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Let $X$ be a random variable with $E[X] > 0$ and $ 0 < E[X^2] < \infty$, prove that $P(X > tE[X]) \geq \frac{(1-t)^2E[X]^2}{E[X^2]}$ for all $0 \leq t \leq 1$.

How do I prove this?

I first used Markov's inequality and get $P(X > tE[X]) \leq \frac{E[X]}{tE[X]}$ and then it will become $ P(X > tE[X]) \geq 1 - \frac{E[X]}{tE[X]}$. But what to do next, am I even on the correct track? I can't think of how to get $E[X]^2$ and $E[X^2]$ into the equality.

user1070420
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    Maybe using the general version of Markov's inequality (see https://en.wikipedia.org/wiki/Markov%27s_inequality) but taking instead a decreasing function $\phi(x)=1/x^2$ might do. Indeed, then you would get $P(X \leq tE[X]) = P(\phi(X)\geq \phi(tE[X]))\leq E[\phi(X)]/\phi(tE[X])$, this might help to find the lower bound since $P(X > tE[X]) = 1-P(X \leq tE[X]) $. To make the $(1-t)^2$ appear, you might have to rewrite $P(X>tE[X])=P(X-E[X]>(t-1)E[X])$. – adrien_vdb Jun 23 '22 at 15:47
  • Use Chebyshev's inequality. It will help you to introduce the square terms. – José Gabriel Astaíza-Gómez Jun 23 '22 at 22:46

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This is in fact the Paley–Zygmund inequality (although you need the additional assumption that $X$ is a non-negative random variable)! You can check it out here. As explained on the Wikipedia page, you can also generalize it by using Hölder's inequality instead of Cauchy-Schwartz in the proof, yielding $P(X>tE[X])\geq (1-t)^{\frac{p}{p-1}}\frac{E[X]^{\frac{p}{p-1}}}{E[X^p]^{\frac{1}{p-1}}}$ for some $p>1$.

adrien_vdb
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