Bayesian choice, incorrect line in the book

Question

Here (Large file) on the page 249, in the Example 5.3.12 isn't there a typo on the last but one line: $$...$$ $$=P^X(|X|>|x|),X \sim\cal N(0,1)$$ $$=1-\Phi(|x|)+\Phi(|x|)=2[1-\Phi(|x|)]$$

Answer 1

Yes, there is a typo. The correction is in red: $$ =1-\Phi(|x|)+\Phi(\color{red}{-}|x|)=2[1-\Phi(|x|)] $$ Reasoning: $|x|$ is a positive quantity (actually non-negative), so $P(|X|>|x|)$ can be written as $$P(|X|>|x|)=P(X>|x|) + P(X<-|x|).$$ Expressed in terms of the cdf $\Phi$, the first entity is $$P(X>|x|)=1-\Phi(|x|)$$ and the second is $$P(X<-|x|)=\Phi(-|x|).$$

Alternatively, $P(|X|>|x|)$ is twice $P(X>|x|)$, which gets you immediately to $2[1-\Phi(|x|)]$, skipping the intermediate step.