0
$\begingroup$

A triply line stochastic matrix (t.l.s.m.) of size $N$ is a 3-dimensional array $(a_{ijk})_{i,j,k=1}^N$ with nonnegative entries, whose any row, column or line sums up to $1$. Their set is $TLS_N$.

Consider the following $0$-$1$ t.l.s.m. of size $2^n$ :

Keep the half-cube (of size $2^{n-1}$) containing $(1,1,1)$, and the 3 neighboring half-cubes "in diagonal", and put $0$ everywhere in the other 4 half-cubes. Repeat this operation in each of the 4 kept cubes of size $2^{n-1}$, keeping 4 cubes of size $2^{n-2}$ (the one nearest to $(1,1,1)$ and its 3 neighbors in diagonal), and putting $0$ everywhere else. An so on, until you are left with $2^{2n}$ points, one in each row, column or line, and put $a_{ijk}=1$ there.

By construction, this t.l.s.m. maximizes the total (sum of the entries) in some cubes containing $(1,1,1)$ i.e. $$\sum_{i,j,k=1}^{m}a_{i,j,k}\ge\sum_{i,j,k=1}^{m}b_{i,j,k}$$ for every t.l.s.m. $B:=(b_{ijk})_{i,j,k=1}^{2^n}\in TLS_{2^n}$ , whenever $m=2^p$, $0\le p\le n$ .

Question (motivated by Bound on the joint distribution of three real random variables with given two dimensional marginals) Does this inequality also hold for other values of $m$, $1\le m\le 2^n$ ?

$\endgroup$
2
  • 1
    $\begingroup$ I'm not happy about using the word "matrix" to describe a three-dimensional array. $\endgroup$ Commented Jan 29, 2017 at 22:08
  • $\begingroup$ I understand. Neither am I. But the notion was given that name (P Fischer, ER Swart - Linear algebra and its applications, 1985 - Elsevier). $\endgroup$ Commented Jan 31, 2017 at 13:25

1 Answer 1

1
$\begingroup$

Maybe for some other values, but not for all. Let $n=3$ (matrices of size $8$) and $m=3$ (comparing the totals of $A$ and of some other t.l.s.m. $B$ on the cube $[1,3]^3$). The total for $A$ on this cube is $7$. We can have as much as $9$ for $B$ if $b_{ijk}=\frac13$ on $[1,3]^3$, $=0$ on $[1,3]\times[1,3]\times[4,8]$, $[1,3]\times[4,8]\times[1,3]$ and $[4,8]\times[1,3]\times[1,3]$ , $=\frac15$ on $[1,3]\times[4,8]\times[4,8]$, $[4,8]\times[1,3]\times[4,8]$ and $[4,8]\times[4,8]\times[1,3]$, and $=\frac2{25}$ on $[4,8]^3$.

This negative result suggests the following modified question : for which coefficients $c_1>\cdots>c_m>\cdots>c_{2^n}>0$ is it true that $$\sum_{m=1}^{2^n}c_m\sum_{i,j,k=1}^m a_{ijk}\ge\sum_{m=1}^{2^n}c_m\sum_{i,j,k=1}^m b_{ijk}$$ for every t.l.s.m. $B$ ? (Is it true when $c_m$ is a convex function of $m$, for example ?)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .