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If I have a random variable $$ X:[0,1] \rightarrow \mathbb{R} \quad \text{so that} \quad X(w)=\min\{w,1-w\} , \quad \text{ where } w \in [0,1]$$ The question is to find the inverse of the random variable ,which is as follows : $$ X^{-1}(]-\infty,x]=`\phi\quad \text{ if } x<0 $$ $$ [0,x]\bigcup[1-x,1] \quad\text {if} 0<=x<=1/2 $$ $$ [0,1] \quad\text {if} x>=1/2 $$ `

Does someone know how did he calculate it?

sara fkh
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1 Answers1

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$X(w) > x$ iff $w > x$ and $1-w > x$ iff $w > x$ and $w < 1-x$. If $x \geq 1-x$ or $x \geq \frac 1 2$ then this is impossible. If $x \in [0,\frac 1 2]$ this is equivalent to $x <w <1- x$. Now just take the complement in $[0,1]$ to get the inverse image of $(-\infty, x]$.

Kavi Rama Murthy
  • 311,013