# [Link] The Bayesian argument against induction.

In 1983 Karl Popper and David Miller published an argument to the effect that probability theory could be used to disprove induction. Popper had long been an opponent of induction. Since probability theory in general, and Bayes in particular is often seen as rescuing induction from the standard objections, the argument is significant.

It is being discussed over at the Critical Rationalism site.

Interesting but extremely unpersuasive.

I agreed with everything up until this point:

This seems to be wordgames. Saying this is a deductive inference misses the whole point that this an inference which can only be used after B has been observed. Otherwise it is just math.

The next paragraph seems to be similarly flawed.

Here I may just be missing the point but I don’t see how logical consequences of A are relevant to the issue of whether induction is occurring.

I have to wonder if some strange notion of induction is occurring where intuitions are simply not shared. I wonder, what would happen if we tabooed induction?

For anyone still following this, I have tried to restate my arguments in a new way here:

http://www.criticalrationalism.net/2011/07/20/more-on-inductive-probability/

There’s no abstract. It is not obvious what is the point of the article is.

I linked to the previous post. That begins with something like an abstract: a statement of intent, at least.

This isn’t an article or a paper: it’s a blog post.

After reading both of your posts I didn’t know what you meant by induction nor what you were arguing against.

In my response I didn’t explicitly urge you to stop using the terms “induction” and “subjective interpretation” (in an attempt to Replace the Symbol with the Substance through the game of Taboo Your Words). But I mentioned that we might find we agree on every important point after stripping away the purely definitional disputes. JoshuaZ and endoself made similar points here, though again they didn’t explicitly

tellyou how to clear up the confusion. I’m asking you now to please find a new way of expressing yourself.Until then I won’t know if this next part actually affects your argument, but it seems worth saying anyway: your mathematical lemma does not deal with the sort of logical implications that scientists care about. If people frequently made statements such as, ‘The Standard Model holds OR we won’t find the Higgs Boson on the 23rd of July 2011,’ then your lemma might seem like a perfectly natural and intuitive description of rational degrees of belief. In other words, your intuition may have misled you just because intuition often fails when it comes to mathematical statements that we can’t interpret using our life experience to date.

Jaynes did point out something about induction which philosophers seem to miss: people perform induction when the data they have gives us information about some sort of

physical mechanismthat allows us to predict the results of that mechanism. For instance, if I have observed that 50 out of the last 50 times you have flipped a certain coin, it has come up heads, I can conclude that it’s probably a two headed coin, and so I predict that that coin will continue to turn up only heads. If, however, you pick up another quarter, induction suddenly becomes a lot less powerful.The other point Jaynes makes is that philosophers forget the biggest advantage induction gives us. Imagine this line of reasoning: Falling apples obey Newtonian mechanics. The moon obeys Newtonian mechanics. The earth obeys Newtonian mechanics. Mars, Jupiter, and Saturn all behave Newtonian mechanics. Neptune...does not. Hmm. Maybe there’s another planet? Oh, there’s Pluto. I guess Neptune was obeying Newtonian mechanics. Venus obeys Newtonian mechanics. Mercury...hmm. Mercury should not be moving like that if the Newtonian theory of gravitation is wrong. I guess we should investigate this.

Most scientific advances have come from times where induction fails. When induction works, it means we haven’t learned anything new.

Well, since I was horribly wrong when I thought I saw a flaw in the math, let me instead look at the

conclusions, and maybe I won’t be horribly wrong :DThis is not what the equation says above. Yes, p(AvB|B)=1. But there’s another term on that side too: -p(AvB), which has to be mentioned. If it could be ignored, then B would increase the probability of A for any choice of A and B!

How’s this: to the extent that B increases the probability of A, p(A v ~B) - p(A v ~B|B) is less than 1 - p(A v B).

Not deductive, I know, but accurate.

This is a bit circular. If something is really independent of B, it will not change at all if we condition on B. Nearly all things that are logical consequences, though, aren’t independent. Maybe the author had some example in mind while writing?

