Confusion about calculating cdf and pdf of random variable Let $X$ be a random variable with pdf:
$$p_X(x)=\begin{cases}
1-|x| & x\in[-1,1] \\
0 & \text{Otherwise}
\end{cases}$$
Let $Y$ be
$$Y=g(X)=\begin{cases}
-X & x<0 \\
2X & x\geq 0
\end{cases}$$
Find cdf and pdf of $Y$.
This is my attempt:
If $0\leq y\leq 1$, then $\mathbb{P}(Y\leq y)=\mathbb{P}(-X\leq y)=\mathbb{P}(X\geq -y)=1-\mathbb{P}(X\leq -y)$. Furthermore, $\mathbb{P}(X\leq -y)=\int_{-1}^{-y}(1+x)dx=\dfrac{1}{2}-y+\dfrac{y^2}{2}$, hence $\mathbb{P}(Y\leq y)=\dfrac{1}{2}+y-\dfrac{y^2}{2}$. On the other hand, $\mathbb{P}(Y\leq y)=\mathbb{P}(2X\leq y)=\mathbb{P}(X\leq \dfrac{y}{2})=\int_0^{y/2}(1-x)dx=\dfrac{y}{2}-\dfrac{y^2}{8}$. So, that would yield
$$P_y(Y)=\begin{cases}
\dfrac{1}{2}+\dfrac{3y}{2}-\dfrac{5y^2}{8} & y\in[0,1] \\
\dfrac{y}{2}-\dfrac{y^2}{8} & y\in(1,2] \\
0 & \text{Otherwise}
\end{cases}$$
Then, I can take the derivative and thus find the pdf. However, I have several problems, because there is a jump at $y=0$, namely $P_y(0^-)=0$, but $P_y(0^+)=\dfrac{1}{2}$; also $P_y(1)=1.375$, which doesn't make sense. I know I am doing something wrong, but I am not sure where or what. Could someone give me some hint? Thanks!
Edit:
Since I know the pdf of $Y$ and I know that the cdf is just the integral and its value final value should be $1$, I calculated:
$$P_y(Y)=\begin{cases}
\dfrac{3y}{2}-\dfrac{5y^2}{8} & y\in[0,1] \\
\dfrac{1}{2}+\dfrac{y}{2}-\dfrac{y^2}{8} & y\in(1,2] \\
0 & \text{Otherwise}
\end{cases}$$
Everything seems to work, but I still can't "formally" get the cdf.
 A: I didn't check your arithmetic, but I got different results, which are consistent.
$P(Y\le y)=\frac{3y}{2}-\frac{5y^2}{8}$ for $0\le y\le 1$.
$P(Y\le y)=\frac{y}{2}-\frac{y^2}{8}+\frac{1}{2}$ for $1\le y\le 2)$.
Note that it is continuous at $y=1$ where $P=\frac{7}{8}$ so getting density is no problem.
Derivation: $Y\le 1$,  $\int_{-y}^0(1-|x|)dx+\int_0^\frac{y}{2}(1-|x|)dx$
$1\le Y\le 2$, $\int_\frac{1}{2}^\frac{y}{2}(1-|x|)dx+\frac{7}{8}$
A: From the other direction: find the Y-pdf from the X-pdf, then integrate to obtain the Y-CDF
(NOTE: that is  CDF for "Cumulative Distribution Function").
We have $Y= -X\,\mathbf 1_{X<0}+ 2X\,\mathbf 1_{X\geqslant 0}$ and $p_{\small X}(x)=(1-\lvert x\rvert)\,\mathbf 1_{x\in[-1..1]}$ , so using the Jacobian Transformation technique:
$$\begin{align}p_{\small Y}(y) &=p_{\small X}(-y)\left\lvert\dfrac{\mathrm d (-y)}{\mathrm d y}\right\rvert\mathbf 1_{y\in(0..1]}+p_{\small X}(y/2)\left\lvert\dfrac{\mathrm d (y/2)}{\mathrm d y}\right\rvert\mathbf 1_{y\in[0..2]}\\[2ex] &=(1-y)\mathbf 1_{y\in(0..1]}+\tfrac 12(1-y/2)\mathbf 1_{y\in[0..2]}\\[2ex]&=\frac {6-5y}4\mathbf 1_{y\in[0..1)}+\frac{2-y}4\mathbf 1_{x\in[1..2]}\\[4ex]F_{\small Y}(y) &=\int_0^y\dfrac{6-5t}{4}\,\mathrm d t\cdot\mathbf 1_{y\in[0..1)}\,+\left(\int_0^1\dfrac{6-5y}{4}\,\mathrm d t+\int_1^y\dfrac{2-y}{4}\,\mathrm d t\right)\mathbf 1_{y\in[1..2]} \\[2ex]&=\left[\dfrac{12t-5t^2}{8}\right]_{t=0}^{t=y}\mathbf 1_{y\in[0..1)}+\left(\left[\dfrac{12t-5t^2}{8}\right]_{t=0}^{t=1}+\left[\dfrac{4t-t}{8}\right]_{t=1}^{t=y}\right)\mathbf 1_{y\in[1..2)}+\mathbf 1_{y\in[2..\infty)}\\[2ex]&=\frac{12y-5y^2}{8}\mathbf 1_{y\in[0..1)}+\frac{4+4y-y^2}{8}\mathbf 1_{y\in[1..2)}+\mathbf 1_{y\in[2..\infty)}\end{align}$$
Which agrees with your answer.

Of course, when given that CDF you may find the pdf by differentiating.
$$\begin{align}p_{\small Y}(y)&=\dfrac{\mathrm d ~~}{\mathrm d y}\left(\frac{12y-5y^2}{8}\mathbf 1_{y\in[0..1)}+\frac{4+4y-y^2}{8}\mathbf 1_{y\in[1..2)}+\mathbf 1_{y\in[2..\infty)}\right)\\[2ex]&=\dfrac{6-5y}{4}\mathbf 1_{y\in[0..1)}+\dfrac{2-y}{4}\mathbf 1_{y\in[1..2]}\end{align}$$
