I have to solve a tricky exercise that requires some ability to work with sets and random variables. I would like your help with this because I'm unable to see how to proceed.

Construction of $\mathcal{A},\mathcal{B}, \mathcal{C}$:

Consider $3$ sets $A,B,C$ with $A\equiv \{a_1,a_2,a_3\}$, $B\equiv \{b_1,b_2,b_3\}$, $C\equiv \{c_1,c_2,c_3\}$.

Let $\mathcal{A},\mathcal{B}, \mathcal{C}$ be the collection of all subsets of $A$,$B$,$C$ excluding the empty set, i.e., $$ \mathcal{A}\equiv \Big\{A, \{a_1\}, \{a_2\}, \{a_3\}, \{a_1,a_2\}, \{a_2, a_3\}, \{a_1,a_3\}\Big\} $$

$$ \mathcal{B}\equiv \Big\{B, \{b_1\},\{b_2\}, \{b_3\},\{b_1,b_2\}, \{b_2, b_3\}, \{b_1,b_3\}\Big\} $$

$$ \mathcal{C}\equiv \Big\{C, \{c_1\},\{c_2\}, \{c_3\},\{c_1,c_2\}, \{c_2, c_3\}, \{c_1,c_3\}\Big\} $$

Construction of $A\times B \times C$:

Let $\times$ denote the Cartesian Product. $A\times B \times C$ is the collection of all ordered $3$-tuples from $A,B,C$, i.e., $$ A\times B \times C\equiv \Big\{(a_1,b_1,c_1),(a_1,b_2,c_1), (a_1,b_3,c_1),... \Big\} $$

Construction of $\mathcal{T}$:

Consider $\mathcal{T}\equiv \mathcal{A}\times \mathcal{B}\times \mathcal{C}$. For example, an element $T\in \mathcal{T}$ is $$ T\equiv \Big(\{a_1,a_2\}, \{b_3\} , \{c_1,c_2\}\Big) $$

Each $T\in \mathcal{T}$ is composed of 3 "components". In the example above, the first component is $\{a_1,a_2\}$, the second is $\{b_3\}$, the third is $\{c_1,c_2\}$.

We rewrite each $T\in \mathcal{T}$ as the set of all possible ordered $3$-tuples that can be constructed from its $3$ components. Continuing the example above, $$ T\equiv \overbrace{\Big(\{a_1,a_2\}, \{b_3\} , \{c_1,c_2\}\Big)}^{\text{Representation 1}} \equiv \overbrace{\Big\{\overbrace{(a_1,b_3, c_1)}^{\text{ordered $3$-tuple}}, (a_2, b_3, c_1), (a_1,b_3, c_2), (a_2,b_3, c_2)\Big\} }^{\text{Representation 2}} $$

Construction of $\mathcal{M}$:

We collect in $\mathcal{M}$ all the subsets of $A\times B\times C$ that cannot be part of $\mathcal{T}$. For example, $$ M\equiv \Big\{(a_1,b_3, c_1), (a_2, b_2, c_1)\Big\} $$ $M$ is not an element of $\mathcal{T}$.

Probability: Suppose that I have a random vector $Y$ with size $3\times 1$ taking value in $A\times B\times C$.

Suppose that I have a random set $S$ taking value in $\mathcal{T}$.

Assume that $$ (\star)\hspace{1cm}\mathbb{P}(Y\in T)\geq \mathbb{P}(S\subseteq T ) \text{ } \forall T\in \mathcal{T} $$ where $\mathbb{P}$ denotes probability measure.

Question: show that $$ \mathbb{P}(Y\in M)\geq \mathbb{P}(S\subseteq M ) \text{ } \forall M\in \mathcal{M} $$

Some clarifications: Random sets are generalisations of random variables to sets.

I believe that to show what I want we should somehow use the fact that $\mathbb{P}(S=M )=0$ $\forall M\in \mathcal{M}$. But I can't see how to proceed.

Also, I realised that every set $M\in \mathcal{M}$ can be written as the union of all the subsets of $M$ that belong to $\mathcal{T}$. But I can't see where to use this.

  • 1
    $\begingroup$ Note that your representation of the typical element $T\in\mathcal{T}$ as $$T\equiv \Big\{\{a_1,a_2\}, \{b_3\} , \{c_1,c_2\}\Big\}$$ is incorrect. $T$ should be ordered, as:$$T\equiv \Big(\{a_1,a_2\}, \{b_3\} , \{c_1,c_2\}\Big)$$ $\endgroup$ – MPW Sep 5 '18 at 20:57
  • $\begingroup$ OK, let me edit, thanks. $\endgroup$ – TEX Sep 5 '18 at 21:02
  • $\begingroup$ I'm not sure of any approach but maybe try brute forcing for $n=2$, i.e. two sets $A=\{a_1,a_2\}$ and $B=\{b_1,b_2\}$, proving the assumption $(\star)$, then the question. (The case for $n=1$ is boring). $\endgroup$ – palmpo Sep 6 '18 at 3:32
  • $\begingroup$ Thanks. The assumption $(*)$ should not be proved. $\endgroup$ – TEX Sep 6 '18 at 6:33
  • $\begingroup$ I try to post my attempt. Step 1: Take any set $M\in \mathcal{M}$ such that the sets $M_1,...,M_k$ are disjoint and belong to $\mathcal{T}$, where $M_1,...,M_k$ are defined as: - consider all the subsets of $M$ that belong to $\mathcal{T}$ - take the union of the subsets that have non-empty intersections - name the list of obtained subsets $M_1,...,M_k$ $\endgroup$ – TEX Sep 19 '18 at 18:14

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