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If I have that $X$ is a random variable satisfying $0\leq X \leq 1$, how can I show that $P\left(X \geq \frac{E(X)}{2}\right) \geq \frac{E(X)}{2}$? I saw a footnote which gave a hint to split up $E(X)$ as two integrals over $\left[X<\frac{E(X)}{2}\right]$ and $\left[X\geq\frac{E(X)}{2}\right]$. However, I am not quite sure how to bound this integral. Would anyone have any ideas?

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Write $$ X = X \mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}} + X \mathbf{1}_{\left\{X < \frac{E[X]}{2}\right\}}, $$ and note that $$ X \mathbf{1}_{\left\{X < \frac{E[X]}{2}\right\}} \leq \frac{E[X]}{2}. $$ Also, since $0 \leq X \leq 1$, we have $$ X \mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}} \leq \mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}}. $$ Then $$ \begin{aligned} E[X] &= E\left[X \mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}}\right] + E\left[X \mathbf{1}_{\left\{X < \frac{E[X]}{2}\right\}}\right] \\ &\leq E\left[\mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}}\right] + E\left[\frac{E[X]}{2}\right] \\ &= P\left(X \geq \frac{E[X]}{2}\right) + \frac{E[X]}{2}. \end{aligned} $$ Therefore $P\left(X \geq \frac{E[X]}{2}\right) \geq \frac{E[X]}{2}$.

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Although the question has been already perfectly answered by Artem Mavrin, I would like to write the answer in a more measure theoretical way, specifically avoiding the introduction of new random variables ad expliciting the integrals appearing.

By definition, since $X \leq 1$, \begin{align*} E(X) & =\int _0^1 P(X > t) dt\\ & =\int _0^{\frac{E(X)}{2}} P(X > t) dt + \int_{\frac{E(X)}{2}}^1 P(X > t) dt. \end{align*}

Since $P(X > t) \leq P(\Omega) = 1$ the first integral is bounded by $\frac{E(X)}{2}$, the second integral is bounded by the sup of the integrand, i.e. $P\left(X\geq \frac{E(X)}{2}\right)$.

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    $\begingroup$ I would like the downvoter to let me know how can I improve my answer, or else at least to explain why it has been downvoted. $\endgroup$ – Giovanni De Gaetano Dec 14 '15 at 11:42
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    $\begingroup$ i agree. anonymous downvotes may satisfy the anonymous downvoter's dislike of people, but usually only discourage, without offering any insight. if persisted in this behaviour must be viewed as a mild form of sociopathy $\endgroup$ – David Holden Dec 14 '15 at 11:52
  • $\begingroup$ @GiovanniDeGaetano Thanks for your answer, I was just wondering how you got that the second integral is bounded by the sup of the integrand? Is this a standard result? $\endgroup$ – user136503 Dec 14 '15 at 11:53
  • $\begingroup$ I'm glad you like it. It is a standard result considering that $1-\frac{E(X)}{2}\leq 1$. Indeed one has in general (under "niceness" hypothesis) $$\int_a^b f(x) dx \leq \sup_{x \in [a,b]} f(x) \cdot |b-a|.$$ I also seize the opportunity to recall you, since looking at your profile you may be unaware of it, that you can (and should) accept one of the answers to your question by clicking the mark next to it. I invite you to do so also for your past questions. $\endgroup$ – Giovanni De Gaetano Dec 14 '15 at 12:02
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    $\begingroup$ @GiovanniDeGaetano Thanks for letting me know, I went through and accepted my answers. I think I got it, so basically in this case because $P(X>t) = 1-P(X \leq t)$, we have a monotonically decreasing function by properties of the cdf (which is increasing). Then the supremum is the left-most point, $\frac{E(X)}{2}$ by this fact and so we have: $$ P\left(Y \geq \frac{E(X)}{2}\right)\left|1-\frac{E(X)}{2}\right| \leq P\left(Y \geq \frac{E(X)}{2}\right)$ $$ Thanks!!!! $\endgroup$ – user136503 Dec 14 '15 at 12:43

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