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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.

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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.

2 Answers 2

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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

  • So we seek $1

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

  • So we also seek $3-t

Observing that in the overlapping interval($0

$$\begin{align}\therefore \quad \mathsf P(Y\leq t) ~&= \mathsf P(\{1

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}