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 $P\{|X^{s}| \geq \epsilon\} \geq 2(1-p)$. $\endgroup$ – Kabo Murphy Mar 9 '18 at 7:58
  • 1
    $\begingroup$ Relevant: math.stackexchange.com/questions/2638117/… $\endgroup$ – Clement C. Mar 12 '18 at 4:16

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 $\varepsilon \in (0,1)$, $$ \mathbb{P}\{\lvert X^s\rvert \geq \varepsilon\} = \frac{1}{2} = p\cdot 1 = p\cdot \mathbb{P}\{\lvert X\rvert > \varepsilon\}\,. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.