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I wonder what the difference between multivariate standard normal distribution and Gaussian copula is since when I look at the density function they seem the same to me.

My issue is why the Gaussian copula is introduced or what benefit the Gaussian copula generates or what its superiority is when Gaussian copula is nothing but a multivariate standard normal function itself.

Also what is the concept behind probability integral transformation in copula? I mean we know that a copula is a function with uniform variable. Why does it have to be uniform? Why not use the actual data like multivariate normal distribution and find the correlation matrix? (Normally we plot the two asset returns to consider their relationships but when it is copula, we plot the Us which are probabilities instead.)

Another question. I also doubt whether the correlation matrix from MVN could be non-parametric or semi-parametric like those of copula (for copula parameter can be kendall's tau, etc.)

I would be very thankful for your help since I'm new in this area. (but I have read a lot of papers and these are the only things that I don't understand)

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  • $\begingroup$ How are you "looking at the density function"? You might not be using a method that is sensitive enough. For instance, the density is assuredly not multivariate normal when the marginals are non-normal! Try this out using a Gaussian copula with a multimodal distribution, such as a Beta$(1/2,1/2)$: that ought to look decidedly non-normal! $\endgroup$ – whuber Jun 17 '13 at 18:45
  • $\begingroup$ equation (6) is bivariate Gaussian copula CDF iopscience.iop.org/2041-8205/708/1/L9/fulltext/… while the first equation of description section is bivariate standard normal CDF roguewave.com/portals/0/products/imsl-numerical-libraries/… and when we compare them together, the functional form is very similar. well they are exactly the same to me. $\endgroup$ – user26979 Jun 17 '13 at 19:22
  • $\begingroup$ You're right: that's why you shouldn't rely on random Internet references, especially those with poorly defined terms and awful typesetting. Consult Nelson (one of the sources for your first link, and eminently readable). $\endgroup$ – whuber Jun 17 '13 at 22:39
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    $\begingroup$ so, if not to mention the sited above, what is the difference in your view? $\endgroup$ – user26979 Jun 18 '13 at 9:38
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One general rule about technical papers--especially those found on the Web--is that the reliability of any statistical or mathematical definition offered in them varies inversely with the number of unrelated non-statistical subjects mentioned in the paper's title. The page title in the first reference offered (in a comment to the question) is "From Finance to Cosmology: The Copula of Large-Scale Structure." With both "finance" and "cosmology" appearing prominently, we can be pretty sure that this is not a good source of information about copulas!

Let's instead turn to a standard and very accessible textbook, Roger Nelsen's An introduction to copulas (Second Edition, 2006), for the key definitions.

... every copula is a joint distribution function with margins that are uniform on [the closed unit interval $[0,1]]$.

[At p. 23, bottom.]

For some insight into copulae, turn to the first theorem in the book, Sklar's Theorem:

Let $H$ be a joint distribution function with margins $F$ and $G$. Then there exists a copula $C$ such that for all $x,y$ in [the extended real numbers], $$H(x,y) = C(F(x),G(y)).$$

[Stated on pp. 18 and 21.]

Although Nelsen does not call it as such, he does define the Gaussian copula in an example:

... if $\Phi$ denotes the standard (univariate) normal distribution function and $N_\rho$ denotes the standard bivariate normal distribution function (with Pearson's product-moment correlation coefficient $\rho$), then ... $$C(u,v) = \frac{1}{2\pi\sqrt{1-\rho^2}}\int_{-\infty}^{\Phi^{-1}(u)}\int_{-\infty}^{\Phi^{-1}(v)}\exp\left[\frac{-\left(s^2-2\rho s t + t^2\right)}{2\left(1-\rho^2\right)}\right]dsdt$$

[at p. 23, equation 2.3.6]. From the notation it is immediate that this $C$ indeed is the joint distribution for $(u,v)$ when $(\Phi^{-1}(u), \Phi^{-1}(v))$ is bivariate Normal. We may now turn around and construct a new bivariate distribution having any desired (continuous) marginal distributions $F$ and $G$ for which this $C$ is the copula, merely by replacing these occurrences of $\Phi$ by $F$ and $G$: take this particular $C$ in the characterization of copulas above.

So yes, this looks remarkably like the formulas for a bivariate normal distribution, because it is bivariate normal for the transformed variables $(\Phi^{-1}(F(x)),\Phi^{-1}(G(y)))$. Because these transformations will be nonlinear whenever $F$ and $G$ are not already (univariate) Normal CDFs themselves, the resulting distribution is not (in these cases) bivariate normal.


Example

Let $F$ be the distribution function for a Beta$(4,2)$ variable $X$ and $G$ the distribution function for a Gamma$(2)$ variable $Y$. By using the preceding construction we can form the joint distribution $H$ with a Gaussian copula and marginals $F$ and $G$. To depict this distribution, here is a partial plot of its bivariate density on $x$ and $y$ axes:

Plot

The dark areas have low probability density; the light regions have the highest density. All the probability has been squeezed into the region where $0\le x \le 1$ (the support of the Beta distribution) and $0 \le y$ (the support of the Gamma distribution).

The lack of symmetry makes it obviously non-normal (and without normal margins), but it nevertheless has a Gaussian copula by construction. FWIW it has a formula and it's ugly, also obviously not bivariate Normal:

$$\frac{1}{\sqrt{3}}2 \left(20 (1-x) x^3\right) \left(e^{-y} y\right) \exp \left(w(x,y)\right)$$

where $w(x,y)$ is given by $$\text{erfc}^{-1}\left(2 (Q(2,0,y))^2-\frac{2}{3} \left(\sqrt{2} \text{erfc}^{-1}(2 (Q(2,0,y)))-\frac{\text{erfc}^{-1}(2 (I_x(4,2)))}{\sqrt{2}}\right)^2\right).$$

($Q$ is a regularized Gamma function and $I_x$ is a regularized Beta function.)

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  • $\begingroup$ Thanks for the edit, @Cardinal: I'm embarrassed about misspelling Nelsen's name, especially when I was looking right at it on the front of the book! (In my defense, I had first noticed it in the bibliography of the OP's referenced paper, where it is also misspelled: that must have stuck with me. :-) $\endgroup$ – whuber Jun 18 '13 at 13:13
  • $\begingroup$ It was such a minor thing, I figured I'd just go ahead and make the edits. The spelling is unusual (at least in English!), especially compared to the more common variant. :-) $\endgroup$ – cardinal Jun 18 '13 at 23:48

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