champagne, not Guinness??? [jatp]

Cauchy Distribution: Evil or Angel?

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

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])
[1] -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])
[1] 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)
[1] 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)
[1] 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

foie gras fois trois

As New Year’s Eve celebrations are getting quite near, newspapers once again focus on related issues, from the shortage of truffles, to the size of champagne bubbles, to the prohibition of foie gras. Today, I noticed an headline in Le Monde about a “huge increase in French people against force-fed geese and ducks: 3% more than last year are opposed to this practice”. Now, looking at the figures, it is based on a survey of 1,032 adults, out of which 47% were against. From a purely statistical perspective, this is not highly significant since

is compatible with the null hypothesis N(0,1) distribution.