## when perfect correlation just means… perfect!

Posted in Statistics with tags , , on February 6, 2018 by xi'an

When looking at an X validated question on generating two perfectly negatively correlated Bernoulli variates last week, my intuition was that one had to be the opposite of the other, which means their parameters had to sum up to one. Intuition that was plain easy to back up by solving the equation

corr(C¹,C²)=-1

in terms of the joint distribution of (C¹,C²). That perfect correlation implies strong constraints on the parameter of the Bernoulli is not highly surprising given its binary support. Although I had no time to pursue the issue, I idly wondered at the generalisation to, say, a Binomial case, i.e., whether or not this case still is the only possible one for the above to hold. But again a perfect correlation can only occur with perfect prediction, i.e., when the Binomial variates have the same number of trials and complementary probability parameters. (Of no particular relevance is the fact that the originator of the question preferred an answer that showed how to simulate two Bernoulli such that C¹+C²=1!)

## correlation matrices on copulas

Posted in R, Statistics, University life with tags , , , , on July 4, 2016 by xi'an  Following my post of yesterday about the missing condition in Lynch’s R code, Gérard Letac sent me a paper he recently wrote with Luc Devroye on correlation matrices and copulas. Paper written for the memorial volume in honour of Marc Yor. It considers the neat problem of the existence of a copula (on [0,1]x…x[0,1]) associated with a given correlation matrix R. And establishes this existence up to dimension n=9. The proof is based on the consideration of the extreme points of the set of correlation matrices. The authors conjecture the existence of (10,10) correlation matrices that cannot be a correlation matrix for a copula. The paper also contains a result that answers my (idle) puzzling of many years, namely on how to set the correlation matrix of a Gaussian copula to achieve a given correlation matrix R for the copula. More precisely, the paper links the [correlation] matrix R of X~N(0,R) with the [correlation] matrix R⁰ of Φ(X) by $r^0_{ij}=\frac{6}{\pi}\arcsin\{r_{ij}/2\}$

A side consequence of this result is that there exist correlation matrices of copulas that cannot be associated with Gaussian copulas. Like $R=\left[\begin{matrix} 1 &-1/2 &-1/2\\-1/2 &1 &-1/2\\-1/2 &-1/2 &1 \end{matrix}\right]$

## another wrong entry

Posted in Books, Kids, R, Statistics, University life with tags , , , , , , on June 27, 2016 by xi'an  Quite a coincidence! I just came across another bug in Lynch’s (2007) book, Introduction to Applied Bayesian Statistics and Estimation for Social Scientists. Already discussed here and on X validated. While working with one participant to the post-ISBA softshop, we were looking for efficient approaches to simulating correlation matrices and came [by Google] across the above R code associated with a 3×3 correlation matrix, which misses the additional constraint that the determinant must be positive. As shown e.g. by the example

> eigen(matrix(c(1,-.8,.7,-.8,1,.6,.7,.6,1),ncol=3))
\$values
 1.8169834 1.5861960 -0.4031794


having all correlations between -1 and 1 is not enough. Just. Not. Enough.

## Gauss to Laplace transmutation interpreted

Posted in Books, Kids, Statistics, University life with tags , , , , , , on November 9, 2015 by xi'an  Following my earlier post [induced by browsing X validated], on the strange property that the product of a Normal variate by an Exponential variate is a Laplace variate, I got contacted by Peng Ding from UC Berkeley, who showed me how to derive the result by a mere algebraic transform, related with the decomposition

(X+Y)(X-Y)=X²-Y² ~ 2XY

when X,Y are iid Normal N(0,1). Peng Ding and Joseph Blitzstein have now arXived a note detailing this derivation, along with another derivation using the moment generating function. As a coincidence, I also came across another interesting representation on X validated, namely that, when X and Y are Normal N(0,1) variates with correlation ρ,

XY ~ R(cos(πU)+ρ)

with R Exponential and U Uniform (0,1). As shown by the OP of that question, it is a direct consequence of the decomposition of (X+Y)(X-Y) and of the polar or Box-Muller representation. This does not lead to a standard distribution of course, but remains a nice representation of the product of two Normals.

