How to find CDF of $Y=|X|\wedge 2$ with $X\sim Laplace(\lambda)$

Question

Given $X$ a bilateral exponential with density $f_X(x)=\frac{1}{2}e^{-|x|}, \forall x\in \mathbb{R}$ and $\lambda=1$, i have to find CDF of $Y=|X|\wedge 2$.

I know that $Y$ is not a monotonic function, so i can't apply the law of transformation of random variables. I know that the support of $Y$ is $[0,2]$, so i can write $F_Y(y)=\left\{\begin{matrix} 0 & if & y<0\\ ? & if & 0\leq y\leq 2\\ 1 & if &y>2 \end{matrix}\right.$

Then:

$F_y(y)=P(Y\leq y)=P(|X|\wedge 2\leq y)=...$

How can I continue?

Answer 1

You are doing this for $0\leq y\leq 2$ so $|X|\wedge2\leq y$ is the same statement as $|X|\leq y$.

That means that you can continue with:$$\cdots=P(|X|\leq y)=\cdots$$