# Probability inequality via symmetrization

Let $X$ and $X'$ be independent and identically distributed random variables. Define the symmetrized version of $X$ as $X^s=X-X'$. If $a \geq 0$ is such that $P(X \leq -a) \leq 1-p$ and $P(X \geq a) \leq 1-p$ then I have to show that, for every $\varepsilon>0$, $$P(\left|X^s\right| \geq \varepsilon) \geq P(\left|X\right|>a+\varepsilon)$$

I do not understand what role does $p$ play in the whole problem. I'd appreciate if anyone can tell me how to proceed. Thank you. • Miscopied?  
– Did
Feb 6, 2018 at 7:24
• @Did : I've attached a screenshot of the problem, sir. Please see part (b). Source : Theorem 7, page 121, An introduction to Probability and Statistics by Rohatgi, Saleh.
– user516379
Feb 6, 2018 at 7:33
• Indeed the statement is absurd. The text mentions that the result is used later on, locating where in the book this happens might help to guess a corrected version of the theorem. (As an aside, one cannot help to note that this is the third edition of a supposedly widely used textbook...)
– Did
Feb 6, 2018 at 8:22