You're looking at Less Wrong's discussion board. This includes all posts, including those that haven't been promoted to the front page yet. For more information, see About Less Wrong.

D_Malik comments on Open Thread, Jun. 29 - Jul. 5, 2015 - Less Wrong Discussion

5 Post author: Gondolinian 29 June 2015 12:14AM

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Comments (210)

You are viewing a single comment's thread.

Comment author: D_Malik 29 June 2015 04:38:23AM 1 point [-]

Random thing that I can't recall seeing on LW: Suppose A is evidence for B, i.e. P(B|A) > P(B). Then by Bayes, P(A|B) = P(A)P(B|A)/P(B) > P(A)P(B)/P(B) = P(A), i.e. B is evidence for A. In other words, the is-evidence-for relation is symmetric.

For instance, this means that the logical fallacy of affirming the consequent (A implies B, and B is true, therefore A) is actually probabilistically valid. "If Socrates is a man then he'll probably die; Socrates died, therefore it's more likely he's a man."

Comment author: Slider 29 June 2015 08:48:11AM 3 points [-]

The surprise comes only to those that try to overload probability with roles it should not have. For example "A implies B" does imply P(B|A)=1 but P(B|A)=1 doesn't imply "A implies B". While it is common that if we know systematically that a certain probability is high it is a promising line of argument for causations and implications it doesn't always carry through (meta-example I am in essence arguing that while an okayish rule of thump, it is actually improper to infer causations from probabilities. I am doing this by pointing out that P(causation|probability) ~ 1 and P(causation|probability)<1 which alone is only suggestive that it is so (or the actual steps are implied)).

Comment author: DanielLC 29 June 2015 09:53:04AM 0 points [-]

but P(B|A)=1 doesn't imply "A implies B".

Can you give a counterexample?

Comment author: gjm 29 June 2015 12:20:13PM 5 points [-]

A: "X is a random number, uniformly distributed on the interval from 0 to 1."

B: "X is irrational."

Then Pr(B|A) = 1 because almost all numbers are irrational (more formally: because the rationals have measure 0), but A doesn't imply B because X could be rational.

Comment author: Slider 29 June 2015 12:47:21PM 0 points [-]

This could be made to not be a counterexample by using a theory of probability that uses surreals.

That is Pr(irrational|random form 0 to 1) being 1 is of the "almost always" kind, which can be separated form the kind of 1 that is of the "always" kind.

for ω that is larger than any surreal that has a real-counter part, there is a ɛ=1/ω.

Taking only finite samples out of a infinite group makes for a probability that is smaller than any real probability that could well be represented by real/natural multiples of ɛ.

Similarly taking only countably infinite samples from a group of uncountably many samples would result in a probability larger than 0 but smaller than any real value.

Thus we could have P(irrational|between 0 and 1)=1-xɛ and P(rational|between 0 and 1)=xɛ that would sum to exactly 1 and yet P(Z|0-1)=xɛ>0 ie a positive probability.

Similarly the probability of a dart landing exactly on a line in a dart board is "almost never" ie 0 yet that place is as probable as any other location on the dart board. It would be possible to find a dart exactly on the line, you would not just expect to encounter it in a finite number of throws.

However there are counterexamples where all As are indeed Bs but no implication is possible.

Comment author: RichardKennaway 29 June 2015 01:26:40PM 4 points [-]

This could be made to not be a counterexample by using a theory of probability that uses surreals.

Surreal numbers do not yet have a good theory of integration. This makes surreal probability theory problematic.

Comment author: Slider 29 June 2015 10:31:23AM 0 points [-]

Coextensive properties that are not the same property. There are some biological facts like these. Probably not remembering correctly but for example B="animal has heart" A="animal is mammal" it can easily be that all mammals in fact have hearts but you couldn't still say that it would be impossible for a mammal to be heartless (and for example have a blood circulation system that is evenly distributed all over the veins (which they kinda partially do but be totally reliant on those kind of mechanism)). The deduction of "It is a mammal, it must have a heart" is false for plenty of reasonable senses of "must". It is true for the probabilistic sense of must but implication has more senses than the probabilistic one.

Comment author: DanielLC 29 June 2015 08:39:59PM 0 points [-]

If it's a given that all mammals have hearts, then being a mammal implies it has a heart. If it's not known that all mammals have hearts, then P(B|A) < 1.

Comment author: MrMind 29 June 2015 07:32:35AM 2 points [-]

Yes, Jaynes talks about this in the first chapter of his book, calling it a "weak syllogism" and using it as a guideline to introduce probability as a kind of extended logic.

Comment author: ZeitPolizei 29 June 2015 03:49:53PM 0 points [-]

See also this.