Distribution of $Y$ derived from standard normal

Question

If $X$ is $ N(0,1)$, then what the distribution of $Y$, where $Y=X$ when $|X|\leq 1$ and $Y=-X$ when $|X| > 1$.

My attempt: when $|Y| \leq 1$, $F(y) = \Phi (y) $. When $|Y| > 1$, $F(y) = 1 - \Phi (y) $. Am I correct?

What bothers me is the fact that this distribution doesn't go to zero (one) as we move to the $- \infty$ ($ \infty$).

Answer 1

Since the distribution of $X$ is symmetric, $Y$ has exactly the same distribution as $X$.

Indeed: If we denote by $p$ the density of $X$, i.e.

$$p(x) = \frac{1}{\sqrt{2\pi}} \exp \left(- \frac{x^2}{2} \right),$$

then $p(x)=p(-x)$. This implies

$$\mathbb{P}(X \in C) = \mathbb{P}(-X \in C) \tag{1}$$

for any Borel set $C$. Consequently,

$$\begin{align*} \mathbb{P}(Y \in B) &= \mathbb{P}(Y \in B \cap [-1,1]) + \mathbb{P}(Y \in B \backslash [-1,1]) \\ &\stackrel{\text{def $Y$}}{=} \mathbb{P}(X \in B \cap [-1,1]) + \mathbb{P}(-X \in B \backslash [-1,1]) \\ &\stackrel{(1)}{=} \mathbb{P}(X \in B \cap [-1,1]) + \mathbb{P}(X \in B \backslash [-1,1]) \\ &= \mathbb{P}(X \in B) \end{align*}$$

for any Borel set $B$; here we have applied $(1)$ for the Borel set $C := B \backslash [-1,1]$.

Answer 2

Although @Saz's answer is correct, I'd like to provide another one that doesn't require knowledge of measure theory:

Consider the density of $X$, $$f(x)=\frac1{\sqrt{2\pi}}e^{-\frac12 x^2}.$$ Since $(-x)^2=x^2$ for all $x\in\mathbb R$, $f$ is an even function. So to compute the distribution of $-X$, we have \begin{align} \mathbb P(-X \leqslant x) &= \mathbb P(X \geqslant -x)\\ &= \int_{-x}^\infty f(t)\ \mathsf dt\\ &= -\int_\infty^{-x} f(t)\ \mathsf dt\\ &= -\int_\infty^x f(-t)\ \mathsf dt\\ &= \int_{-\infty}^x f(t)\ \mathsf dt\\ &= \mathbb P(X\leqslant x). \end{align}