Consider a multivariate logistic-normal variable $z \sim \mathcal {LN}(\mu,{\Sigma})$, where ${\Sigma}$ is and $n$-by-$n$ positive definite matrix. I mean, for $x = (x_1,\ldots,x_n)\sim \mathcal N(\mu,\Sigma)$ define the vector $z = (z_1,...,z_n, z_{n+1})$ by $$z_j = \dfrac{\exp(x_j)}{\sum_{l=1}^{n+1} \exp(x_l)}, $$ with $x_{n+1} := 0$.

Observations: It is well-known that in general, the moments of $z$ admit not a closed-form solution. There are some trivial exceptions though. For example if $n = 1$ and $\mu = 0$, then it's easy to see that probability density of $z$ is symmetric about the point line $z = 1/2$ and $z$ has mean $1/2$, irrespective of the variance $\sigma^2$ of $x$.

Question 1: In a general dimension $n \ge 2$, if $\mu = 0 \in \mathbb R^n$, what can be said about the expected value $\mathbb E[z]$ of $z$ ?

A completely blind guess is the centroid $(1/n,\ldots,1/n)$ of the $n$-dimensional simplex (at least in the spherical case $\Sigma = \operatorname{I}$).

Question 2: Say I knew before hand that $\mathbb E [z] = (1/n,\ldots,1/n)$. What can I say about the mean $\mu$ of the underlying Gaussian ?

Question 3: If $\mu = \epsilon^{-1}\tilde{\mu}$, and $\Sigma = \epsilon^{-2}\tilde{\Sigma}$, what is (an approximation of) the value of the limit $\lim_{\epsilon \rightarrow 0^+}\mathbb E[z]$ (as a function of $\tilde{\mu}$ and $\tilde{\Sigma}$)?


1 Answer 1


Take $n=2$ as an example. If $\mu={\bf 0}$, the density of $Z$ is $$ \frac{1}{|2\pi\Sigma|^{1/2}z_1z_2z_3} exp\{ -\frac{1}{2} {\bf z^T_{-3}} \Sigma^{-1}{\bf z_{-3}}\}, $$

where ${\bf z_{-3}} = (log(z_1/z_3), log(z_2/z_3))^T$ and $\sum_{i=1}^3 z_i =1$.

For question #1, if $\Sigma = I_2$, $E({\bf z})$ is not $(1/3,1/3,1/3)^T$ by numerical evaluation, and $E(z_1)=E(z_2)\ne E(z_3)$. Note the exponent part of the density $exp\{-1/2 [(log(z_1/z_3))^2+ (log(z_2/z_3))^2]\}$ is not symmetric for $z_1,z_2,z_3$. However, it is symmetric for $z_1,z_2$. To the best of my knowledge, there is no closed form for the expectation.

For question #2, to make the expectations equal for each component, one sufficient condition is to make the exponent part in the density symmetric for $z_1,z_2,z_3$. Note that if $$ \Sigma = c\pmatrix {1&0.5\\0.5&1}, c>0, $$ the exponent part becomes proportional to $$ (log(z_1/z_3))^2+ (log(z_2/z_3))^2-log(z_1/z_3)log(z_2/z_3) = [log(z_1)]^2+[log(z_2)]^2+[log(z_3)]^2-log(z_1)log(z_2)-log(z_1)log(z_3)-log(z_2)log(z_3), $$ which is symmetric for $z_1,z_2,z_3$. Thus the expectation $E(z_1)=E(z_2)= E(z_3)=1/3$. This is generalizable to any $n \ge 2$. If $\mu=0$ , the diagonal elements of $\Sigma$ are the same and all the off-diagonal elements of $\Sigma$ are half of the diagonal, the density would be symmetric for $z_1,...,z_{n+1}$, which implies $ E(\bf{z}) = \bf{1/(n+1)}$. I am not sure if this is a necessary condition and I am interested if there exist other examples either when $\mu=0$ or $\mu \ne 0$.

I do not understand the idea of question #3. Are you thinikng what the expectation would be when $\mu \rightarrow \infty$ ?

  • $\begingroup$ Thanks for the response. Those are some really nice ideas. About Q#3, not quite. For example, take $c = \epsilon^{-1}$ and $\mu = \epsilon^{-1}(a,b)$, for some real scalars $a$ and $b$ (e.g take $a=b=0$ for purpose of illustration) in your example above and let $\epsilon \rightarrow 0^+$. I'm interested in the limiting value of expectations $\mathbb E[z_1]$ and $\mathbb E[z_2]$ $\endgroup$
    – dohmatob
    Dec 8, 2017 at 15:11
  • $\begingroup$ I think it depends on $a,b$ and the variance. If variance is equal and $c=\epsilon^{-2}$, then $E(z_i)$ with highest mean $\mu_i$ will tend to 1 no matter what $\sigma_{ij}$ is. If variances are not equal, I am not sure. I guess it will depend on $\sigma_{ij}, \mu$. $\endgroup$
    – Statisfun
    Dec 8, 2017 at 16:23
  • $\begingroup$ Ya sure, i seek the limit as a function of $\tilde{\mu}$ and $\tilde{\Sigma}$ as in the statement of the question. I'll make this more explicit in the post. $\endgroup$
    – dohmatob
    Dec 8, 2017 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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