## simulating correlated random variables [cont’ed]

Posted in Books, Kids, Statistics with tags , , , , on May 28, 2015 by xi'an Following a recent post on the topic, and comments ‘Og’s readers kindly provided on that post, the picture is not as clear as I wished it was… Indeed, on the one hand, non-parametric measures of correlation based on ranks are, as pointed out by Clara Grazian and others, invariant under monotonic transforms and hence producing a Gaussian pair or a Uniform pair with the intended rank correlation is sufficient to return a correlated sample for any pair of marginal distributions by the (monotonic) inverse cdf transform.  On the other hand, if correlation is understood as Pearson linear correlation, (a) it is not always defined and (b) there does not seem to be a generic approach to simulate from an arbitrary triplet (F,G,ρ) [assuming the three entries are compatible]. When Kees pointed out Pascal van Kooten‘s solution by permutation, I thought this was a terrific resolution, but after thinking about it a wee bit more, I am afraid it is only an approximation, i.e., a way to return a bivariate sample with a given empirical correlation. Not the theoretical correlation. Obviously, when the sample is very large, this comes as a good approximation. But when facing a request to simulate a single pair (X,Y), this gets inefficient [and still approximate].

Now, if we aim at exact simulation from a bivariate distribution with the arbitrary triplet (F,G,ρ), why can’t we find a generic method?! I think one fundamental if obvious reason is that the question is just ill-posed. Indeed, there are many ways of defining a joint distribution with marginals F and G and with (linear) correlation ρ. One for each copula. The joint could thus be associated with a Gaussian copula, i.e., (X,Y)=(F⁻¹(Φ(A)),G⁻¹(Φ(B))) when (A,B) is a standardised bivariate normal with the proper correlation ρ’. Or it can be associated with the Archimedian copula

C(u; v) = (u + v − 1)-1/θ,

with θ>0 defined by a (linear) correlation of ρ. Or yet with any other copula… Were the joint distribution perfectly well-defined, it would then mean that ρ’ or θ (or whatever natural parameter is used for that copula) do perfectly parametrise this distribution instead of the correlation coefficient ρ. All that remains then is to simulate directly from the copula, maybe a theme for a future post…

## Cauchy Distribution: Evil or Angel?

Posted in Books, pictures, Running, Statistics, Travel, University life, Wines with tags , , , , , , , , , , , , on May 19, 2015 by xi'an Natesh Pillai and Xiao-Li Meng just arXived a short paper that solves the Cauchy conjecture of Drton and Xiao [I mentioned last year at JSM], namely that, when considering two normal vectors with generic variance matrix S, a weighted average of the ratios X/Y remains Cauchy(0,1), just as in the iid S=I case. Even when the weights are random. The fascinating side of this now resolved (!) conjecture is that the correlation between the terms does not seem to matter. Pushing the correlation to one [assuming it is meaningful, which is a suspension of belief!, since there is no standard correlation for Cauchy variates] leads to a paradox: all terms are equal and yet… it works: we recover a single term, which again is Cauchy(0,1). All that remains thus to prove is that it stays Cauchy(0,1) between those two extremes, a weird kind of intermediary values theorem!

Actually, Natesh and XL further prove an inverse χ² theorem: the inverse of the normal vector, renormalised into a quadratic form is an inverse χ² no matter what its covariance matrix. The proof of this amazing theorem relies on a spherical representation of the bivariate Gaussian (also underlying the Box-Müller algorithm). The angles are then jointly distributed as $\exp\{-\sum_{i,j}\alpha_{ij}\cos(\theta_i-\theta_j)\}$

and from there follows the argument that conditional on the differences between the θ’s, all ratios are Cauchy distributed. Hence the conclusion!

