# Upper bounds for the copula density?

The Fréchet–Hoeffding upper bound applies to the copula distribution function and it is given by

$$C(u_1,...,u_d)\leq \min\{u_1,..,u_d\}.$$

Is there a similar (in the sense that it depends on the marginal densities) upper bound for the copula density $c(u_1,...,u_d)$ instead of the CDF?

Any reference would be greatly appreciated.

• What kind of a bound are you looking for? A description of your actual problem might help. Technically, the answer is "no" in two different ways: (i) there may not be a density(!) and (b) if there were, we could change it on a set of measure zero to be as large as we'd like. We do know something, though. In particular, suppose $c$ exists and let $R = [a_1,b_1] \times \dots \times [a_n,b_n] \subset \mathbb [0,1]^d$ be any (hyper)rectangle with side lengths $w_i = b_i - a_i$. Then, certainly $$\mathrm{ess\,inf}_{x \in R} \,c(x) \leq (\min_i w_i) / \prod_i w_i \>.$$ Nov 27, 2012 at 17:57
• Since you can easily construct examples that satisfy this bound, I suspect there is not too terribly much more that can be said. But, I haven't thought about that carefully. Nov 27, 2012 at 17:58
• @cardinal Thank you for your comments. Indeed, I am assuming that the density exists in order to avoid the trivial case. I was looking for an upper bound in terms of the marginal densities. I am particularly interested on the Gaussian copula. Nov 27, 2012 at 18:03
• If it's a copula, all the marginal densities are uniform, i.e., a constant function. :) Nov 27, 2012 at 18:09
• @cardinal Pardon my French. Let me rephrased my question. The Gaussian copula (which I am particularly interested in) is given by $s(x_1,...,x_d;R)=\dfrac{1}{\mbox{det}(R)^{1/2}}\exp(-0.5 u^T(R^{-1}-I)u) \prod_{j=1}^d f_j(x_j)$. Where $u=(u_1,...,u_d)$ and $u_j=\Phi^{-1}(F_j(x_j))$. This, for instance, cannot be bounded by the product $\prod_{j=1}^n f_j(x_j)$. So, I was looking for another upper bound that involves only the marginals. And, of course, I was trying to ask the question in a more general manner, relating it with the aforementioned bounds. Apologies for my vague words. Nov 27, 2012 at 18:17

• @whuber The density of a gaussian copula isn't a standard gaussian. If you look at coppola's comment above, you'll notice the gaussian copula density has an $$R^{-1}-I$$ where you'd expect just the inverse covariance matrix. The inverse covariance matrix should be symmetric positive semi definite, but the -I allows for non positive-definiteness, and therefore a saddle point. It's presence is due to the change of variables when converting from the $$\mathbb{R}^n$$ to $$[0,1]^n$$ May 5, 2017 at 17:46
• Yes, I am aware of that--but that's not what your answer implies. This copula is parameterized by the correlation matrix $R$, but for any such $R$ it's a function of the $x_i$ only. As such it doesn't ever "explode to infinity". There are no valid correlation matrices $R$ (that is, nondegenerate ones) for which this copula is unbounded. Those are the reasons I was requesting a clarification of your answer.