4
$\begingroup$

Suppose $X$ is sampled from a symmetric Dirichlet distribution with arbitrary shape and $n$ dimensions. Equivalently we can independently sample $z_i \sim \text{Gamma}(\alpha, 1)$ and then set $x_i=\frac{z_i}{\sum z_i}$

I want to show that the coordinates $x_i$ are negatively associated. Intuition tells me that, of course they are, since occupancy numbers (balls-in-bins) are negatively associated.

I'm considering using a scaling argument: assume that each $x_i$ is rational, and then let $\lambda$ be the least common multiple of the denominators. Then we can phrase this problem as a balls-in-bins problem where we have $1/\lambda$ balls and $n$ bins.

This isn't as elegant as I'd like, and it seems that this is the type of thing results must exist for. Perhaps there's a nice application of the FKG inequality?

$\endgroup$
2
$\begingroup$

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

Let $x:=(x_1,\dots,x_n)$ and $z:=(z_1,\dots,z_n)$, and assume that $\al\ge1$. It is easy to see that $x$ is independent of $\sum z_i$, so that the distribution of the random vector $x$ is the same as the conditional distribution of the random vector $z$ given $\sum z_i=1$. Also, the density of each $z_i$ is log concave. Therefore, by the following theorem by Joag-Dev and Proschan, $x$ is negatively associated (NA):

Theorem 2.8 Let $z_1,\dots,z_n$ be independent random variables with PF${}_2$ (log-concave) densities. Then the joint conditional distribution of $z_1,\dots,z_n$ given $\sum z_i$ is NA (almost surely).

$\endgroup$
  • $\begingroup$ Perfect! This is exactly the result I was looking for. Thanks! $\endgroup$ – bucket Apr 10 '18 at 16:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.