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For $i=1,2,3$, consider a random variable $Y_i$ taking value in $$ \mathcal{Y}:=\{(1,1), (1,0), (0,1), (0,0)\} $$ and a random closed set $S_i$ taking value in $\mathcal{S}$ that is the power set of $\mathcal{Y}$ (without the empty set), i.e. $$ \mathcal{S}:=\{\{(1,1)\}, \{(1,0)\}, \{(0,1)\}, \{(0,0)\},\\ \{(1,1), (1,0)\}, \{(1,1), (0,1)\}, \{(1,1), (0,0)\}, \{(1,0), (0,1)\}, \{(1,0), (0,0)\}, \{(0,1), (0,0)\},\\ \{(1,1), (1,0), (0,1)\}, \{(1,1), (1,0), (0,0)\}, \{(1,1), (0,1), (0,0)\}, \{(1,0), (0,1), (0,0)\},\\ \{(1,1), (1,0), (0,1), (0,0)\}\} $$ $Y_i,S_i$ are defined on the same probability space $(\Omega, \mathcal{F}, P)$.

Also, $Y_1, Y_2, Y_3$ are independent, $S_1, S_2, S_3$ are independent.


Suppose that $$ P(Y_i\in K)\leq P(S_i\cap K\neq \emptyset) \text{ } \forall K \in \mathcal{S} \text{ for } i=1,2,3 $$

For example, for $K=\{(1,1), (0,1)\}$ and $i=1$ $$ P(Y_1=(1,1))+P(Y_1=(0,1))\leq \\ P(S_1=\{(1,1)\})+P(S_1=\{(0,1)\})\\+P(S_1=\{(1,1), (1,0)\})+P(S_1= \{(1,1), (0,1)\})\\+P(S_1=\{(1,1), (0,0)\})+P(S_1=\{(1,0), (0,1)\})+P(S_1=\{(0,1), (0,0)\})\\+P(S_1= \{(1,1), (1,0), (0,1)\})+P(S_1=\{(1,1), (1,0), (0,0)\})\\+P(S_1= \{(1,1), (0,1), (0,0)\})+P(S_1= \{(1,0), (0,1), (0,0)\})\\+P(S_1= \{(1,1), (1,0), (0,1), (0,0)\}) $$


I would like your help to show that $$ (\star) \hspace{1cm} P(Y_1=(1,1))\times P(Y_2=(1,1))\times P(Y_3=(1,1)) +\\P(Y_1=(0,0))\times P(Y_2=(0,0))\times P(Y_3=(0,0))\leq\\ P(S_1\cap \{(1,1)\}\neq \emptyset \text{ and } S_2\cap \{(1,1)\}\neq \emptyset \text{ and } S_3\cap \{(1,1)\}\neq \emptyset \text{ OR }\\ S_1\cap \{(0,0)\}\neq \emptyset \text{ and } S_2\cap \{(0,0)\}\neq \emptyset \text{ and } S_3\cap \{(0,0)\}\neq \emptyset) $$


My attempt

(A) I take the inequalities referred to $K=\{(1,1), (0,0)\}$ for $i=1,2,3$ and multiply them across $i$: $$ [P(Y_1=(1,1))+P(Y_1=(0,0))]\times [P(Y_2=(1,1))+P(Y_2=(0,0))]\times [P(Y_3=(1,1))+P(Y_3=(0,0))]\leq\\ [P(S_1\cap \{(1,1),(0,0)\}\neq \emptyset)]\times [P(S_2\cap \{(1,1),(0,0)\}\neq \emptyset)]\times [P(S_3\cap \{(1,1),(0,0)\}\neq \emptyset)] $$

