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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

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  • $\begingroup$ I doubt you need countability here. ${}\qquad{}$ $\endgroup$ Commented Mar 15, 2015 at 20:52
  • $\begingroup$ If $Y=f(X)$ then you can say $Y:\Omega\to\mathbb R$ and $\Pr(0\le Y\le\alpha)=1$ and then say you want to prove that for $0\le a\le\alpha$ you have $\Pr(Y\ge a)\ge\dfrac{\operatorname{E}(Y)-a}{\alpha-a}$. There doesn't seem to be any point in writing it as $f(X)$ instead of just $Y$. ${}\qquad{}$ $\endgroup$ Commented Mar 15, 2015 at 20:56
  • $\begingroup$ Hint: Use $$ P( Y \geq y ) \leq \frac{ E(Y)}{y} $$ with $$Y=\alpha-f(X)\ge0\ \text{almost surely},\qquad y=\alpha-a>0.$$ $\endgroup$ Commented Mar 15, 2015 at 21:01

1 Answer 1

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Set $A = \alpha$ and $g(x) := A - f(x)$. Notice that $g$ is a nonnegative function. Therefore $$\operatorname{Pr}(f(X)<a) = \operatorname{Pr}(g(X) > A-a)\leq \frac{\operatorname{E}[g(X)]}{A-a}=\frac{A-\operatorname{E}[f(x)]}{A-a}.$$ Your result follows immediately.

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