A question that stems from reading this version of the paper is whether this property extends to other formats of non-independent Cauchy variates. Somewhat connected to my recent post about generating correlated variates from arbitrary distributions: using the inverse cdf transform of a Gaussian copula shows this is possibly the case: the following code is meaningless in that the empirical correlation has no connection with a “true” correlation, but nonetheless the experiment seems of interest…

> ro=.999999;x=matrix(rnorm(2e4),ncol=2);y=ro*x+sqrt(1-ro^2)*matrix(rnorm(2e4),ncol=2)
> cor(x[,1]/x[,2],y[,1]/y[,2])
 -0.1351967
> ro=.99999999;x=matrix(rnorm(2e4),ncol=2);y=ro*x+sqrt(1-ro^2)*matrix(rnorm(2e4),ncol=2)
> cor(x[,1]/x[,2],y[,1]/y[,2])
 0.8622714
> ro=1-1e-5;x=matrix(rnorm(2e4),ncol=2);y=ro*x+sqrt(1-ro^2)*matrix(rnorm(2e4),ncol=2)
> z=qcauchy(pnorm(as.vector(x)));w=qcauchy(pnorm(as.vector(y)))
> cor(x=z,y=w)
 0.9999732
> ks.test((z+w)/2,"pcauchy")

One-sample Kolmogorov-Smirnov test

data:  (z + w)/2
D = 0.0068, p-value = 0.3203
alternative hypothesis: two-sided
> ro=1-1e-3;x=matrix(rnorm(2e4),ncol=2);y=ro*x+sqrt(1-ro^2)*matrix(rnorm(2e4),ncol=2)
> z=qcauchy(pnorm(as.vector(x)));w=qcauchy(pnorm(as.vector(y)))
> cor(x=z,y=w)
 0.9920858
> ks.test((z+w)/2,"pcauchy")

One-sample Kolmogorov-Smirnov test

data:  (z + w)/2
D = 0.0036, p-value = 0.9574
alternative hypothesis: two-sided


## arbitrary distributions with set correlation

Posted in Books, Kids, pictures, R, Statistics, University life with tags , , , , , , , , , , on May 11, 2015 by xi'an A question recently posted on X Validated by Antoni Parrelada: given two arbitrary cdfs F and G, how can we simulate a pair (X,Y) with marginals  F and G, and with set correlation ρ? The answer posted by Antoni Parrelada was to reproduce the Gaussian copula solution: produce (X’,Y’) as a Gaussian bivariate vector with correlation ρ and then turn it into (X,Y)=(F⁻¹(Φ(X’)),G⁻¹(Φ(Y’))). Unfortunately, this does not work, because the correlation does not keep under the double transform. The graph above is part of my answer for a χ² and a log-Normal cdf for F amd G: while corr(X’,Y’)=ρ, corr(X,Y) drifts quite a  lot from the diagonal! Actually, by playing long enough with my function

tacor=function(rho=0,nsim=1e4,fx=qnorm,fy=qnorm)
{
x1=rnorm(nsim);x2=rnorm(nsim)
coeur=rho
rho2=sqrt(1-rho^2)
for (t in 1:length(rho)){
y=pnorm(cbind(x1,rho[t]*x1+rho2[t]*x2))
coeur[t]=cor(fx(y[,1]),fy(y[,2]))}
return(coeur)
}


Playing further, I managed to get an almost flat correlation graph for the admittedly convoluted call

tacor(seq(-1,1,.01),
fx=function(x) qchisq(x^59,df=.01),
fy=function(x) qlogis(x^59)) Now, the most interesting question is how to produce correlated simulations. A pedestrian way is to start with a copula, e.g. the above Gaussian copula, and to twist the correlation coefficient ρ of the copula until the desired correlation is attained for the transformed pair. That is, to draw the above curve and invert it. (Note that, as clearly exhibited by the graph just above, all desired correlations cannot be achieved for arbitrary cdfs F and G.) This is however very pedestrian and I wonder whether or not there is a generic and somewhat automated solution…