(B) On the lhs the terms "in excess" with respect to $(\star)$ are $$ P(Y_1=(1,1))\times P(Y_2=(0,0))\times P(Y_3=(0,0))+\\ P(Y_1=(0,0))\times P(Y_2=(1,1))\times P(Y_3=(0,0))+\\ P(Y_1=(0,0))\times P(Y_2=(0,0))\times P(Y_3=(1,1))+\\ P(Y_1=(1,1))\times P(Y_2=(1,1))\times P(Y_3=(0,0))+\\ P(Y_1=(1,1))\times P(Y_2=(0,0))\times P(Y_3=(1,1))+\\ P(Y_1=(0,0))\times P(Y_2=(1,1))\times P(Y_3=(1,1)) $$

(C) On the rhs the terms "in excess" with respect to $(\star)$ are $$ P(S_1\cap \{(1,1)\}\neq \emptyset \text{ and } S_1\cap \{(0,0)\}=\emptyset)\times P(S_2\cap \{(0,0)\}\neq \emptyset \text{ and } S_2\cap \{(1,1)\}=\emptyset)\times P(S_3\cap \{(0,0)\}\neq \emptyset \text{ and } S_3\cap \{(1,1)\}=\emptyset)+\\ P(S_1\cap \{(0,0)\}\neq \emptyset \text{ and } S_1\cap \{(1,1)\}=\emptyset)\times P(S_2\cap \{(1,1)\}\neq \emptyset \text{ and } S_2\cap \{(0,0)\}=\emptyset)\times P(S_3\cap \{(0,0)\}\neq \emptyset \text{ and } S_3\cap \{(1,1)\}=\emptyset)+\\ P(S_1\cap \{(0,0)\}\neq \emptyset \text{ and } S_1\cap \{(1,1)\}=\emptyset)\times P(S_2\cap \{(0,0)\}\neq \emptyset \text{ and } S_2\cap \{(1,1)\}=\emptyset)\times P(S_3\cap \{(1,1)\}\neq \emptyset \text{ and } S_3\cap \{(0,0)\}=\emptyset)+\\ P(S_1\cap \{(1,1)\}\neq \emptyset \text{ and } S_1\cap \{(0,0)\}=\emptyset)\times P(S_2\cap \{(1,1)\}\neq \emptyset \text{ and } S_2\cap \{(0,0)\}=\emptyset)\times P(S_3\cap \{(0,0)\}\neq \emptyset \text{ and } S_3\cap \{(1,1)\}=\emptyset)+\\ P(S_1\cap \{(1,1)\}\neq \emptyset \text{ and } S_1\cap \{(0,0)\}=\emptyset)\times P(S_2\cap \{(0,0)\}\neq \emptyset \text{ and } S_2\cap \{(1,1)\}=\emptyset)\times P(S_3\cap \{(1,1)\}\neq \emptyset \text{ and } S_3\cap \{(0,0)\}=\emptyset)+\\ P(S_1\cap \{(0,0)\}\neq \emptyset \text{ and } S_1\cap \{(1,1)\}=\emptyset)\times P(S_2\cap \{(1,1)\}\neq \emptyset \text{ and } S_2\cap \{(0,0)\}=\emptyset)\times P(S_3\cap \{(1,1)\}\neq \emptyset \text{ and } S_3\cap \{(0,0)\}=\emptyset)+\\ P(S_1\cap \{(0,0)\}\neq \emptyset \text{ and } S_1\cap \{(1,1)\}\neq\emptyset)\times P(S_2\cap \{(1,1)\}\neq \emptyset \text{ and } S_2\cap \{(0,0)\}=\emptyset)\times P(S_3\cap \{(0,0)\}\neq \emptyset \text{ and } S_3\cap \{(1,1)\}=\emptyset)+\\ P(S_1\cap \{(0,0)\}\neq \emptyset \text{ and } S_1\cap \{(1,1)\}\neq\emptyset)\times P(S_2\cap \{(0,0)\}\neq \emptyset \text{ and } S_2\cap \{(1,1)\}=\emptyset)\times P(S_3\cap \{(1,1)\}\neq \emptyset \text{ and } S_3\cap \{(0,0)\}=\emptyset)+\\ P(S_1\cap \{(1,1)\}\neq \emptyset \text{ and } S_1\cap \{(0,0)\}=\emptyset)\times P(S_2\cap \{(1,1)\}\neq \emptyset \text{ and } S_2\cap \{(0,0)\}\neq \emptyset)\times P(S_3\cap \{(0,0)\}\neq \emptyset \text{ and } S_3\cap \{(1,1)\}=\emptyset)+\\ P(S_1\cap \{(0,0)\}\neq \emptyset \text{ and } S_1\cap \{(1,1)\}=\emptyset)\times P(S_2\cap \{(1,1)\}\neq \emptyset \text{ and } S_2\cap \{(0,0)\}\neq \emptyset)\times P(S_3\cap \{(1,1)\}\neq \emptyset \text{ and } S_3\cap \{(0,0)\}=\emptyset)+\\ P(S_1\cap \{(1,1)\}\neq \emptyset \text{ and } S_1\cap \{(0,0)\}=\emptyset)\times P(S_2\cap \{(0,0)\}\neq \emptyset \text{ and } S_2\cap \{(1,1)\}= \emptyset)\times P(S_3\cap \{(1,1)\}\neq \emptyset \text{ and } S_3\cap \{(0,0)\}\neq\emptyset)+\\ P(S_1\cap \{(0,0)\}\neq \emptyset \text{ and } S_1\cap \{(1,1)\}=\emptyset)\times P(S_2\cap \{(1,1)\}\neq \emptyset \text{ and } S_2\cap \{(0,0)\}= \emptyset)\times P(S_3\cap \{(1,1)\}\neq \emptyset \text{ and } S_3\cap \{(0,0)\}\neq\emptyset) $$

