# Equivalence relation and different sample spaces

gregthenovelist
TL;DR Summary
Equivalence Relation fails when two propositions are not in the same sample space. Why?
It is a theorem that: two propositions implying each other, in the sense that the set of outcomes making one true is the same as the one making the other true) have the same probability. this comes from the fact that if p --> q, the P(p&q) = P(p), we have that if p <-> q, then P(p&q) = P(p)= P(q). but this is only so if p and q dwell in one sample space.

Question: what is the problem when they are not in the same sample space?

Mentor
2021 Award
Summary:: Equivalence Relation fails when two propositions are not in the same sample space. Why?

the set of outcomes making one true is the same as the one making the other true
How could this even hold if they are in different sample spaces? Do you have an example?

Homework Helper
Gold Member
2021 Award
More generally, you always need an implied or explicit universal set. For example, consider the non-negative integers ##\{0, 1, 2 \dots \}## and the non-positive integers ##\{\dots -2, -1, 0 \}##. It only makes sense to form the intersection or union of these sets if we take the integers (or rationals or reals) as our universal set, of which both are subsets.

gregthenovelist
gregthenovelist
How could this even hold if they are in different sample spaces? Do you have an example?
This is exactly my question. why can we not even ask this question if they are in different sample spaces?

Mentor
2021 Award
This is exactly my question. why can we not even ask this question if they are in different sample spaces?

Because you cannot compare things that are different by nature. How would you define reflexivity or symmetry?

gregthenovelist
gregthenovelist
More generally, you always need an implied or explicit universal set. For example, consider the non-negative integers ##\{0, 1, 2 \dots \}## and the non-positive integers ##\{\dots -2, -1, 0 \}##. It only makes sense to form the intersection or union of these sets if we take the integers (or rationals or reals) as our universal set, of which both are subsets.
Great, that helps a lot. So, if we cannot combine them into a universal set (in this case rationals or reals), we cannot get an intersection. Thus, the equivalence principle would fail as it is in logical terms the same as the union of the two sets. Is my reasoning correct here?

gregthenovelist
Because you cannot compare things that are different by nature. How would you define reflexivity or symmetry?
great, makes sense!

Homework Helper
Gold Member
2021 Award
Great, that helps a lot. So, if we cannot combine them into a universal set (in this case rationals or reals), we cannot get an intersection. Thus, the equivalence principle would fail as it is in logical terms the same as the union of the two sets. Is my reasoning correct here?
It's just the same technical point that your overall or universal sample space must include all events. Implicitly or explicity, both sample spaces must be subsets of an overall universal sample space under consideration.

gregthenovelist
Mentor
2021 Award
great, thanks! How exactly is reflexivity and symmetry important for the equivalence relation?
An equivalence relation ##\sim## is defined to be
a) reflexive ##a\sim a##,
b) symmetric ##a\sim b \Longrightarrow b\sim a## and
c) transitive ##a\sim b \wedge b\sim c \Longrightarrow a\sim c##
This is its definition.

gregthenovelist