Lower bound on the probability that a random variable is greater than half of its mean

Question

Consider a random variable $X$ that takes values between $-1$ and $1$. What is a non-trivial lower bound on the probability that the outcome of $X$ is greater than half of its mean?

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EDIT:

I am looking for a bound that is a function of the mean (and so would cover the case where the mean is positive).

I.e.

$\text{Pr} \left[ X > {\mathbb{E}[X] \over 2} \right] \geq f(\mathbb{E}[X]).$

Answer 1

Let $\mu=E[X]$ and $Z = \mathbb{1}_{X > \mu/2} $ (indicator variable), $a=P(Z=1)$

We are seeking for the maximal $g(\mu)$ such that $a \ge g(\mu)$

If $\mu < 0 $ then we cannot do better than the trivial $g(\mu)=0$.

(That can be seen by considering a Dirac delta on $x=\mu$)

For $\mu >0$:

$$ \mu = E[E[X | Z]]= a E[X| Z=1] + (1-a) E[X|Z=0] \tag1 $$

but $E[X| Z=1] \le 1$ (because $X\le 1$) and $E[X|Z=0] \le \mu/ 2$. Hence

$$ \mu \le a + (1-a) \mu /2 \implies a \ge \frac{\mu}{2-\mu} \tag2$$

This bound is attained by two Dirac deltas on $x=\mu/2$ and $x=1$ with weights $a$ and $1-a$ resp. Hence he desired bound is $$g(\mu)=\begin{cases} 0 & \mu\le 0\\ \frac{\mu}{2-\mu} & \mu>0 \end{cases}\tag3$$

Answer 2

For this distribution the mean is near $-1$, so half the mean is $-1/2$, and clearly the probability of finding $x>-1/2$ can be made arbitrarily small.