(D) One strategy could be to show that (B) $\geq $ (C), and, hence, because of (A), $(\star)$ holds. However, I am unable to do it.

(E) What I have shown, instead, is that $$ 1-P(Y_1=(0,0))=P(Y_1=(1,1))+P(Y_1=(1,0))+P(Y_1=(0,1))\geq\\ P(S_1\cap \{(1,1)\}\neq \emptyset \text{ and } S_1\cap \{(0,0)\}=\emptyset) $$ and $$ P(Y_1=(1,1))\geq P(S_1=\{(1,1)\}) $$ and $$ P(Y_1=(1,1))+P(Y_1=(0,0))\geq P(S_1\cap\{(1,1)\}\neq \emptyset \text{ and } S_1\cap\{(0,0)\}\neq \emptyset\}) $$ and $$ P(Y_1=(1,1))+P(Y_1=(0,0))\geq P(S_1\cap\{(0,1),(1,0)\}=\emptyset) $$ which, however, do not seem to be useful.

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    $\begingroup$ I will spare you your hard-earned reputation by restoring the bounty and migrating this to the Math site where it is more likely to be answered. $\endgroup$
    – whuber
    May 4, 2017 at 16:59
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    $\begingroup$ Can we assume independence between the various random variables? $\endgroup$
    – N.Bach
    May 4, 2017 at 17:34
  • $\begingroup$ Yes, in the sense that $P(Y_1\in K_1\text{ and } Y_2\in K_2 \text{ and } Y_3\in K_3)=P(Y_1\in K_1) \times P(Y_2\in K_2)\times P(Y_3\in K_3)$ and $P(S_1\cap K_1\neq \emptyset \text{ and } S_2\cap K_2\neq \emptyset \text{ and } S_3\cap K_3\neq \emptyset)=P(S_1\cap K_1\neq \emptyset) \times P(S_2\cap K_2\neq \emptyset)\times P(S_3\cap K_3\neq \emptyset)$. $\endgroup$
    – Star
    May 4, 2017 at 17:37
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    $\begingroup$ Thanks for your question. I have added it to the main body. $\endgroup$
    – Star
    May 4, 2017 at 17:37
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    $\begingroup$ Question edited because some terms were missing in C) $\endgroup$
    – Star
    May 5, 2017 at 10:37

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