How to compute a non-trivial lower bound on $P \left[|X| > \frac{|\alpha|}{2} \right]$?

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.

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.

• The key to this solution is the initial characterization of the minimum. How do you demonstrate that?
– whuber
Sep 22 '21 at 15:15
• @Henry, Can you please provide any reference to the proof, how did you get $P \left[|X| > \frac{|\alpha|}{2} \right] \ge \frac{\alpha^2}{4\beta-3\alpha^2}$? Sep 28 '21 at 10:47
• @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 Sep 28 '21 at 11:04