# Generating an analytical copula for an example problem

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!

• There is a typographical error in your formula $(2).$ Perhaps that's the source of your difficulty?
– whuber
May 19 at 13:31
• That was a mistranslation. I did not have the error when I was writing it down. Thank you for pointing it out! I have corrected it here. Any thoughts on other potential issues/approaches I should consider. May 20 at 14:12

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 $$x_1 and the second and third equality follows from $$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 $$x_1 by our assumption and so the probability does not change. Remember that if $$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.

• Thank you for this detailed answer! I am will got through it in more detail and post, but overall it looks really good and has been very enlightening. May 31 at 17:42