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This is a follow-up question to A question about copulas and directional derivatives. Since no answer was given, I am going to precise the definition of copula. I am interested in proving (or disproving) that \begin{align*} \langle\nabla C(a,a),(a,a)\rangle \geq C(a,a) \end{align*}

where $0\leq a \leq 1$ and $C$ is a bivariate copula. A bivariate copula is a function $C:[0,1]^{2}\rightarrow[0,1]$ such that \begin{align*} \begin{cases} C(1,t) = C(t,1) = t\\\\ C(0,t) = C(t,0) = 0\\\\ \displaystyle\frac{\partial^{2}C}{\partial u\partial v} = \frac{\partial^{2} C}{\partial v\partial u} \geq0 \end{cases} \end{align*}

It also satisfies the properties \begin{align*} \begin{cases} \max\{u+v-1,0\} \leq C(u,v) \leq \min\{u,v\}\\\\ \displaystyle 0 \leq \frac{\partial C}{\partial u} \leq 1\\\\ \displaystyle 0 \leq \frac{\partial C}{\partial v} \leq 1 \end{cases} \end{align*}

This is all that I know for the moment. I tested the given property for $C(u,v) = \min\{u,v\}$, $C(u,v) = uv$, $C(u,v) = \frac{uv}{1-(1-u)(1-v)}$ and $C(u,v) = u + v - 1 + \theta(1-u)(1-v)$ and it has worked quite well so far. Based on such considerations, could someone provide a partial or full answer to my question? It is worthy emphasizing that it is not an exercise. It makes part of my research on the theory of copulas.

EDIT

Sorry gentlemen, but I forgot to mention that $C$ must be symmetric.

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  • $\begingroup$ Isn't $\max\{u+v-1,0\}$ that you have mentioned already a counter-example? $\endgroup$ – Mateusz Kwaśnicki Oct 4 '19 at 6:49
  • $\begingroup$ @MateuszKwaśnicki: how so? where $C$ doesn't vanish $\nabla C = (1,1)$ so the inner product is $2 a$. There $C(a,a) = 2a - 1$. $\endgroup$ – Willie Wong Oct 4 '19 at 13:13
  • $\begingroup$ Quick note: $u+v - 1 + (1-u)(1-v) = uv$. $\endgroup$ – Willie Wong Oct 7 '19 at 14:03
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Consider $C(u,v)$ be such that

  • On the subset $\{ u \leq 1/2\} \cup \{v \leq 1/2\}$, we set $C(u,v) = \min(u,v)$.
  • On the subset $\{ u,v > \frac12\}$, set $ C(u,v) = \max(u+v - 1, \frac12)$

(One can actually summarize this to be $C(u,v) = \max(u+v - 1, \min(\frac12, u, v))$. )

One easily checks that

  • $C(0,v) = C(u,0) = 0$
  • $C(1,v) = v$, $C(u,1) = u$.
  • $C$ is continuous across the lines $u = 1/2$ and $v = 1/2$.
  • Fixed $u,v$, the functions $s\mapsto C(u,s)$ and $s\mapsto C(s,v)$ are both monotone (but not strict).
  • Obviously the comparison $\max(u+v-1,0) \leq C(u,v) \leq \min(u,v)$ holds.

The derivatives of $C$ satisfy

  • On the subset $\{u < \min(v,1/2)\}$, we have $C(u) = u$ so $C_u = 1$ and $C_v = 0$;
  • similarly on $\{v < \min(u,1/2)\}$, we have $C_v = 1$ and $C_u = 0$.
  • On the subset $\{u,v > 1/2, u + v < 3/2\}$, we have that $C(u,v) = 1/2$ so that $C_u = C_v = 0$.
  • Finally, on the subset that $\{u,v > 1/2, u+v > 3/2\}$ we have that $C(u,v) = u+v - 1$ so that $C_u = C_v = 1$.

Diagramming simply one see that $C_u$ is increasing in $v$ and $C_v$ is increasing in $u$. So except for the non-differentiability along the segments $$ \{(s,s) : s \in [0,1/2]\}, \{ (s,1/2) : s\in [1/2,1]\}, \{ (1/2,s) : s\in [1/2,1]\}, \{ (s,3/2-s): s \in [1/2,1] \} $$ $C$ verifies all the conditions of your copula. Along those points of non-differentiability the differential inequalities you quote are also satisfied in a weak sense so that you can smooth the function slightly to get differentiability still satisfied.

