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Let $X$ be a random variable such that $E[X] = \alpha$, $\alpha \in \mathbb{R}$ and $E[X^2] = \beta$. The problem is to find a lower bound on the following probability $$ P \left[|X| > \frac{|\alpha|}{2} \right]. $$

Any help would be appreciated.

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You are going to minimise the probability in question if you put as much of the probability as possible at $\frac \alpha 2$ and the rest at a single other point, though you may need two other points when $\alpha=0$.

This happens when $X=\frac{\alpha}{2}$ with probability $\frac{4\beta-4\alpha^2}{4\beta-3\alpha^2}$ and $X=\frac{2\beta-\alpha^2}{\alpha}$ with probability $\frac{\alpha^2}{4\beta-3\alpha^2}$, and these give the desired mean and second moment, noting $\beta\ge \alpha^2$ since $E[X^2] \ge (E[X])^2$ so all the probabilities are between $0$ and $1$ and the "other point" is larger in magnitude than $\alpha$. Thus

$$P \left[|X| > \frac{|\alpha|}{2} \right] \ge \frac{\alpha^2}{4\beta-3\alpha^2}$$

When $\alpha=0$, for some arbitrarily small $\delta$, let $X=+\sqrt{\frac \beta\delta}$ with probability $\frac{\delta}2$, $X=-\sqrt{\frac \beta\delta}$ with probability $\frac{\delta}2$ and $X=0$ with probability $1 -{\delta}$. So the lower bound in this situation is $0$, as suggested by the earlier expression.

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    $\begingroup$ The key to this solution is the initial characterization of the minimum. How do you demonstrate that? $\endgroup$
    – whuber
    Sep 22 '21 at 15:15
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    $\begingroup$ @Henry, Can you please provide any reference to the proof, how did you get $P \left[|X| > \frac{|\alpha|}{2} \right] \ge \frac{\alpha^2}{4\beta-3\alpha^2} $? $\endgroup$
    – Bhisham
    Sep 28 '21 at 10:47
  • $\begingroup$ @Bhisham It is an immediate consequence of my first point. The explanation for that point (not proof) is that if you have any probability in $\left[-\frac{|\alpha|}{2},\frac{|\alpha|}{2}\right]$ not at $\frac{\alpha}{2}$ then you can shift some of it to $\frac{\alpha}{2}$ and the rest outside the interval while maintaining the first two moments; then if you have any probability outside the interval but not at $\frac{2\beta-\alpha^2}{\alpha}$ you can shift it and some of the probability at $\frac{\alpha}{2}$ to $\frac{2\beta-\alpha^2}{\alpha}$ while maintaining the first two moments $\endgroup$
    – Henry
    Sep 28 '21 at 11:04

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