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:

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

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

\begin{equation} \int_0^1 c(u, v) \, du = 1 \end{equation}

this can be equivalently stated as

\begin{equation} \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} \end{equation}

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

\begin{equation} - \gamma g'(v) \exp(-\alpha - \gamma g(v)) \tag{2} \end{equation}

The derivative of the right-hand side is

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

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

\begin{equation} \int_0^1 \left[ \gamma g'(v) + \beta f(u) f'(v) \right] \exp(\beta f(u) f(v) + \gamma g(u)) \, du= 0 \end{equation}

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

\begin{equation} \gamma g'(v) + \beta f(u) f'(v) = 0 \end{equation}

which cannot be satisfied.

  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ – gung Oct 3 '16 at 12:24
  • $\begingroup$ Thanks & done. I might add that this is a question not related to a course nor any specific book, script, e.t.c. $\endgroup$ – R.G. Oct 3 '16 at 12:32
  • $\begingroup$ 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? $\endgroup$ – Kiran K. Oct 4 '16 at 12:10
  • $\begingroup$ Of course. The marginals of a copula are uniform distributions on $[0, 1]$, hence their marginal density is 1. $\endgroup$ – R.G. Oct 4 '16 at 12:18

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