Let $V\equiv (V_1,..., V_M)$ be a random vector defined on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$.

Consider the map $\mu$ such that $$ a\equiv \begin{pmatrix} a_1\\ a_2\\ ...\\ a_M \end{pmatrix} \in \mathbb{R}^M \mapsto \mu(a)\equiv \begin{pmatrix} \mu_1(a)\\ \mu_2(a)\\ ...\\ \mu_M(a)\\ \end{pmatrix} \equiv \begin{pmatrix} \mathbb{P}(V_1+a_1\geq V_y+a_y \text{ }\forall y \neq 1)\\ \mathbb{P}(V_2+a_2\geq V_y+a_y \text{ }\forall y \neq 2)\\ ...\\ \mathbb{P}(V_M+a_M\geq V_y+a_y \text{ }\forall y \neq M)\\ \end{pmatrix}\in [0,1]^M $$ (A.1) Assume that the map $\mu$ is continuous and strictly increasing in each of the $M$ dimensions, where strict monotonicity in each dimension means that, for any $k\in \{1,...,M\}$ and $\forall h \in (0,\infty)$ $$ \mu_k(a_1,a_2,...,a_k+h, ..., a_M)>\mu_k(a_1,a_2,...,a_k, ..., a_M) $$

Question: under which sufficient conditions on the distribution of $V$ is A.1 satisfied (the weakest sufficient conditions that you can think of)?

My idea is that if

  • the support of $V$ is $\mathbb{R}^M$, and

  • the distribution of $V$ is absolutely continuous with respect to Lenesgue measure with everywhere strictly positive density on $\mathbb{R}^M$

then A.1 is satisfied. Is this correct? Can you think of weaker conditions?


1 Answer 1


Your $\mu$ is symmetric in $a$ and related only to differences in $V$, so adding any random X to all elements of V has no effect on $mu$ - that is one relaxation of $V$.$$\mu_1(a) = \mathbb{P}[a_1 - a_y \geq max_{y\neq1}(V_y) - V_1]$$ You constrain each $\mu_i$ to be "continuous and strictly increasing", $V_1$ is constrained relative to next largest $V_i$ say $V_2$.

The difference must meet "absolute continuous" probability relative to Lebesgue Measure on $\mathbb{R}$, i.e. non-zero probability on sets of positive measure for the difference of values, the gap.

Considering each element we obtain your constraint, but only applied on the gap between largest to second largest element of $V$.

$V$ has a distribution s.t. defining i, j as the index of the maximum and second greatest elements then the derived distribution of Vi-Vj is absolutely continuous w.r.t Lebesgue measure.

  • $\begingroup$ Thanks, but the conclusion is not clear to me. Could you write explicitly your suggested set of sufficient conditions? $\endgroup$
    – Star
    Commented May 23, 2018 at 17:55
  • $\begingroup$ I've tried to formalise the constraint, note in particular $V$ doesn't require $\mathbb{R}^M$ as support, e.g. $V_1$ could be constant. $\endgroup$ Commented May 23, 2018 at 19:50
  • $\begingroup$ Are my conditions implying your conditions? $\endgroup$
    – Star
    Commented May 23, 2018 at 20:27
  • $\begingroup$ The condition flows from $\mu$ being strictly increasing. The equality translates to $\mathbb{P}$ and thus to the form I've tried to explain - a constraint on distribution of V only in the gap between max and second greatest elements. $\endgroup$ Commented May 23, 2018 at 22:43
  • 1
    $\begingroup$ no your condition is very restrictive on $V$ you suggested "support of $V$ is $\mathbb{R}^M$" .. this is not required, e.g. $V_1$ can be constant and meet my conditions. you required "distribution of $V$ is absolutely continuous" .. this is much stronger than just the gap being absolutely continuous $\endgroup$ Commented May 24, 2018 at 13:18

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