# Existence of copula bound pointwise strictly smaller than the Fréchet-Hoeffding upper bound

Consider bivariate copulas $$C_1$$ and $$C_2$$ with $$\max\{C_1(u,v), C_2(u,v)\}< M_2(u,v)$$ for all $$u,v \in(0,1)$$, where $$M_2(u,v) := \min\{u,v\}$$ is the Fréchet-Hoeffding upper bound. Is there a copula $$D$$ with $$\max\{C_1(u,v), C_2(u,v)\}\leq D(u,v) < M_2(u,v)$$ for all $$u,v \in(0,1)$$?

The problem here is that $$\max\{C_1(u,v), C_2(u,v)\}$$ is in general only a quasi-copula and I need a bound strictly below $$M_2$$. I think such a bound should exist in full generality but I don't have a reference/proof for it.

You can prove more. Let $$F(u,v)$$ be any $$1$$-Lipschitz function on $$[0,1]^2$$ such that $$F(u,v)<\min(u,v)$$ inside the square. Then there exists a copula $$D(u,v)$$ such that $$F(u,v)\le D(u,v)<\min(u,v)$$ everywhere inside the square.

The second inequality will be immediate if we just construct the corresponding joint distribution measure $$\mu$$ on $$[0,1]^2$$ with full support. Indeed, then $$D(u,v)=\mu([0,u]\times [0,v])=\mu([0,u]\times [0,1])-\mu([0,u]\times(u,1]) \\ =u-\mu([0,u]\times(u,1]) and similarly for $$v$$.

It remains to make the following two observations.

Observation 1: There is a copula $$D_0(u,v)>F(u,v)$$ inside the unit square such that the associated $$\mu$$ has density $$p_0(u,v)\ge q_0>0$$ separated away from $$0$$ in some open neighborhood $$\Omega$$ of $$\{(x,x):0. To see it, just take a sufficiently fine countable partition of $$(0,1)$$ into disjoint intervals $$I_j$$, consider the density $$p=\sum_j|I_j|^{-1}\chi_{I_j\times I_j}$$ and mix two such distributions to take care of the corners. "Sufficiently fine" just means that the length of each $$I_j$$ is much less than $$u-F(u,u)$$ for $$u\in I_j$$. Now take any sequence $$q_0>q_1>\dots \ge q_0/2$$ and

Observation 2: Suppose we have a copula $$D_n$$ for which $$D_n(u,v)>F(u,v)$$ inside the unit square and the density $$p_n$$ of $$\mu_n$$ is at least $$q_n$$ in $$\Omega$$. Let $$(u_n,v_n)\in(0,1)^2$$ be any point outside the diagonal. Then, for every $$\delta_n\ge 0$$, there exists a copula $$D_{n+1}$$ with $$D_{n+1}>F$$ inside the unit square, such that $$p_{n+1}\ge p_n$$ outside $$\Omega$$, $$p_{n+1}\ge q_{n+1}$$ in $$\Omega$$, and $$(u_n,v_n)$$ is in the support of $$\mu_{n+1}$$. To do it, just choose two very short intervals $$U,V$$ of equal length containing $$u_n$$ and $$v_n$$ respectively and set $$p_{n+1}=p_n+t(-\chi_{U\times U}-\chi_{V\times V}+\chi_{V\times U}+\chi_{U\times V})$$ with very small $$t$$. Notice that this changes $$D_n(u,v)$$ only on a compact subset of the open unit square (say, $$I\times I$$ where $$I$$ is a closed interval such that $$U\cup V\subset I\subset (0,1)$$) and there the difference $$D_n-F$$ is separated from $$0$$, so this surgery leaves it positive for small $$t>0$$.

Now just take any sequence $$(u_n,v_n)\in (0,1)^2\setminus{\rm diag}$$ dense in the square and run this recursion. Then take either the weak limit of $$\mu_n$$, or the $$L^1$$-limit of $$p_n$$, whichever you are more comfortable with, to get $$D$$.

• This is very nice. I have fixed some typos here. Also, to have $D(u,v)<\min(u,v)$ for all $(u,v)\in(0,1)^2$ for a copula $D$, it is already sufficient (and also necessary) that the support of the distribution corresponding to $D$ contain the points $(1.0)$ and $(0,1)$ (rather than the entire unit square); I am not sure, though, if this observation can simplify the proof; apparently not. Jan 17 at 16:13
• Wow, this is extremely nice - thank you very much! Jan 18 at 13:14