Given scenario: Jim invites two friends J and W to a party. The arrival times of the two friends are independent and each is uniformly distributed over an hour starting from 19:00. Define the following:
Y1 = time elapsed since 19:00 until the first of the two friends arrives.
Y2: the time elapsed since 19:00 until the second of the two friends arrives.
How do you show that (over some region) the joint cdf of Y1 and Y2 is given by $2y_{1}y_{2} - (y_{1})^2$
On
Note that $Y_1=\min\{Y_J,Y_W\}$ and $Y_2=\max\{Y_J,Y_W\}$ where $Y_J$ is the time elapsed since 19:00 when J arrives and $Y_W$ is the time elapsed since 19:00 when W arrives (both measured in hours). Thus, $(Y_J,Y_W)$ is uniform on $(0,1)^2$, and, for every $x\lt y$ in $(0,1)$, $$ P(x\lt Y_1,Y_2\lt y)=P(x\lt Y_J\lt y,x\lt Y_W\lt y)=(y-x)^2. $$ Differentiating twice yields the density $f_{Y_1,Y_2}$ of $(Y_1,Y_2)$ as $$ f_{Y_1,Y_2}(y_1,y_2)=2\,\mathbf 1_{0\lt y_1\lt y_2\lt 1}. $$ Thus, for every $0\lt y_1\lt y_2\lt1$, $$ P(Y_1\lt y_1,Y_2\lt y_2)=\int_0^{y_2}\int_0^{y_1}f_{Y_1,Y_2}(u,v)\mathrm du\mathrm dv=\int_0^{y_2}\int_0^{\min\{y_1,v\}}2\mathrm du\mathrm dv, $$ that is, $$ P(Y_1\lt y_1,Y_2\lt y_2)=\int_0^{y_2}2\min\{y_1,v\}\mathrm dv=\int_0^{y_1}2v\mathrm dv+\int_{y_1}^{y_2}2y_1\mathrm dv, $$ that is, $$ P(Y_1\lt y_1,Y_2\lt y_2)=y_1^2+2y_1(y_2-y_1)=2y_1y_2-y_1^2. $$
Sorry Did is right I wrote an answer before which was completely wrong but I think I found an alternative method to find the pdf which is just the first step.
Since $Y_1$ and $Y_2$ are order statistics the joint pdf is given by $$f_{(1),(2)}(y_1,y_2)=\frac{2!}{(1-1)!(2-1-1)!(2-2)!}(y_1)^0(y_2-y_1)^0(y_2)^0f_{(1)}(y_1)f_{(2)}(y_2)=2$$