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2020 11.26: I finished the derivation but haven't posted it here. I leave this open so that those willing to try their own version can if they want

Question

How can we derive a closed-form analytical solution for $\mathbb{E}\left[ -\log c(u,v)\right] $?

$c(u,v)$ is the bivariate copula density for Student's $t$-copula, whose degrees of freedom is $\nu$ and dependence parameter is $\rho\in (-1,1)$:

\begin{aligned} c(u,v) &= \frac{1}{2\pi \sqrt{1-\rho^2}} \frac{1}{dt(x_1; \nu) dt(x_2; \nu)} \Bigg(1+\frac{x_1^2 + x_2^2 -2\rho x_1 x_2}{\nu (1-\rho^2)} \Bigg)^{-\frac{\nu +2}{2}} \end{aligned}

where $$dt(x_i; \nu) = \frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2}) \sqrt{\pi\nu}} \Bigg( 1+\frac{x_i^2}{\nu} \Bigg)^{-\frac{\nu+1}{2} } \enspace, i=1,2 $$

or equivalently, $$\displaystyle dt(x_i;\nu) = \frac{1}{\sqrt{\nu}B \left(\frac{1}{2}, \frac{\nu}{2} \right)} \Bigg( 1+\frac{x_i^2}{\nu} \Bigg)^{-\frac{\nu+1}{2} }, i=1,2 $$ where $\Gamma(\cdot)$ is the gamma function, and $B(\cdot)$ is the beta function.

First Attempt

\begin{align} h(c(u,v)) &= -\int_{[0,1]^2} c(u,v) \ln c(u,v) \hspace{1mm} du \hspace{1mm} dv \\ &= \mathbb{E}\left[ -\log c(u,v)\right] \\ &= \mathbb{E}\left[ -\log \frac{1}{2\pi \sqrt{1-\rho^2}} \frac{1}{dt(x_1; \nu) dt(x_2; \nu)} \left(1+\frac{x_1^2 + x_2^2 -2\rho x_1 x_2}{\nu (1-\rho^2)} \right)^{-\frac{\nu +2}{2}}\right] \\ &= -\mathbb{E}\left[\log \frac{1}{2\pi \sqrt{1-\rho^2}} + \log \frac{1}{dt(x_1; \nu) dt(x_2; \nu)} +\log \left(1+\frac{x_1^2 + x_2^2 -2\rho x_1 x_2}{\nu (1-\rho^2)} \right)^{-\frac{\nu +2}{2}}\right] \\ &= -\mathbb{E}\left[-\log \left(2\pi \sqrt{1-\rho^2}\right) + \log \frac{1}{dt(x_1; \nu) dt(x_2; \nu)} -\left(\frac{\nu +2}{2}\right) \log \left(1+\frac{x_1^2 + x_2^2 -2\rho x_1 x_2}{\nu (1-\rho^2)} \right)\right] \\ &= \log \left(2\pi \sqrt{1-\rho^2}\right) -\mathbb{E}\left[\log \frac{1}{dt(x_1; \nu) dt(x_2; \nu)} -\left(\frac{\nu +2}{2}\right) \log \left(1+\frac{x_1^2 + x_2^2 -2\rho x_1 x_2}{\nu (1-\rho^2)} \right)\right] \\ & = \dots ? \end{align}

Hint

Why do I think an analytical solution of copula entropy can be found? Because there is one for the entropy of the Normal distribution's pdf, which I copy and paste now from its derivation here: \begin{align}\label{equation:hN} h(X_\mathrm{Gauss}) = &-\int_{-\infty}^{\infty} \left(2\pi \sigma^2\right)^{-\frac{1}{2}} e^{-(x-\mu)^2 / 2\sigma^2} \ln\left[ \left(2\pi \sigma^2\right)^{-\frac{1}{2}} e^{-(x-\mu)^2 / 2\sigma^2} \right] \mathrm{d} x\\ = &\frac{1}{2} \ln(2\pi \sigma^2) \int_{-\infty}^{\infty} (2\pi \sigma^2)^{-\frac{1}{2}} e^{-(x-\mu)^2 / 2\sigma^2} \mathrm{d} x\\ &+ \frac{1}{2\sigma^2} \left(2\pi \sigma^2\right)^{-\frac{1}{2}} (x-\mu)^2 e^{-(x-\mu)^2 / 2\sigma^2} \mathrm{d} x\\ =& \frac{1}{2} \ln (2\pi\sigma^2) + \frac{1}{2}\\ =& \frac12\ln(2\pi\sigma^2) + \frac12\ln e\\ =& \frac12\left(\ln(2\pi\sigma^2) + \ln e\right)\\ =& \frac{1}{2} \ln (2\pi e \sigma^2) \end{align}

Link to someone's attempt at the Clayton copula entropy

develarist
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  • I guess $x_{1}=u$ and $x_{2}=v$? – Tobsn Nov 06 '20 at 15:13
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    So $x=(x_1,x_2)\in[0,1]^2=D$, and the question wants to find "in full generality", i.e. for $\nu$ general (i suppose among $0,1,2,3,4,\dots$) and for $\rho\in(-1,1)$ general, an explicit formula for$$\iint_D f(x);\ln f(x); dx\ ,$$where $f(x)$ is a product of four factors. One factor is a constant, and the other three are:$$(1+ax_1^2)^p\ ,\ (1+ax_2^2)^p\ ,\ (1+b(x_1^2+x_2^2-2\rho x_1x_2))^q\ ,\ p=(\nu+1)/2\ ,\ q=-(\nu+2)/2\ .$$This is a completely different situation compared to the linked normal distribution case, where $f(x) = \exp g(x)$ with a nice $g$, there's no log inside the integral. – dan_fulea Nov 06 '20 at 16:37
  • The $\log f(x)$ splits then as a sum of four log factors, but there is no easy way to solve the issue. Just take a look at integrals of "somehow similar" (but else quite different) shape: https://math.stackexchange.com/questions/2953815/evaluating-int-01-ln1x2-lnx2x3-fracdx1x2 - Here, there appear logarithms of "simple polynomials" , also just one logarithmic factor, the other factor is a "simple" rational function, and in the denominators we also have "one simple piece" like $(1+x^2)$ (with no parameter). For a very special value of $\nu$ (or $\rho$) i would look closer. But there is in my... – dan_fulea Nov 06 '20 at 16:47
  • ... eyes no chance for a formula in the given generality. This is a quick look. If i would have a reason, at least, i would maybe try to force some first step... – dan_fulea Nov 06 '20 at 16:49
  • @Tobsn No, they are the inverse of the standard univariate Gaussian distribution functions: $x_1 = \phi^{-1} (u)$ and $x_2 = \phi^{-1} (v)$ – develarist Nov 25 '20 at 07:35
  • @dan_fulea I made a first attempt after receiving a hint from someone that the integration is equivalent to $\mathbb{E}\left[ -\log c(u,v)\right]$. what do you think of the edit? the log of the middle factor, containing the gamma/beta function, is tough – develarist Nov 25 '20 at 07:44

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