Knowing that…
$Y:=\left\{\begin{matrix} X & if & |X|<a\\ -X & if & |X| \geq a \end{matrix}\right.$;
$X \sim N(0,1)$;
$a>0$;
…and obviously $|X|:=\left\{\begin{matrix} X & if & X \geq 0\\ -X & if & X <0 \end{matrix}\right.$, I have to prove that $Y$ is normal.
I got to get there myself, but I'm stuck how to broach the problem. Can you give me a hint to go on?
EDIT:
Looking the hints below I wrote:
$F_Y(y)=\mathbb{P}(Y \leq y)=\mathbb{P}([|X|<a \cap X \geq 0] \cup[|X| \geq a \cap X<0])=\mathbb{P}(|X|<a \cap X \geq 0) + \mathbb{P}(|X| \geq a \cap X<0)=\mathbb{P}(-a<X<a \cap X \geq 0)+\mathbb{P}(-a \geq X \geq a \cap X<0)=2\mathbb{P}(0<X<a \cap X \geq 0)+\underbrace{2\mathbb{P}(0 \geq X \geq a \cap X<0)}_{=0}=2\mathbb{P}(0<X<a \cap X \geq 0)=2\mathbb{P}(0<X<a)=2[\mathbb{P}(X<a)-\mathbb{P}(X<0)]=2[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{a}e^{-\frac{x^2}{2}}dx-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{-\frac{x^2}{2}}dx]$
Obviously it's about two CDF of $X$, so I can conclude that $Y$ is normal. Is it correct?