# Gaussian Bivariate Copulas Inconsistent Reasoning

While studying Gaussian copulas, I have stumbled accross a question which seems to result from wrong reasoning. In the arguments below, where have I gone wrong?

Let $c(u, v)$ denote the density of the bivariate Gaussian copula:

$$c(u, v) = \exp(\alpha + \beta \Phi^{-1}(u) \Phi^{-1}(v) + \gamma (\Phi^{-1}(u))^2 + \gamma (\Phi^{-1}(v))^2)$$

where $\Phi^{-1}(\cdot)$ denotes the inverse of the cumulative Gaussian distribution and $\alpha, \beta, \gamma$ are chosen according to some correlation matrix $R$.

Since a copula has uniform mariginals on $[0, 1]$, we require

$$\int_0^1 c(u, v) \, du = 1$$

this can be equivalently stated as

$$\exp(-\alpha - \gamma (\Phi^{-1}(v))^2) = \int_0^1 \exp(\beta \Phi^{-1}(u) \Phi^{-1}(v) + \gamma (\Phi^{-1}(u))^2) \, du \tag{1}$$

To keep notation compact, we now let $f = \Phi^{-1}$ and $g = f^2$. The above equation implies equivalence of the derivatives with respect to $v$. The derivative of the left-hand side is easily evaluated as

$$- \gamma g'(v) \exp(-\alpha - \gamma g(v)) \tag{2}$$

The derivative of the right-hand side is

$$\int_0^1 \beta f(u) f'(v) \exp(\beta f(u) f(v) + \gamma g(u)) \, du \tag{3}$$

Equating (2) and (3) and plugging in (1) $\to$ (2) finally yields:

$$\int_0^1 \left[ \gamma g'(v) + \beta f(u) f'(v) \right] \exp(\beta f(u) f(v) + \gamma g(u)) \, du= 0$$

Since $\forall x : \exp(x) \geq 0$ we subsequently require

$$\gamma g'(v) + \beta f(u) f'(v) = 0$$

which cannot be satisfied.

• Please add the [self-study] tag & read its wiki. – gung Oct 3 '16 at 12:24
• Thanks & done. I might add that this is a question not related to a course nor any specific book, script, e.t.c. – R.G. Oct 3 '16 at 12:32
• Can you explain why $\int_0^1 c(u,v) du = 1$ ? The double integral of the copula density is the copula function, why does integrating over one variable yield a number? – Kiran K. Oct 4 '16 at 12:10
• Of course. The marginals of a copula are uniform distributions on $[0, 1]$, hence their marginal density is 1. – R.G. Oct 4 '16 at 12:18