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I want to prove the following result: Let $X$ be a nonnegative random variable defined on a probability space $\Omega$. Then

$$\sum_{\omega\in\Omega:\ X(\omega)\geq\frac{\mathbb{E}X}{2}}\mathbb{P}(\omega)X(\omega) \geq \frac{\mathbb{E}X}{2}$$

I need it in another proof, but I am not sure whether it is true at all. Can anyone help me with the proof? Thanks

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What's $\mathbb{P}(\omega)$? – saz Dec 12 '12 at 19:19
The accepted answer shows that this is true if you replace the central $>$ with $\ge$. The strict inequality isn't true; e.g., if $X(\omega)=0$ everywhere, both sides are zero. – mjqxxxx Dec 12 '12 at 20:38

Assume not, and call this assumption (*). Then: 

$E[X] = \sum_{\omega : X(\omega ) < \frac{E[X]}{2}} P(\omega) X(\omega)  + \sum_{\omega : X(\omega ) \ge \frac{E[X]}{2}} P(\omega) X(\omega)$  $< \sum_{\omega : X(\omega ) < \frac{E[X]}{2}} P(\omega)  \frac{E[X]}{2} + \frac{E[X]}{2}$ $\le \frac{E[X]}{2} + \frac{E[X]}{2} = E[X]$. Contradiction. 

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ok, thanks, I understood – werther Dec 12 '12 at 20:52

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