Show $P(h(X)\ge a)\ge\frac{E(h(X))-a}{b-a}$

Question

Show that, if $h:\mathbb R\to[0,b]$ and $0\le a< b$ then, $\displaystyle P(h(X)\ge a)\ge\frac{E(h(X))-a}{b-a}$

So $h$ is nonnegative and bounded. If $a=0$ then the inequality holds.

because $P(h(X)\ge 0)=1$ and $\displaystyle\frac{E(h(X))}{b}\le 1$

but I don't know how to show the general case, the only inequalities I know, are Markov and Chebyshev, but how can I use them ?

Answer 1

Divide the range of $h(x)$ into two parts, below $a$ and above $a$.

$$E[h(X)] = E[h(X)I(h(X)<a)]+E[h(X)I(h(X)\ge a)] \le aP(h(X)<a)+ bP(h(X)\ge a).$$

But $P(h(X)<a)=1-P(h(X)\ge a)$. Substitute it into the above inequality and rearrange, you get the desired inequality.