Consider the following subset of the unit cube in $\mathbb R^n$: $$ \mathcal D = \{ p = (p_1,p_2,\dots,p_n) \in [0,1]^n:\; p_1 \le p_2 \le \cdots \le p_n\}. $$ We would like to construct a probability measure supported on $\mathcal D$, continuous w.r.t. to the $n$-dimensional Lebesgue measure and "somehow" diffuse for large $n$. By this last statement, I roughly mean that the variances of the components of the vector $p$ stay bounded away from 0 under that measure. The problem is quite open-ended, but I will try to make this a bit more precise towards the end.

To see the difficulty, consider the general approach of starting with a probability distribution on $[0,1]^n$ and conditioning it to lie in $\mathcal D$. If we start with the uniform distribution on $[0,1]^n$ and perform this conditioning, we are effectively looking at the order statistics of a sample of size $n$ from the uniform distribution on $[0,1]$. The resulting marginal distributions of the components of $p$ are well-known: $p_k \sim \text{Beta}(k, n+1-k)$ whose variance rapidly goes to zero as $n \to \infty$.

In general, there seems to be a repulsion among the coordinates of $p$, induced by the order constraint, that forces their variances to shrink. The open-ended question is whether we can fight this repulsion and come up with a distribution whose marginal coordinates have more or less constant variances as $n\to \infty$?

We can also make the problem more precise by asking this (though I am afraid it might be too much to ask):

Is there a distribution on $[0,1]^n$ ($n \ge 2$) supported on $\mathcal D$ and absolutely continuous w.r.t. to the Lebesgue measure whose marginal distributions are all uniform on $[0,1]$?

A second more relaxed question could be something like this: For a distribution $Q$ on $[0,1]^n$, let $\text{var}_i(Q)$ be the variance of its $i$ coordinate (i.e., the variance of $p_i$ where $p \sim Q$).

Is there a sequence of distributions $Q_{n}$ on $[0,1]^n$ ($n \ge 2$) supported on $\mathcal D$ and absolutely continuous w.r.t. to the Lebesgue measure, with the propery that the sequence $n \mapsto \min_{1 \le i\le n} \text{var}_i(Q_n)$ is bounded away from zero as $n\to \infty$?

We can also relax the minimum to, say, the arithmetic average if the above is still too stringent.

A bit more generally, I am interested in the answer to the higher-order versions of this problem, for example, the case where we have a collection of points $(p_{ij})_{i,j=1}^{n_1,n_2}$ arranged in the $2$-dimensional lattice $[n_1] \times [n_2]$ with the constraints being imposed by the natural partial order, $p_{i,j} \le p_{i',j'}$ if $i \le i'$ and $j \le j'$. You can then imagine the 3-dimensional version and so on.

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    $\begingroup$ Do you have a solution for the case $n=2$, such that the probability measure is not concentrated on the diagonal? For all corresponding discrete versions of your problem in case $n = 2$ with $\mathcal{D} = \{(x,y) \in \{0,1/k,2/k,\ldots,1\}^2 \colon x \leq y\}$ you always have that the probability measure is concentrated on the diagonal of $\mathcal{D}$. $\endgroup$ Jan 4 at 14:12
  • $\begingroup$ @DieterKadelka, I hadn't given the discrete case much thought, but this is an interesting example. $\endgroup$
    – passerby51
    Jan 6 at 5:15
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    $\begingroup$ My original (intuitive) idea was partioning the interval $[0,1]$ into $n$ subintervals and showing by the same argument as above that there can be only mass on the "diagonal cubes". But I didn't elaborate this idea any further. Iosif Pinelis idea is better, I think. $\endgroup$ Jan 6 at 9:49

Let $X$ be any nondegenerate random variable with values in the interval $[0,1]$. Let $p:=(p_1,\dots,p_n)$, where $p_1=\dots=p_n=X$. Then $p$ takes values in $\mathcal D$ and $\min_i Var\,p_i=Var\,X>0$ for all $n$.

This answers positively your second question. Letting $X$ be uniformly distributed on $[0,1]$, we almost get a positive answer to your first question, except that the joint distribution of $p_1,\dots,p_n$ is not absolutely continuous here.

On the other hand, if you insist on the absolutelte continuity, then it is also clear now that the answer to your first question is negative. Indeed, suppose that $p_1,\dots,p_n$ are random variables (r.v.'s) such that each of them has the uniformly distribution on $[0,1]$ (or any other distribution) and $p_1\le\dots\le p_n$. Then, for each $j\in\{1,\dots,n-1\}$, the r.v. $p_j$ is stochastically no greater than $p_{j+1}$. Therefore and because $p_j$ has the same distribution as $p_{j+1}$, we have $p_j=p_{j+1}$ almost surely (a.s.), again for each $j\in\{1,\dots,n-1\}$; details on this conclusion are given by Lemma 1 below. Thus, $p_1=\dots=p_n$ a.s. and hence the joint distribution of $p_1,\dots,p_n$ is, necessarily, not absolutely continuous.

Lemma 1: Let $Y$ and $Z$ be identically distributed r.v.'s such that $Y\le Z$ a.s. Then $Y=Z$ a.s.

$\newcommand\Q{\mathbb Q}$Proof: Suppose the contrary: that $Y$ and $Z$ are identically distributed r.v.'s such that $Y\le Z$ a.s. but $P(Y=Z)<1$. Then, denoting by $\Q$ the set of all rational real numbers, we get $$0<P(Y<Z)=P\Big(\bigcup_{r\in\Q}\{Y\le r<Z\}\Big),$$ so that for some $r\in\Q$ we have $$0<P(Y\le r<Z)=P(Y\le r)-P(Z\le r),$$ which contradicts the condition that $Y$ and $Z$ are identically distributed; the equality in the latter display holds because $Y\le Z$ a.s. $\Box$

Somehow, previously I missed the absolute continuity condition in your second question. It can still be answered positively, with the following modification of the previous answer. For $i=1,\dots,n$, let $p_i:=Y+i/(3n)$, where $Y$ is uniformly distributed on $[1/4,1/2]$. Then $0<p_1<\cdots<p_n<1$ and $Var\,p_i=Var\,Y>0$ for all $n,i$. Also, $0<1/4<p_i\le5/6<1$ for $i=1,\dots,n$ and $p_{i+1}-p_i=1/(3n)>0$ for $i=1,\dots,n-1$. So, convolving the distribution of $p=(p_1,\dots,p_n)$ with any absolutely continuous distribution supported on the $n$-cube of a small enough size centered at the origin, we get the desired result.

  • $\begingroup$ Nice application of stochastic order I didn't know! $\endgroup$ Jan 4 at 15:05
  • $\begingroup$ Great, thanks! I was wondering whether you implicitly had convolutions in mind for the second question. The edit clears it up. $\endgroup$
    – passerby51
    Jan 6 at 5:13
  • $\begingroup$ @Iosif Pinelis, that lemma is amazing, by the way. I wonder if it was (well) known or you just proved it. $\endgroup$
    – passerby51
    Jan 6 at 5:35
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    $\begingroup$ @passerby51 : I think Lemma 1 should be known, but I don't recall seeing it elsewhere. $\endgroup$ Jan 6 at 17:30

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