*Bounty: 100*

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 $mathcal{Y}$, 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 $(Omega, mathcal{F}, P)$.

Suppose that

$$

P(Y_iin K)leq P(S_icap Kneq 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), (1,0)})+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_1cap {(1,1)}neq 0 text{ and } S_2cap {(1,1)}neq 0 text{ and } S_3cap {(1,1)}neq 0 text{ OR }\

S_1cap {(0,0)}neq 0 text{ and } S_2cap {(0,0)}neq 0 text{ and } S_3cap {(0,0)}neq 0)

$$

**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_1cap {(1,1),(0,0)}neq emptyset)]times [P(S_2cap {(1,1),(0,0)}neq emptyset)]times [P(S_3cap {(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_1cap {(1,1)}neq emptyset text{ and } S_1cap {(0,0)}=emptyset)times P(S_2cap {(0,0)}neq emptyset text{ and } S_2cap {(1,1)}=emptyset)times P(S_3cap {(0,0)}neq emptyset text{ and } S_3cap {(1,1)}=emptyset)+\

P(S_1cap {(0,0)}neq emptyset text{ and } S_1cap {(1,1)}=emptyset)times P(S_2cap {(1,1)}neq emptyset text{ and } S_2cap {(0,0)}=emptyset)times P(S_3cap {(0,0)}neq emptyset text{ and } S_3cap {(1,1)}=emptyset)+\

P(S_1cap {(0,0)}neq emptyset text{ and } S_1cap {(1,1)}=emptyset)times P(S_2cap {(0,0)}neq emptyset text{ and } S_2cap {(1,1)}=emptyset)times P(S_3cap {(1,1)}neq emptyset text{ and } S_3cap {(0,0)}=emptyset)+\

P(S_1cap {(1,1)}neq emptyset text{ and } S_1cap {(0,0)}=emptyset)times P(S_2cap {(1,1)}neq emptyset text{ and } S_2cap {(0,0)}=emptyset)times P(S_3cap {(0,0)}neq emptyset text{ and } S_3cap {(1,1)}=emptyset)+\

P(S_1cap {(1,1)}neq emptyset text{ and } S_1cap {(0,0)}=emptyset)times P(S_2cap {(0,0)}neq emptyset text{ and } S_2cap {(1,1)}=emptyset)times P(S_3cap {(1,1)}neq emptyset text{ and } S_3cap {(0,0)}=emptyset)+\

P(S_1cap {(0,0)}neq emptyset text{ and } S_1cap {(1,1)}=emptyset)times P(S_2cap {(1,1)}neq emptyset text{ and } S_2cap {(0,0)}=emptyset)times P(S_3cap {(1,1)}neq emptyset text{ and } S_3cap {(0,0)}=emptyset)

$$

(D) One strategy could be to show that

$$

P(Y_1=(1,1))geq P(S_1cap {(1,1)}neq emptyset text{ and } S_1cap {(0,0)}=emptyset)

$$

and, similarly, for the other terms, so 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

$$

P(Y_1=(1,1))+P(Y_1=(1,0))+P(Y_1=(0,1))geq\ P(S_1cap {(1,1)}neq emptyset text{ and } S_1cap {(0,0)}=emptyset)

$$

and that

$$

P(Y_1=(1,1))geq P(S_1={(1,1)})

$$

which, however, do not seem to be useful.

Get this bounty!!!