Equivalence relation and different sample spaces

[gregthenovelist]

TL;DR

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?

 

[Dale]

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?

 

[PeroK]

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]

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?

 

[fresh_42]

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]

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!

 

[PeroK]

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.

 

[fresh_42]

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.

 

[Stephen Tashi]

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.

What theorem are you talking about?

In the first place we should distinguish between a proposition versus a propositional function. In the usual terminology, a proposition is a statement that is either true or false and not both. For example, in mathematics, $2 < 5$ is a proposition. A propositional function is a function that maps some set of things into the set $\{True, False\}$. For example, $x < 5$ is a propositional function.

So if you want to talk about a logical function whose domain is a set of outcomes, then you should talk about a propositional function. This brings up the question of what it means for one propositional function to imply another propositional function.

Presumably, we interpret that like $\forall x ( P(x) \implies Q(x) )$. Even before we introduce the idea of probability, we have to decide if that interpretation implies that $P$ and $Q$ are functions with the same domain.