If $P(|X-E(X)| \geq a \sigma ) = \frac{1}{a^2}$, show the following properties

Question

If $X_1$ and $X_2$ have this property, $E(X_1) = E(X_2)$, $Var(X_1) = Var(X_2)$, then $P(X_1 = t) = P(X_2 = t)$ for all $t$

How do I show this?

Answer 1

Recall how the Chebyshev's inequality is derived. Denoting $\mu = \mathbb{E}X$, it comes from the following inequality

$$ (a\sigma)^2 \mathbf{1}_{\{|X - \mu| \geq a\sigma\}} \leq |X - \mu|^2 $$

which holds pointwise. Since the Chebyshev's inequality is saturated, it implies that in fact

$$ (a\sigma)^2 \mathbf{1}_{\{|X - \mu| \geq a\sigma\}} = |X - \mu|^2 $$

holds $\mathbb{P}$-a.s. In particular, $X$ can take only three values $\mu$ and $\mu\pm a\sigma$ with probability 1. Denoting $p = \mathbb{P}(X = \mu + a\sigma)$ and $q = \mathbb{P}(X = \mu - a\sigma)$, computing $\mu = \mathbb{E}X$ and $\sigma^2 = \operatorname{Var}(X)$ gives

$$ p = q \quad\text{and}\quad p+q = \frac{1}{a^2}. $$

This completely determines the distribution $X$ modulo constants $\mu \in \mathbb{R}$, $\sigma > 0$, $a \geq 1$ as follows:

$$ \mathbb{P}(X = \mu + a\sigma) = \mathbb{P}(X = \mu - a\sigma) = \frac{1}{2a^2}, \qquad \mathbb{P}(X = \mu) = 1 - \frac{1}{a^2}. $$

In your case, the first two parameters coincide, so $X_1$ and $X_2$ have the same distribution exactly when the associated constant $a$ (which a priori depends on $X$) also coincides.