Exercise 2.1: Tightness of Inequalities
(a)
Recall that Markov’s inequality can be derived as follows: assuming that \(X\ge0\) almost surely, \begin{equation} \P(X \ge a) = \frac1a \E[a\ind_{\set{X \ge a}}] \le \frac1a\E[X\ind_{\set{X \ge a}}] \le \frac1a \E[X]. \end{equation} Hence, we have equality if \(a \ind_{\set{X \ge a}} = X\) almost surely. For example, this is the case when \(X \in \set{0, a}\).
(b)
Similarly, for Chebyshev’s inequality, we have equality if \((X - \E[X])^2 \in \set{0, a}\). For example, this is the case when \begin{equation} X = \begin{cases} 2 \sqrt{a} & \text{with probability $p$}, \newline \sqrt{a} & \text{with probability $1 - 2p$}, \newline 0 & \text{with probability $p$}. \end{cases} \end{equation} Then \(\E[X] = \sqrt{a}\), so \((X - \E[X])^2 \in \set{0, a}\).