Finding CDF of $Y = \min(X, X^2)$ and $Z=\max(X,X^2)$ when $X$ is Uniform on $[0,2]$ Let X be a random variable with uniform distribution on $[0,2]$. Find CDF of $Y = \min({X, X^2})$ and $Z = \max(X, X^2)$.
$F_Y(t) = \mathbb{P}(Y \leq t) = \mathbb{P}(\min(X, X^2) \leq t) = \dots$
If $t < 0$ then $\mathbb{P}(\min(X, X^2) \leq t) = \mathbb{P}(X \leq t) = 0 $
If $t \geq 2$ then $\mathbb{P}(\min(X, X^2) \leq t) = 1 $ because $ \mathbb{P}(X \leq t) = 1 $
What about when $t \in [0, 2)$? How can I find $F_Z(t)$?
 A: If $0 \le t \le 2$, we have
$\mathbb{P}(\max(X,X^2) \le t)=\mathbb{P}(X \le t, X^2 \le t)=\mathbb{P}(X \le t, X \le \sqrt{t})=\mathbb{P}(X \le \min(t,\sqrt{t}))=F_X(\min(t,\sqrt{t})).$
Now when is $t \le \sqrt{t}$?
For $\min(X,X^2)$, try to look for $\mathbb{P}(\min(X,X^2) \ge t)$
A: Observing that $
X^2 \leq X \iff X \leq 1
$
For $t \in [0,1)$ we obtain
$$
F_Y(t) = \mathbb{P}(\min(X,X^2) \leq t) = \mathbb{P}(X^2 \leq t) = \mathbb{P}(X \leq \sqrt{t}) = \frac{\sqrt{t}}{2}
$$
On the other hand if $t \in [1,2)$ then
$$
F_Y(t) = \mathbb{P}(\min(X,X^2) \leq 1) + \mathbb{P}(1 < \min(X,X^2) \leq t) = \\
\mathbb{P}(X^2 \leq 1) + \mathbb{P}(X \leq t) - \mathbb{P}(X\leq1) = \mathbb{P}(X \leq t) = \frac{t}{2}
$$
Our CDF is thus 
$$
F_Y(t) = \begin{cases}
0 & : t< 0 \\ 
\frac{\sqrt{t}}{2} & : 0\leq t < 1 \\ 
\frac{t}{2} & : 1\leq t < 2 \\
1 & : 2\leq t 
\end{cases}
$$
Using pretty much the same logic with CDF of Z we get:
$$
F_Z(t) = \begin{cases}
0 & : t< 0 \\ 
\frac{t}{2} & : 0\leq t < 1 \\ 
\frac{\sqrt{t}}{2} & : 1\leq t < 4 \\
1 & : 4\leq t 
\end{cases}
$$
