Lower bound for tail probability I need help to prove the following lower bound for tail probability. I have tried using well-known inequalities like Chebyshev and Paley-Zygmund, but cannot get the required bound.
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and $X$ be a random variable with $\mathbb{P}(X\in [-M,M])=1$ for some positive constant $M$. Show that $\mathbb{P}\left(|X-\mu|\geq \frac{\sigma^2}{4M} \right)\geq \frac{\sigma^2}{8M^2}$, where $\mu= \mathbb{E}(X)$ and $\sigma^2=\text{Var}(X)$.
Thank you.
Additional question: How can I also prove that $\mathbb{P}\left(X\geq \mu+ \frac{\sigma^2}{4M} \right)\geq \frac{\sigma^2}{8M^2}$ ?
 A: The function $f(t):= \mathbb P(|X-\mu|\ge t)-\frac t{2M}$ is strictly decreasing on $[0,\infty)$ and we have $f(0)=1$, $f(2M)=-1$.
Let $s=\inf\{t\ge0\mid f(t)\le 0\}$.
Then
$$ \sigma^2 = \int (x-\mu)^2 d\mu =\int_0^\infty 2r\mathbb P(|X-\mu|\ge r) dr\\
\le\int_0^s2r\cdot1\,dr+\int_s^{2M}2r\cdot\frac s{2M}dr\\
=s^2+2Ms-\frac{s^3}{2M}.
$$
If $s=0$, then this shows $\sigma^2=0$ and thus the claim we want to show is the trivial statement $\mathbb P(|X-\mu|\ge0)\ge0$.
Therefore, we may assume  $0< s\le 2M$ and conclude
$$ \sigma^2\le s^2+2Ms-\frac{s^3}{2M}< 2Ms+2Ms-0=4Ms.$$
But then $\frac{\sigma^2}{4M}<s$ implies $f\left(\frac{\sigma^2}{4M}\right)>0$, i.e.
$$ \mathbb P\left(|X-\mu|\ge \frac{\sigma^2}{4M}\right)>\frac{\sigma^2}{8M^2}.$$
Apparently, the only case where $\le$ canot be replaced with $>$ is the case of an (almost surely) constant variable.
A: The fourth moment bound says:
$$
E[|X|] \ge \frac{E[X^2]^{3/2}}{E[X^4]^{1/2}}.\\
$$
Let $s = \max\{|\mu+M|, |\mu-M|\} \le 2M$, that is the furthest the mean can be from $X$.
Then we have $E[(X-\mu)^4] \le s^2 E[(X-\mu)^2]$.
Thus, by Paley–Zygmund's inequality:
$$
\Pr[|X-\mu| \ge \alpha\sigma^2]
\ge (1-\alpha)^2\frac{E[|X-\mu|]^2}{E[|X-\mu|^2]}
\ge (1-\alpha)^2\frac{E[|X-\mu|^2]^2}{E[|X-\mu|^4]}
\ge (1-\alpha)^2\frac{E[|X-\mu|^2]}{s^2}.
$$
Setting $\alpha=1/(4M)$ we have
$$
\Pr[|X-\mu| \ge \frac{\sigma^2}{4M}]
\ge (1-1/(4M))^2\frac{\sigma^2}{s^2}
\ge \frac{\sigma^2}{8M^2}
,
$$
where the last bound requires $M\ge 1$. However, that requirement could probably be removed by using a slightly stronger version of Paley–Zygmund.
