Find probability distribution Random variable $X$ has continuous probability distrubtion with density given by the formula $f(x)=x^{-2}1{\hskip -2.5 pt}\hbox{l}_{\left(1;\infty\right)}\left(x\right)$. Find the distribution for random variable $Y=1-|X-2|$.
Step 1. $F_Y(t)=\mathbb{P}(Y\le t)=\mathbb{P}(1-|X-2|\le t)=\star$
Step 2. Make a graph.

Step 3. Find $t$ from formulas in two intervals.
1) $x<2\rightarrow y=3-x\rightarrow x=1+t$
2) $x\ge2\rightarrow y=1+t\rightarrow x=3-t$
Step 4. Find proper intervals. Since density equals $0$ for $x\le1$, thus:
$t<0: \ <3-t;\infty)$
$0\le t<1: \ <1;1-t>, \ <3-t;\infty)$
$t=1: \text{ does not matter}$
$t>1 \text{ probability equals 1}$
Step 5. Write the formula for $F_Y(t)$
$\star= \begin{cases} \int_{3-t}^\infty x^{-2}\text{ d}x &\text{ for } t<0\\ \int_{1}^{1+t}x^{-2}\text{ d}x + \int_{3-t}^{\infty}x^{-2}\text{ d}x   &\text{ for } 0\le t < 1\\ 1 &\text{ for } t\ge 1 \end{cases} $
Step 6. Calculate integrals
$\star= \begin{cases} \frac{1}{3-t} &\text{ for } t<0\\ \frac{1}{3-t}+\frac{t}{t+1}  &\text{ for } 0\le t < 1\\ 1 &\text{ for } t\ge 1 \end{cases} $
Is this solution correct? I always have doubts about those intervals. 
 A: Yes, that seems okay.   Here's my work through
0) You've identify the critical point $X=2$ as the Y-maximum, $Y=1$.


*

*And the CDF(of Y) is $1$ when $t\geq 1$.


1) $1<X<2 ~~\to~~ 0<Y<1$ and $ 1-(2-X)\leq t ~~\to~~X\leq t+1$


*

*So we seek $1<X\leq t+1$ when $0<t<1$


2) $X> 2~~\to~~ Y< 1$ and $  1-(X-2)\leq t ~~\to~~X\geq 3-t$


*

*So we also seek $3-t<X$ when $t<1$


Observing that in the overlapping interval($0<t<1$) these events are disjoint for $X$, so:
$$\begin{align}\therefore \quad \mathsf P(Y\leq t) ~&= \mathsf P(\{1<X\leq t+1\}\cup \{3-t\leq X\}) 
\\[2ex] &=~ \mathsf P(1\leq X\leq t+1)\mathbf 1_{t\in[0;1)}~+~\mathsf P(3-t\leq X)\mathbf 1_{t\in(-\infty;1)} + \mathbf 1_{t\in[1;\infty)} 
\\[2ex] &=~ \mathbf 1_{x\in(0;1)}\cdot \int_1^{t+1} x^{-2}\operatorname d x~+~\mathbf 1_{t\in(-\infty;1)}\cdot\int_{3-t}^\infty2 x^{-2}|\operatorname d x~+~\mathbf 1_{t\in[1;\infty)}
\\[2ex] &=~ \mathbf 1_{t\in(0;1)}\cdot \left.\left(\frac {-1}{2x}\right)\right\rvert_{x=1}^{x=t+1}~+~\mathbf 1_{t\in(-\infty;1)}\cdot \left.\left(\frac {-1}{2x}\right)\right\rvert_{x=3-t}^{x\to\infty}~+~\mathbf 1_{t\in[1;\infty)} \\[1ex] &=~ \dfrac t{t+1}\mathbf 1_{t\in(0;1)}~+~\frac 1{3-t}\mathbf 1_{t\in(-\infty;1)}~+~\mathbf 1_{t\in[1;\infty)} 
\\[2ex] &=~ \frac 1{3-t}\mathbf 1_{t\in(-\infty;0]}~+~\left(\frac{1}{3-t}+\frac t{t+1}\right)\mathbf 1_{t\in(0;1)}~+~\mathbf 1_{t\in [1;\infty)} \end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\int_{1}^{\infty}{\delta\pars{y - 1 + \verts{x - 2}} \over x^{2}}\,\dd x =
\int_{1}^{2}{\delta\pars{y + 1 - x} \over x^{2}}\,\dd x +
\int_{2}^{\infty}{\delta\pars{y -3 + x} \over x^{2}}\,\dd x
\\[5mm] = &\
{\bracks{1 < y + 1 < 2} \over \pars{1 + y}^{2}} + {\bracks{3 - y > 2} \over \pars{3 - y}^{2}} =
{\bracks{0 < y < 1} \over \pars{y + 1}^{2}} +
{\bracks{y < 1} \over \pars{y - 3}^{2}}
\\[5mm] = &\
\left\{\begin{array}{lcl}
\ds{1 \over \pars{y - 3}^{2}} & \mbox{if} & \ds{y < 0}
\\[2mm]
\ds{{1 \over \pars{y + 1}^{2}} + {1 \over \pars{y - 3}^{2}}} &
\mbox{if} & \ds{0 < y < 1}
\\[2mm]
\ds{0} & \mbox{if} & \ds{y > 1}
\end{array}\right.
\end{align}
