Suppose that $X_1$ and $X_2$ are independent. $Y_1=\max\{X_1, X_2\}$ and $Y=\min\{X_1, X_2\}$. Find the joint pdf of $(Y_1, Y_2)$. Suppose that $X_1$ and $X_2$ are independent with common pdf $f(x)$ and cdf $F(x)$. $Y_1=\max\{X_1, X_2\}$ and $Y=\min\{X_1, X_2\}$. Find the joint pdf of $(Y_1, Y_2)$.

My solution: I have no idea about the pdf of $(Y_1, Y_2)$ but
I can find the cdf and pdf of $Y_1$, $Y_2$:
$$
P(Y_1\le y_1)=F(y_1)^2
$$
and then $f_{Y_1}(y_1)=2F(y_1)f(y_1)$ for $y_1\in A$.
Similarly,
$$
P(Y_2\le y_2)=1-(1-F(y_2))^2
$$
and then $f_{Y_2}(y_2)=2(1-F(y_2))f(y_2)$ for  $y_2\in A$.
How to go the next step?

Update:
$$
P(\max\{X_1, X_2\}\le x_1, \min\{X_1, X_2\}\le x_2)=1-P(\max\{X_1, X_2\}\le x_1, \min\{X_1, X_2\}> x_2)
$$
$$
=1-P(X_1\le x_1, X_2\le x_1, X_1>x_2, X_2> x_2)
$$
When $x_1>x_2$, we get
$$
=1-(F(x_1)-F(x_2))^2
$$
 A: Let $X_1$ and $X_2$ iid continuous random variables. I will write $Y_1:=\min(X_1,X_2)$ and $Y_2:=\max(X_1,X_2)$ since  this notation it's more classic and preserves the idea of an order statistic (we could be more precise and write $(Y_{(1)},Y_{(2)})$ instead $(Y_1,Y_2)$). Recall that $Y_1\leqslant Y_2$ by definition. We want $P(Y_1\leqslant x_1,Y_2\leqslant x_2)$.
Casework:

*

*If $x_1\leqslant x_2$, then $$[Y_1\leqslant x_1,Y_2\leqslant x_2]=[(X_1\leqslant x_1 , X_2\leqslant x_2)]\cup [(X_2\leqslant x_1,X_1\leqslant x_2)]$$
Thus, additive law of the probability give $$P(Y_1\leqslant x_1,Y_2\leqslant x_2)=P(X_1\leqslant x_1,X_2\leqslant x_2)+P(X_2\leqslant x_1,X_1\leqslant x_2)-P(X_1\leqslant x_1,X_2\leqslant x_1)$$
Independence give $$P(Y_1\leqslant y_1,Y_2\leqslant y_2)=F(x_1)F(x_2)+F(x_1)F(x_2)-F(x_1)F(x_1)$$


*If $x_1>x_2$, then since $Y_1\leqslant Y_2$ we have $$P(Y_1\leqslant x_1,Y_2\leqslant x_2)=P(Y_1\leqslant x_2,Y_2\leqslant x_2)=F(x_2)F(x_2)$$
End casework.
Thus, the cdf of $(Y_1,Y_2)$ is just $$F_{Y_1,Y_2}(x_1,x_2)=\begin{cases}2F(x_1)F(x_2)-[F(x_1)]^2,\quad x_1\leqslant x_2,\\ [F(x_2)]^2,\quad x_1>x_2\end{cases}$$
Therefore, the pdf of $(Y_1,Y_2)$ is the derivative then $$f_{Y_1,Y_2}(x_1,x_2)=\begin{cases}2f(x_1)f(x_2),\quad x_1\leqslant x_2\\ 0,\quad \text{otherwise}\end{cases}$$
