Hmm. But in the envelope experiment, once Alice commits to a decision (e.g. choose A), her probabilities are well-defined. So in Sleeping Beauty, if we make it so the day is automatically disclosed to Alice at 5pm let’s say, it seems like her probabilities about it should be well-defined from the get go. Or at least, the envelope experiment doesn’t seem to shed light why they should be undefined. Am I missing something?
Hmm. But in the envelope experiment, once Alice commits to a decision (to e.g. choose A), the probabilities are well-defined. So in Sleeping Beauty, if the day is automatically disclosed to Alice at 5pm let’s say, it seems like her probabilities about it should be well-defined from the get go.
Do you mean that conditional probabilities should be well defined? They indeed are.
P(Heads|Monday) = 1⁄2; P(Heads|Tuesday)=1. But as P(Monday) and P(Tuesday) are not defined you can’t use them to arrive to P(Heads&Monday) and P(Heads&Tuesday) via Bayes theorem.
If you say things like “P(X|Y) is defined but P(Y) isn’t”, doesn’t that call for a reformulation of all probability theory? Like, if I take the interpretation of probability theory based on sigma-algebras (which is quite popular), then P(Y) gotta be defined, no way around it. The very definition of P(X|Y) depends on P(X∧Y) and P(Y). You can say “let’s kick out this leg from this table”, but the math tells me pretty insistently that the table can’t stand without that particular leg. Or at least, if there’s a version of probability theory where P(Y) can be undefined but P(X|Y) defined, I’d want to see more details about that theory and how it doesn’t trip over itself. Does that make sense?
This equation for a conditional probability, although mathematically equivalent, may be intuitively easier to understand. It can be interpreted as “the probability of B occurring multiplied by the probability of A occurring, provided that B has occurred, is equal to the probability of the A and B occurrences together, although not necessarily occurring at the same time”. Additionally, this may be preferred philosophically; under major probability interpretations, such as the subjective theory, conditional probability is considered a primitive entity. Moreover, this “multiplication rule” can be practically useful in computing the probability of
and introduces a symmetry with the summation axiom for Poincaré Formula
Not sure I understand. My question was, what kind of probability theory can support things like “P(X|Y) is defined but P(Y) isn’t”. The snippet you give doesn’t seem relevant to that, as it assumes both values are defined.
The kind of probability theory that defines P(X|Y) axiomatically as a primitive entity and only then defines P(X&Y) as a multiplication of P(X|Y) and P(Y), instead of defining conditional probability as a ratio between P(X&Y) and P(Y).
While it’s mathematically equivalent, the former method is more resembling the way people deal with probabilities in practice—usually conditional probability is known and probability of an intersection isn’t—and formally allows us to talk about conditional probabilities, even when the probability of an event we condition on is not defined.
Hmm. But in the envelope experiment, once Alice commits to a decision (e.g. choose A), her probabilities are well-defined. So in Sleeping Beauty, if we make it so the day is automatically disclosed to Alice at 5pm let’s say, it seems like her probabilities about it should be well-defined from the get go. Or at least, the envelope experiment doesn’t seem to shed light why they should be undefined. Am I missing something?
Do you mean that conditional probabilities should be well defined? They indeed are.
P(Heads|Monday) = 1⁄2; P(Heads|Tuesday)=1. But as P(Monday) and P(Tuesday) are not defined you can’t use them to arrive to P(Heads&Monday) and P(Heads&Tuesday) via Bayes theorem.
If you say things like “P(X|Y) is defined but P(Y) isn’t”, doesn’t that call for a reformulation of all probability theory? Like, if I take the interpretation of probability theory based on sigma-algebras (which is quite popular), then P(Y) gotta be defined, no way around it. The very definition of P(X|Y) depends on P(X∧Y) and P(Y). You can say “let’s kick out this leg from this table”, but the math tells me pretty insistently that the table can’t stand without that particular leg. Or at least, if there’s a version of probability theory where P(Y) can be undefined but P(X|Y) defined, I’d want to see more details about that theory and how it doesn’t trip over itself. Does that make sense?
Sure. But this has already been done and took much less trouble than you might have though. Citing Wikipedia on Conditional Probability:
Not sure I understand. My question was, what kind of probability theory can support things like “P(X|Y) is defined but P(Y) isn’t”. The snippet you give doesn’t seem relevant to that, as it assumes both values are defined.
The kind of probability theory that defines P(X|Y) axiomatically as a primitive entity and only then defines P(X&Y) as a multiplication of P(X|Y) and P(Y), instead of defining conditional probability as a ratio between P(X&Y) and P(Y).
While it’s mathematically equivalent, the former method is more resembling the way people deal with probabilities in practice—usually conditional probability is known and probability of an intersection isn’t—and formally allows us to talk about conditional probabilities, even when the probability of an event we condition on is not defined.