Uniform distribution on the [0,2]

Question

Random variable $X$ has uniform distribution on the $[0,2]$.How to find distribution radnom variable $Y=min \left\{X,X^2 \right\}$?

Answer 1

$$\min(X,X^2)=\begin{cases}X^2&\text{ if }&X\leq1\\X&\text{ if }&X>1.\end{cases}$$

Then

$$P(\min(X,X^2)<x)=P(X^2<x\cap X<1)+P(X<x\cap X\geq 1).$$

And

$$P(X^2<x\cap X<1)=\begin{cases}0&\text{ if }& x\leq0\\P(X<\sqrt x)&\text{ if }&0<x\leq 1\\P(X<1)&\text{ otherwise}\end{cases}=$$$$=\begin{cases}0&\text{ if }& x\leq0\\\frac12\sqrt x&\text{ if }&0<x\leq 1\\\frac12&\text{ otherwise,}\end{cases}$$ $$P(X<x\cap X\geq1)=\begin{cases}P(1\leq X<x)&\text{ if }& 1\leq x\leq2\\\frac12&\text{ if }& x>2\\0&\text{ otherwise}\end{cases}=$$$$=\begin{cases}\frac12(x-1)&\text{ if }&1\leq x\leq2\\\frac12&\text{ if }& x>2\\0&\text{ otherwise.}\end{cases}$$

The sum of the two piecewise given function is depicted below:

Answer 2

The key is that either $X^2{<}X{<}1$, $1{=}X{=}X^2$, or $1{<}X{<}X^2$.

So when $x{<}1$ we wish to know the probability that $X^2{\leqslant}x$, and when $1{\leqslant}x$ we wish to know the probability that $X{<}x$.

$$\mathsf P(\min\{X,X^2\}{\leqslant} x) ~{=~\mathsf P(X^2{\leqslant} x)\mathbf 1_{x < 1}+\mathsf P(X{\leqslant}x)\mathbf 1_{1\leqslant x}\\ = \underline{\phantom{\tfrac 12 \surd x}}\,\mathbf 1_{0\leqslant x < 1}+\underline{\phantom{\tfrac 12\,x\quad}}\,\mathbf 1_{1\leqslant x< 2}+\mathbf 1_{2\leqslant x}\\=\begin{cases}0 &:& x< 0\\ \underline{\phantom{\tfrac 12\surd x}}&:&0\leqslant x< 1\\ \underline{\phantom{\tfrac 12 x\quad}} &:& 1\leqslant x<2\\ 1 &:& 2\leqslant x \end{cases}}$$

Just fill in the blanks as appropriate, is all.