Definition of symmetrized random variable symm

The theorem that I want to prove : enter image description here

Source : Rohatgi, Saleh (3rd edition), p.$121$

Part $(a)$ is an easy consequence of triangle inequality. I'm having trouble with part $(b)$. The right-hand side of the inequality is estimated as follows.

$$P\left\{\left|X\right|>a+\epsilon\right\}=P\left\{X<-a-\epsilon\right\}+P\left\{X>a+\epsilon\right\} \leq P\left\{X \leq -a\right\}+P\left\{X \geq a\right\} \leq 2(1-p) \tag{1}$$

Thus, if we can show that $P\left\{\left|X^s\right| \geq \epsilon\right\} \geq 2(1-p)$, then we are done. This is where I'm stuck. Is it possible to show this? Or are we loosing too much in $(1)$? Also, I'd be grateful if anyone can provide me a direct solution/hint. Thank you.

  • $\begingroup$ You are loosing too much in (1); nothing is known about the value of $p$ except that $0<p<1$. If $p<1/2$ then $2(1-p)>1$ so you cannot show that $P\{|X^{s}| \geq \epsilon\} \geq 2(1-p)$. $\endgroup$ Mar 9, 2018 at 7:58
  • 1
    $\begingroup$ Relevant: math.stackexchange.com/questions/2638117/… $\endgroup$
    – Clement C.
    Mar 12, 2018 at 4:16

1 Answer 1


Note that it is not hard to show $$\mathbb{P}\{\lvert X^s\rvert \geq \varepsilon\} \geq \mathbb{P}\{\lvert X\rvert > a+\varepsilon\}\cdot p\,.\tag{$\dagger$}$$ Indeed, for $a,p$ as in the statement and any $\varepsilon>0$, we have $$ \{ \lvert X\rvert > a+\varepsilon \} = \{ X > a+\varepsilon \}\cup \{ X < -(a+\varepsilon) \} $$ and $$ \left(\{ X > a+\varepsilon\}\cap \{ X' \leq a \}\right)\cup \left(\{ X < -(a+\varepsilon)\}\cap \{ X' \geq -a \}\right) \subseteq \{ \lvert X-X'\rvert \geq \varepsilon\}\,. $$ Therefore, since $\mathbb{P}\{ X' \leq a \} \geq p$ and $\mathbb{P}\{ X' \geq -a \} \geq p$ by assumption, we get (here I'll detail a lot) $$\begin{align} \mathbb{P}\{\lvert X^s\rvert \geq \varepsilon\} &= \mathbb{P}\{\lvert X-X'\rvert \geq \varepsilon\} \geq \mathbb{P}\left( \left(\{ X > a+\varepsilon\}\cap \{ X' \leq a \}\right)\cup \left(\{ X < -(a+\varepsilon)\}\cap \{ X' \geq -a \}\right) \right)\\ &= \mathbb{P}\left(\{ X > a+\varepsilon\}\cap \{ X' \leq a \}\right) + \mathbb{P}\left(\{ X < -(a+\varepsilon)\}\cap \{ X' \geq -a \}\right) \\ &= \mathbb{P}\{ X > a+\varepsilon\} \mathbb{P}\{ X' \leq a \} + \mathbb{P}\left(\{ X < -(a+\varepsilon)\} \mathbb{P}\{ X' \geq -a \}\right) \\ &\geq \mathbb{P}\{ X > a+\varepsilon\} \cdot p + \mathbb{P}\{ X < -(a+\varepsilon)\} \cdot p \\ &= p\cdot \mathbb{P}\{ X > a+\varepsilon\} \cup \{ X < -(a+\varepsilon)\}\\ &= p\cdot \mathbb{P}\{ \lvert X\rvert > a+\varepsilon \}\,. \end{align} $$

Without the extra $p$, however, the result is simply false (the fact that the end inequality does not depend on $p$ is a big clue). As a counter example, take $X$ to be a Rademacher r.v., i.e. uniform on $\{-1,1\}$; then, for $a=0$ and $p=1/2$, we have $$ \mathbb{P}\{X\geq 0\} = 1-p, \qquad \mathbb{P}\{X\leq 0\} = 1-p $$ (so a fortiori the inequalities hold). Now, the symmetrized r.v. $X^s$ satisfies $$ \mathbb{P}\{X^s=0\} = \frac{1}{2},\qquad \mathbb{P}\{X^s=2\} =\mathbb{P}\{X^s=-2\} = \frac{1}{4} $$ so, for any $\varepsilon \in (0,1)$, $$ \mathbb{P}\{\lvert X^s\rvert \geq \varepsilon\} = \frac{1}{2} = p\cdot 1 = p\cdot \mathbb{P}\{\lvert X\rvert > \varepsilon\}\,. $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .