Copulas, implication

Question

Let $C$ be a copula function. Prove that $C(t,1-t)=0$ for all $t\in[0,1]$ implies that $C(u,v)=\max(u+v-1,0)$.

I think the implication other way around is easy to see, however I can't see why the "upper diagonal" part of the copula function could not be some type of a different function with $C(u,1)=u$ and $C(1,v)=v$.

See the image below - the leftmost plot is the Frechet-Hoeffding lower bound. I need to prove that $C$ is equal to that.

Answer 1

the condition on C implies $U \ge 1-V$. Since $U$ and $1-V$ are both uniform equality must hold, and $U = 1-V$. See also frechet-hoeffding lower bound.

Answer 2

You're saying the probability of being below the diagonal $u+v=1$ is zero.

Now consider a point $(u,v)$ exactly on the diagonal, so that $u+v=1$. What probability is assigned to the closed triangle with vertices $(u,v)$, $(1,v)$, and $(1,0)$? By the definition of copula conjoined with the fact that the probability below the diagonal is $0$, it must be $1-u$. And the probability assigned to that triangle must also be equal to the length of its projection onto the $v$-axis, and that length is also $1-u$. But the probability of falling within that part of the $u$-axis is equal to the probability assigned to the closed trapezoid with vertices $(u,v)=(u,1-u)$, $(1,0$), $(u,1)$, and $(1,1)$. That means the probability assigned to the rectangle with vertices $(u,v)=(u,1-u)$, $(1,v)$, $(u,1)$ and $(1,1)$ must be zero! This holds for every value of $u$, thus for all such rectangles.

Conclusion: The probability of being above the diagonal must be $0$.

That means all the probability is concentrated on the diagonal $u+v=1$. And there's only one distribution on the diagonal that gives the right marginal distributions to satisfy the definition of a copula.

The way I thought of this was by thinking first of the discrete case, the set of pairs $(u,v)$ of integers in the set $\{1,\ldots,n\}$. Just do the $2\times2$ and $3\times3$ cases and you see quickly why all the probability must be concentrated on that one diagonal, and must be uniformly distributed there.