# How to show the Joint CDF

I have a such example in my text book

Let Y and Z be two IID random variables each with CDF, $$F(\cdot)$$. Let $$X_1:=min(Y,Z)$$ and $$X_2:=max(Y,Z)$$ with marginals $$F_1(\cdot)$$ and $$F_2(\cdot)$$, respectively. We then have \begin{align} P\left(X_{1} \leq x_{1}, X_{2} \leq x_{2}\right)=2 F\left(\min \left\{x_{1}, x_{2}\right\}\right) F\left(x_{2}\right)-F\left(\min \left\{x_{1}, x_{2}\right\}\right)^{2}. \end{align} We can derive this by considering separately the two cases $$\text { (i) } x_{2} \leq x_{1} \text { and (ii) } x_{2}>x_{1}$$

We would like to compute the copula, $$C\left(u_{1}, u_{2}\right) \text {, of }\left(X_{1}, X_{2}\right)$$. Towards this end we rst note the two marginals satisfy

\begin{align} \begin{aligned} F_{1}(x) &=2 F(x)-F(x)^{2} \\ F_{2}(x) &=F(x)^{2} \end{aligned} \end{align} But Sklar's Theorem states that$$C(\cdot,\cdot)$$ satisfies $$C\left(F_{1}\left(x_{1}\right), F_{2}\left(x_{2}\right)\right)=F\left(x_{1}, x_{2}\right)$$ so if we connect the pieces we will obtain $$C\left(u_{1}, u_{2}\right)=2 \min \left\{1-\sqrt{1-u_{1}}, \sqrt{u_{2}}\right\} \sqrt{u_{2}}-\min \left\{1-\sqrt{1-u_{1}}, \sqrt{u_{2}}\right\}^{2}$$

My question is how does the Joint CDF $$P\left(X_{1} \leq x_{1}, X_{2} \leq x_{2}\right)$$ is equal to $$2 F\left(\min \left\{x_{1}, x_{2}\right\}\right) F\left(x_{2}\right)-F\left(\min \left\{x_{1}, x_{2}\right\}\right)^{2}$$ by using $$\text { (i) } x_{2} \leq x_{1} \text { and (ii) } x_{2}>x_{1}$$

When $$x_2 \leq x_1$$ we have $$X_1 \leq x_1$$ and $$X_2 \leq x_2$$ if and only if $$Y \leq x_2$$ and $$Z \leq x_2$$. [Note that $$Y \leq x_2$$ implies $$Y \leq x_1$$ and so $$X_1 \leq x_1$$ also]. Hence, $$P(X_1 \leq x_1$$ and $$X_2 \leq x_2)=P(Y \leq x_2,Z \leq x_2)=F(x_2)^{2}$$. Also, $$2 F\left(\min \left\{x_{1}, x_{2}\right\}\right) F\left(x_{2}\right)-F\left(\min \left\{x_{1}, x_{2}\right\}\right)^{2}=2F(x_2)F(x_2)-(F(x_2))^{2}=(F(x_2))^{2}$$.
Now let $$x_1 . Then $$P(X_1 \leq x_1, X_2\leq x_2)$$ can be split as $$P(Y\leq x_1,Z \leq x_2)+P(Y>x_1, Z\leq x_1, Y\leq x_2)$$. So we get $$F(x_1)F(x_2)+(F(x_2)-F(x_1)) (F(x_1))=2F(x_1)F(x_2)-(F(x_1))^{2}$$.
• Thank you for your response. I have now tried to do the proof for $x_1 < x_2$. I find that evaluating the formula yields $$P(X_1\leq x_1, X_2\leq x_2)=2F(x_1)F(x_2)-F(x_1)^2$$ I can reason that $x_2>x_1$ allow for the events $(Y\leq x_1) \cap (Z\leq x_2)$ and $(Y\leq x_2) \cap (Z\leq x_1)$ that yields the $2F(x_1)F(x_2)$, but I cannot reason how to get the $-F(x_1)^2$. Sep 14 at 13:29
• Thanks for your answer, an last question, why we need $Y>x_1$ but not $Z>x_1$?? thanks alot Sep 16 at 8:18
• @QiYao I am splitting $(X_1 \leq x_1, X_2\leq x_2)$ into $(Y \leq x_1,X_1 \leq x_1, X_2\leq x_2)$ and $(Y > x_1,X_1 \leq x_1, X_2\leq x_2)$. When $Y >x_1$ we need $Z \leq x_1$ so that $X_1 \leq x_1$. Note that $Z \leq x_2$ follows from $Z \leq x_1$ since $x_1<x_2$. Sep 16 at 8:38