Generating an analytical copula for an example problem

Question

I am currently doing research that requires me to understand dependence modeling. As a first step, I am reading An Introduction to Copulas. I am, stuck on the first example problem which I have re-written below and would like some well earned wisdom to help unstick me.

Example 1

Let $Y$ and $Z$ be two IID random variables each with CDF, $F()$. Let $X_1 = \min(Y, Z)$ and $X_2 = \max(Y, Z)$ with marginals $F_1$ and $F_2$ respectively. Use Sklar's Theorem to find an expression that satisfies $C(F_1(X_1)$, $F_2(X_2)) = F(x_1, x_2)$.

To solve this we need to start by finding $F(x_1,x_2)$, and then find the marginal distributions F_1(X_1), $F_2(X_2)$. After playing with it, and using the hint given in the notes, I understand(ish) how they arrived at the following:

1. $F(x_1, x_2) = 2F(\min(x_1,x_2))F(x_2) - F(\min(x_1,x_2))$

2. $F_1(X_1) = 2F(x_1) - F(x_1)^2$

3. $F_2(X_2) = F(x_2)^2$

Now is where it gets hard for me. The copula derived is

$C(u_1, u_2) = 2 \min(1 - \sqrt{1 - u_1}, \sqrt{u_2})\sqrt{u_2} - \min(1 - \sqrt{1 - u_1}, \sqrt{u_2})^2 = F(x_1, x_2)$

I understand how equation (3) was used to get $\sqrt{u_2} = F(x_2)$. But, I am having trouble with how to derive $F(\min(x_1,x_2)) = \min(1 - \sqrt{1 - u_2}, \sqrt{u_2})$.

Any help would be appreciated and let me know if there is anything unclear about my question!

Answer 1

I know you said you understand how to get the marginals here but I figured I would start from the beginning for any newcomers. The first thing to notice is that by the law of total probability we have,

$$F_1(x_1) = \Pr[X_1 \leq x_1] = \Pr[X_1\leq x_1, X_2\leq X_2]+\Pr[X_1\leq x_1, X_2>x_2]$$

And,

$$F_2(x_2) = \Pr[X_2 \leq x_2] = \Pr[X_1\leq x_1, X_2\leq X_2]+\Pr[X_1> x_1, X_2\leq x_2]$$

To determine the marginal CDFs and the joint CDF we take advantage of the fact we know that $X_1 = \min(Y,Z), X_2=\max(Y,Z)$ for iid $Y,Z\sim F$. In particular notice the following equivalences,

$$F_2(x_2)=\Pr[X_2 \leq x_2]=\Pr[\max(Y,Z) \leq x_2]=\Pr[Y\leq x_2, Z\leq x_2]\underbrace{=}_{\text{iid}}\Pr[Y\leq x_2]^2=F(x_2)^2$$

The argument for the min is a little longer because we need to check the possibility that either variable is larger than the other,

$$F_1(x_1)=\Pr[X_1 \leq x_1]=\Pr[\min(Y,Z)\leq x_1]=\Pr[Y\leq x_1, Z> x_1] + \Pr[Y> x_1, Z\leq x_1]\underbrace{=}_{\text{iid}}F(x_1)(1-F(x_1))+F(x_1)(1-F(x_1))=2F(x_1)-F(x_1)^2$$

Now in order to find the joint CDF we need to determine the final term in the expansion of $F_2$,

$$\Pr[X_1> x_1, X_2\leq x_2]$$

First, suppose that $x_1 \geq x_2$ then this probability is zero since it would be saying the minimum is larger than the maximum. So we are left to consider the case $x_1 < x_2$. Here we have,

$$\Pr[X_1> x_1, X_2\leq x_2] = \Pr[x_2\geq X_1 \geq x_1, x_1< X_2\leq x_2]=(\Pr[x_2\geq X_1 \geq x_1)^2=[F(x_2)-F(x_1)]^2$$

where the first inequality follows from our assumption $x_1 <x_2$ and the second and third equality follows from $Y,Z\sim F$ iid. Putting all this together we get the joint PDF,

$$\Pr[X_1\leq x_1, X_2\leq x_2]=F_2(x_2)-[F(x_2)-F(x_1)]^2=F(x_2)^2-[F(x_2)-F(x_1)]^2=2F(x_1)F(x_2)-F(x_1)^2$$

Notice we can also plug in $x_1=\min(x_1,x_2)$ here because $x_1<x_2$ by our assumption and so the probability does not change. Remember that if $x_1\geq x_2$ then $\Pr[X_1\leq x_1, X_2\leq x_2]=\Pr[X_2\leq x_2]$ because the minimum could not be greater than the maximum. So we have,

$$\Pr[X_1\leq x_1, X_2\leq x_2]=2F(\min(x_1,x_2))F(x_2)-F(\min(x_1,x_2))^2$$

The next thing that we must do before we can apply Sklar's theorem is to find the inverses of the marginals. Luckily this is straightforward in this case,

$$F_1^{-1}(u_1)=F^{-1}(1-\sqrt{(1-u_1)})=x_1$$ $$F_2^{-1}(u_1)=F^{-1}(\sqrt{(u_2)})=x_2$$

Where the final equalities simply follow from $f^{-1}(f(x))=x$. Now we are ready to apply Sklar's theorem which states that any 2-dimensional CDF, $F(x_1,x_2)=\Pr[X_1\leq x_1, X_2\leq x_2]$, can be expressed with a copula, $C$, and univariate CDFs, $F_1,F_2$ such that,

$$F(x_1,x_2)=C(F_1(x_1),F_2(x_2))$$

Using the probability integral transformation we want to convert the marginals to uniform r.v.s so that $u_i = F_i(x_i)$ or $x_i = F^{-1}(u_i)$ so we have,

$$F(x_1,x_2)=F(F^{-1}(u_1),F^{-1}(u_2))=F(F^{-1}(1-\sqrt{(1-u_1))},F^{-1}(\sqrt{(u_2)}))$$

And we have,

$$F(x_1,x_2)=\Pr[X_1\leq x_1, X_2\leq x_2]=2F(\min(x_1,x_2))F(x_2)-F(\min(x_1,x_2))^2$$

So putting these together we have,

$$C(u_1, u_2)=F(F_1^{-1}(u_1),F_2^{-1}(u_2))=2F(\min(F_1^{-1}(u_1),F_2^{-1}(u_2)))F(F_2^{-1}(u_2))-F(\min(F_1^{-1}(u_1),F_2^{-1}(u_2)))^2=2(\min(1-\sqrt{(1-u_1)},\sqrt{(u_2)}))(\sqrt{(u_2)})-(\min(1-\sqrt{(1-u_1)},\sqrt{(u_2)}))^2$$

Which is the expression found in the notes. The point of this is of course that by plugging in the inverse values we essentially have found a function that is completely known (since $F$ and $F^{-1}$ have dropped out) and fully describes the joint structure of our random variables.