpha)=1$ because of the assumption.

However, I'm not sure how to use this all together or if I'm even on the right track.