# How to prove a multivariate r.v. does not follow the nonparanormal distribution?

## Background

You may find the definition of the non-paranormal distribution at the 2nd paragraph in p.2296 of this paper.

In short, $$(X_1, \ldots, X_p)$$ is non-paranormal if there exists a set of strictly increasing functions such that $$(f_1(X_1), \ldots, f_p(X_p))$$ is multivariate Gaussian.

# Question

I'm trying to think of an example of a multivariate random variable that is not non-paranormal. For example, how do I check if the multivariate t-distribution (with some d.f.) follows the non-paranormal distribution or not?

Any comments are very much welcome!

## 2 Answers

First, a discrete distribution (or a distribution with atoms) cannot be transformed such to a multinormal, so assume an absolutely continuous distribution. Then, if $$p=1$$, we can always transform to normal. So assume $$p>1$$. Then, if the components are independent, again, we can use the univariate solution. So there must be some dependencies between the components. Then, this is essentially the same question as Can any probability distribution be transformed into a normal distribution?, which however do not give a very concrete example.

One wonders why the authors of the paper you quoted choose the strange name non-paranormal. Their requirement is essentially that $$X_1, \dotsc,X_p$$ have a distribution with a Gaussian copula. So your question is really about giving examples of distributions with a non-gaussian copula (since copulas are unique.) That also makes it clear that there must be very many examples. That answers your question for the multivariate $$t$$, which defines a $$t$$-copula, not a Gaussian one. Some papers are this and this other.

One simple class of examples will be uniform distributions on polygons in the plane, with (at least some) edges not parallel to the $$x$$- or $$y$$-axes. To see why, use a diamond in the plane, a polygon with corners at $$(0,1),(1,0),(0,-1), (-1,0)$$. To have any chance at all of transforming to binormal, we must have that $$\lim_{x \to 1} f_i(x) =+\infty$$, and likewise $$\lim_{x \to -1} f_i(x) =-\infty$$, for if these limits where finite, we would get a finite range. But the image of the diamond cannot be all of the plane! so the distribution cannot be binormal. Note that here only the geometry plays a role, not the uniformity.

• Thanks for clarifying my question. Also, would you give me more details about the uniform distribution on polygons? Why can't it be transformed to the multivariate normal? – inmybrain Apr 1 '20 at 1:27
• Did my edits resolve your doubts? If so, you can accept&upvote. If not, please explain. – kjetil b halvorsen Apr 1 '20 at 16:05
• Thanks a lot for your additional explanation, and sorry about my late decision. – inmybrain Apr 26 '20 at 1:50

I think the following example could do it :

Let $$X_1 \sim \mathcal{N}(0, 1)$$ and $$B$$ be a binary variable which is equal to $$1$$ with probability $$0.5$$ and $$-1$$ with probability $$0.5$$. Define $$X_2 = B\times X_1$$.

Then the marginal distributions of $$X_1$$ and $$X_2$$ are obviously $$\mathcal{N}(0, 1)$$ but $$(X_1; X_2)$$ is not a multivariate Gaussian. An easy way to see it is that $$\mathrm{cov}(X_1, X_2) = 0$$ whereas $$X_1$$ and $$X_2$$ are not independent, which would be impossible if $$(X_1, X_2)$$ was a multivariate Gaussian.

The only increasing functions $$f_1$$, $$f_2$$ that with conserve $$X_1$$ and $$X_2$$ as a normal random variables are affine functions with positive coefficient, but they wont transform $$(X_1, X_2)$$ into a multivariate Gaussian (the argument stated before still applies). Thus there is no increasing functions $$f_1$$, $$f_2$$ such that $$(f_1(X_1), f_2(X_2))$$ is a multivariate Gaussian.

This is a bit of an artificial example, sorry for that. I would expect the multivariate $$t$$ distribution not to be non-paranormal, but derivation of that seems a bit tedious..

• Thanks for your example! That gives me insights. I have two questions. 1. Does your argument still valid with affine functions having non-zero y-intercepts? 2. Is it right that you used a fact "If $X$ and $f(X)$ are both normal, then $f$ should be affine?. Also, is this observation true? – inmybrain Apr 1 '20 at 1:22
• 1. Yes if $X_1$ and $X_2$ are defined as before, then $cov(a_1X_1 +b_1, a_2X_2 +b_2)= \mathbb{E}[a_1X_1a_2X_2]= a_1a_2\mathbb{E}[B]\mathbb{E}[X_1^2] =0$ (using independance of $B$ and $X_1$. – Pohoua Apr 1 '20 at 7:35
• 2. Yes I think this is the case for monotonous functions. If $X$ is normal and $f(X)$ also with f increasing and differentiable, then $f$ must be affine. – Pohoua Apr 1 '20 at 7:37
• Thanks for your answers! The first one is obvious, shame on me :\ For the second, what if $f$ is not differentiable? There is no such condition on $f$ when defining non-paranormal. – inmybrain Apr 1 '20 at 11:13
• In the end, no need for differentiability. If $X\sim \mathcal{N}(0, 1)$ and $f(X) \sim \mathcal{N}(\mu, \sigma)$, note $g(x) = (f(x) - \mu) / \sigma$ so $g(X)\sim\mathcal{N}(0, 1)$. One can prove that $g$ is continuous (otherwise there would be gaps in the domain of $f(X)$), so $g$ is bijective (stricly increasing + continuous). Then $P(g(X) < y) = \Phi(y)$ (by hypothesis) so $P(X < g^{-1}(y)) = \Phi(y)$ but as $X$ is std normal $P(X < g^{-1}(y)) = \Phi(g^{-1}(y))$ thus $\Phi(y) = \Phi(g^{-1}(y))$ so, applying $\Phi^{-1}$ to last equation, gives $g^{-1}(y) = y$ so $g(x) = x$. So $f$ is affine! – Pohoua Apr 1 '20 at 13:14