But obviously we have that at $a = 5/8$, the directional derivative $$ \langle \nabla C(a,a), (a,a) \rangle = 0 \not\geq \frac12 = C(a,a)$$

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  • $\begingroup$ Very nice! It may help visualize your idea to note that your copula is the (joint) cdf of the uniform distribution on the union of straight line segments $AB$ and $CD$, where $A:=(0,0),B:=(1/2,1/2),C:=(1,1/2),D:=(1/2,1)$. $\endgroup$ – Iosif Pinelis Oct 4 '19 at 15:01
  • $\begingroup$ I forgot to mention that $C$ has to be uniformly continuous. Does it affect the proof? $\endgroup$ – user1337 Oct 4 '19 at 15:31
  • $\begingroup$ @user1337: this $C$ is piecewise linear. So no. Not even after smoothing. $\endgroup$ – Willie Wong Oct 4 '19 at 15:31
  • $\begingroup$ @IosifPinelis: I will have to take your word for it (that what I wrote is that cdf.) $\endgroup$ – Willie Wong Oct 4 '19 at 15:32
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    $\begingroup$ @user1337 : You asked: "could you suggest additional restrictions on $C$ which make the original inequality true?" -- There is nothing in the definition of a copula that may have to do with partial derivatives of the copula. Therefore, your conjecture appeared hopeless to begin with. I think there hardly are additional conditions to imply your conjecture other than conditions very similar (if not identical) to the conjecture itself. $\endgroup$ – Iosif Pinelis Oct 4 '19 at 18:51
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No, the inequality $$ \langle\nabla C(a,a),(a,a)\rangle\ge C(a,a) \quad\forall a\in[0,1] $$ does not hold in general.

Indeed, a copula is just the (joint) cdf of a bivariate distribution on the unit square $[0,1]^2$ with the uniform marginals. That is, a map $C\colon[0,1]^2\to\mathbb R$ is a copula iff $C(x,y)=\mu([0,x]\times[0,y])$ for all $(x,y)\in[0,1]^2$, where $\mu$ is any probability measure on $[0,1]^2$ such that $\mu([0,x]\times[0,1])=\mu([0,1]\times[0,x])=x$ for all $x\in[0,1]$.

Let now $\mu$ be the uniform distribution on the intersection of the union of the straight lines $x+y=1/2$ and $x+y=3/2$ with the square $[0,1]^2$. This intersection is shown here:

enter image description here

Let then $C$ be the corresponding copula.

$\Big($One might note at this point that for this copula $C$ we have $$C(x,y)=\max \left(0,\min \left(\tfrac{1}{2},x\right)+\min \left(\tfrac{1}{2},y\right)-\tfrac{1}{2}\right)+\max \left(0,\max \left(\tfrac{1}{2},x\right)+\max \left(\tfrac{1}{2},y\right)-\tfrac{3}{2}\right) \tag{1} $$ for $x$ and $y$ in $[0,1]$; this is straightforward but a bit tedious -- and, in fact, unnecessary -- to establish.$\Big)$

Directly from the above picture or, alternatively, from (1), we can see that for any $a\in(1/2,3/4)$ we have $C(a,a)=1/2$ and hence $$ \langle\nabla C(a,a),(a,a)\rangle=0\not\ge1/2= C(a,a). $$

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  • $\begingroup$ @WillieWong : I have now given a completely different answer, inspired by a geometric interpretation of your answer. $\endgroup$ – Iosif Pinelis Oct 4 '19 at 16:49
  • $\begingroup$ Oh Dang! This is where knowing the probability theory behind the term actually is helpful (and in fact make the answer obvious). (I just looked it up on Wikipedia now.) Great answer and thank you for helping me learn something new today! $\endgroup$ – Willie Wong Oct 4 '19 at 17:38
  • $\begingroup$ @WillieWong : Thank you for your kind comment -- and for your answer, which inspired the above answer. I have now added a link to the Copula article on Wikipedia. $\endgroup$ – Iosif Pinelis Oct 4 '19 at 18:53

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