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*} \frac{\partial C(a,a)}{\partial u} + \frac{\partial C(a,a)}{\partial v} \geq \frac{C(a,a)}{a} \end{align*}

where $C$ is a copula and $0 < a \leq 1$. 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\\\\ \displaystyle \frac{\partial^{2}C}{\partial u\partial v} = \frac{\partial^{2} C}{\partial v\partial u} \end{cases} \end{align*}

This is all that I know for the moment. I tested the given property for $C(u,v) = \min\{u,v\}$, $C(u,v) = uv$ and $C(u,v) = \frac{uv}{1-(1-u)(1-v)}$ and it has worked well so far. Based on such considerations, could someone provide a partial or full answer to my question?

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  • $\begingroup$ I believe the definition of copula is well-known and it's certainly easy to look up. The angle brackets, though, could mean a lot of things: they deserve a definition. $\endgroup$ – whuber Oct 3 '19 at 15:35
  • $\begingroup$ The angle brackets mean the euclidean inner product. Could you please remove the "on hold"? $\endgroup$ – user1337 Oct 3 '19 at 17:03
  • $\begingroup$ After you include that information in your post, the community will be made aware of the change and will vote on reopening it. It would help to provide specific quantifiers for the scope of the variable $a.$ $\endgroup$ – whuber Oct 3 '19 at 17:23
  • $\begingroup$ How do you define the derivatives when the Copula is not differentiable? (E.g. the Copula for $\rho =1$ full correlation) $\endgroup$ – Sextus Empiricus Oct 4 '19 at 5:46
  • $\begingroup$ It is not the case that $\partial C/\partial u \le 1$ for all copulas: is this a special condition you are imposing, or could it just reflect a misunderstanding of the properties of copulas? $\endgroup$ – whuber Oct 4 '19 at 14:52

A bivariate copula is the distribution function $C$ of a random variable $(X,Y)$ on $[0,1]\times[0,1]$ where the marginal distributions of $X$ and $Y$ are both uniform (Sklar's Theorem).

That is, there exists a random variable $(X,Y)$ for which (for all $0\le u,v\le 1$)

$$C(u,v) = \Pr(X\le u,\, Y\le v) \tag{1}$$


$$\Pr(X\le u) = u,\ \Pr(Y\le v)=v.$$

The question compares a derivative of $C$ to the values of $C.$ The former is local information while the latter is global (in the sense that $C(u,v)$ accumulates all the probability density of smaller values of $X$ and $Y$ but the first derivatives provide the rates at which $C$ is changing at the point $(u,v)$). We should therefore expect no such inequality to hold generally.

A counterexample would be a case where $C(u,u)$ is large but its derivatives are small. In all copulas, $C(u,u)$ is small for small $u$ and many of them have small derivatives for small $u,$ too. One approach, then, is somehow to shift the smaller parts of a copula over to large values of $u,$ without changing its derivatives, because the accumulated probability at large $u$ will produce the counterexample.

This motivates the following construction of the "$a$-sum" $C\oplus_a D$ of two copulas $C$ and $D.$ For any $0\lt a\lt 1$, we will rescale $C$ to the rectangle $[0,a]^2$ and shift and rescale $D$ to the rectangle $[1-a,1]^2.$ Provided we extend the domain of definition of the copulas to include all real numbers (which formula $(1)$ automatically does), linear transformations easily accomplish this:

$$(C\oplus_a D)(u,v) = a C\left(\frac{u}{a}, \frac{v}{a}\right) + (1-a) D\left(\frac{u-a}{1-a}, \frac{v-a}{1-a}\right).$$

It is straightforward to check that this defines a copula and, if all first partial derivatives of $C$ and $D$ are less than $1$ (an additional condition given in the question), then all first partials of the $a$-sum are less than $1,$ too.

To illustrate, let $C(u,v)=D(u,v)=uv$ be the "independence copula" (representing independent uniform random variables). Here is a filled contour plot of $C\oplus_{0.78}D:$


(The vertical and horizontal lines at coordinate $a=0.78$ show where one of the first partial derivatives is undefined. The remaining black curves are the contours of $C.$)

To compare $C$ to the sum of its first partial derivatives, I have shaded the colors according to that sum, as shown in the legend, so that darker areas have smaller derivatives.

Note the region near $(a,a)=(0.78,0.78):$ above and to the right of that point, the first partial derivatives are reset to zero, as indicated by the dark shading. However, the value of $C$ in that region cannot be any less than $a,$ because a proportion $a$ of all the probability lies (by construction) in the square $[0,a]^2$ below and to the left.

In the next plot I have shown both the values of $C(u,u)$ (in red) and the sum of the first partial derivatives (in black).

Figure 2

Right at $a=0.78$ the sum of first partials drops from $2$ back to $0,$ but (of course) the value of $C$ cannot decrease. In the region from $0.78$ to approximately $0.87,$ the sum of first partials is less than $C(u,u),$ so a fortiori it is less than $C(u,u)/u.$ That's a counterexample.

Note that the partial derivatives in this counterexample are undefined at some points. Since the question explicitly entertains such possibilities as $C(u,v)=\min(u,v),$ which have undefined first partials, I have understood that not to be a concern.

An interesting counterexample is afforded by $C=W\oplus_{1/2}W$ where $W$ is the Fréchet–Hoeffding minimum copula

$$W(u,v) = \max(0, u+v-1).$$

$W$ is the distribution function of the perfectly anticorrelated uniform random variable $(X,1-X).$ There is an entire triangular area where $C$ has attained a value of $1/2$ but is flat (both partial derivatives are zero):


This example is noteworthy because the corresponding random variable $(X,Y)$ is positively correlated: the Pearson correlation coefficient is $1/2.$


Roger B. Nelsen (2005), An Introduction to Copulas. Second Edition, Springer.

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positive correlation copulas

I guess that most copulas with sufficient positive correlation should work

Geometric intuition

These large correlation distributions look like a sort of mountain where the line along $a,a $ is a sort of rib. There is a steep rise to $a,a $ after which it is sort of flat.

You can also see it as: you do not increase much probability when moving either $u$ or $v$ because most mass is on the axis. When the correlation is highly positive then $$P (u \le a,v \le a) \approx P (u \le a) = P (u \le a, v \le 1)$$ So once you reach $a,a$ there is not much increase (small slope) from $P(u \le a, v\le a)$ to $P (u \le a, v\le 1)$

See this image from Wikimedia


The image on the right, fully correlated, has even zero derivatives in direction $u$ and $v$ (with a bit of imagination and ignoring that the derivative is not well defined there).

For instance, take the Ali-Mikhail-Haq copula

$$C (u,v) = \frac {uv}{1-\theta (1-u)(1-v)} $$

with derivatives

$$C (u,v)_u = \frac{\theta(v-1)v+v}{(1-\theta (1-u)(1-v))^2}$$

$$C (u,v)_v = \frac{\theta(u-1)u+u}{(1-\theta (1-u)(1-v))^2}$$

and say we take the point $a=u=v=0.5.$


$$C (0.5,0.5)_u = \frac {C (0.5,0.5)}{0.5} \times \left (0.5 \frac {2+\theta}{1-0.25 \theta} \right)$$

and when $\theta <-0.8$ then you have that $C (0.5,0.5)_u +C (0.5,0.5)_v > \frac {C (0.5,0.5)}{0.5}.$

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