Sufficient conditions for strict monotonicity and continuity of multidimensional cdf

Question

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 $\forall h \in (0,\infty)$ $$ \mu_k(a_1,a_2,...,a_k+h, ..., a_M)>\mu_k(a_1,a_2,...,a_k, ..., a_M) $$

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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?

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.