Suppose $f: \mathbb{R} \to [0, \alpha] $ is a non-negative and bounded function. For $0 \leq a < \alpha$, we have

$$ P ( f(X) \geq a ) \geq \frac{ \mathbb{E}(f(X) ) - a }{\alpha -a } $$

where $X : \Omega \to \mathbb{R}$ is a random variable. ($\Omega$ is countable).

##Try: I believe this is just a corollary of the following inequality I proved: for $f: \mathbb{R} \to [0, \infty)$ nonnegative function, then

$$ P( f(X) \geq a ) \leq \frac{ \mathbb{E}(f(X))}{a} $$

for $a > 0$

However, I am stuck on trying to use this inequality to prove my claim. Is there a trick to solve this problem? thanks

Suppose $f: \mathbb{R} \to [0, \alpha] $ is a non-negative and bounded function. For $0 \leq a < \alpha$, we have

$$ P ( f(X) \geq a ) \geq \frac{ \mathbb{E}(f(X) ) - a }{\alpha -a } $$

where $X : \Omega \to \mathbb{R}$ is a random variable. ($\Omega$ is countable).

Try:

I believe this is just a corollary of the following inequality I proved: for $f: \mathbb{R} \to [0, \infty)$ nonnegative function, then

$$ P( f(X) \geq a ) \leq \frac{ \mathbb{E}(f(X))}{a} $$

for $a > 0$

However, I am stuck on trying to use this inequality to prove my claim. Is there a trick to solve this problem? thanks