But the decrease in probability of A v ~B is not “purely deductive” because ~(A v ~B) is not a logical consequence of B. So the net change in the probability of A is not entirely deductive.

EDIT: This attacks the argument on its own terms, but in fact I think the argument given does not define induction well enough to say anything about it.

Ey acknowledges that P(A|B) != P(A), ey just disputes that this can be called induction. This seems pointless to me; Bayesians can just define induction as cases where P(A|B) != P(A) and avoid this kind of word game.

Whoops—v is apparently intended to mean mean “or” or “union.” It’s just that the author incorrectly translates P(A v B) & P(A v ~B) into P(A v B)

+P(A v ~B).Hi,

I am the author. It wasn’t a mistranslation. The logical equivalence was not translated into anything. It was merely intended to break down A according to its logical consequences shared with B. I never wrote “P(A v B) + P(A v ~B),” because that would be irrelevant.

In the very next equation after “A = (A v B) & (A v ~B)”, you write:

This is the equation where you put in the plus signs. Additionally, you can break things down like that

insidethe P() operator, but you can’t just move that tooutsidethe P() operator, because things might be correlated (and, in the case of B and ~B, certainly are).Well, it wasn’t actually an equation. That’s why I used the =||= symbol. It was a bientailment. It asserts logical equivalence (in classical logic), and it means something slightly different than an equals symbol. The equation with the plus signs and the logical equivalence shouldn’t be confused.

I’m back and there’s been no response, so I’ll be specific. Starting from

Using p(X v Y) = p(X) + p(Y) - p(XY), we get

.15 = p(A|B) + p(B|B) - p(AB|B) - p(A) - p(B) + p(AB) + p(A|B) + p(~B|B) - p(A~B|B) - p(A) - p(~B) + p(A~B)

= p(A|B) + 1 - p(A) - p(A) - p(B) + p(AB) + p(A|B) - p(A) - p(~B) + p(A~B)

= 2 p(A|B) − 2 p(A) = twice the thing you started from, which is bad.

It looks like you simplified p(AB|B) as p(A), but in fact p(AB|B)=p(ABB)/p(B)=p(AB)/p(B)=p(A|B). (I made a similar mistake earlier.)

I get p(A|B) + p(B|B) - p(AB|B) - p(A) - p(B) + p(AB) + p(A|B) + p(~B|B) - p(A~B|B) - p(A) - p(~B) + p(A~B)

= p(A|B) + 1 - p(A|B) - p(A) - p(B) + p(AB) + p(A|B) + 0 − 0 - p(A) − 1 + p(B) + p(A~B)

= - p(A) - p(B) + p(AB) + p(A|B) - p(A) + p(B) + p(A~B)

= - p(A) + p(AB) + p(A|B) - p(A) + p(A~B)

= p(A|B) −2p(A) +p(AB) + p(A~B)

= p(A|B) −2p(A) + p(A)

= p(A|B) - p(A)

But this is

quod erat demonstrandum.Okay. Yay, a use for the retract button where it’s still good to have the text visible!

Manfred,

I calculated the result for about three different sets of probabilities before making the original post. The equation was correct each time. I could have just been mistaken, of course, but even Zack (the commenter above) conceded that the equation is true.

EDIT: Oh, I see now. You have changed all my disjunctions into conjunctions. Why?

Ah, I’m sorry. Do you agree that the equation I quoted above is incorrect, though? I’m going to have to leave now, but the relevant basic equations of probability are P(X v Y) = P(X) + P(Y) - P(X&Y), and P(X&Y) = P(X) * P(X | Y)

I don’t know what this is trying to mean: “On the inductive view of probabilistic inference, B is amplified to imply that A is more probably true. This would would mean the logical consequences of A which are not also logical consequences of B should be more probable given B.”

Edit: there was random other stuff here. Ignore it, I’m confused about the above.

[Correction: In the original version of this comment, I claimed that the linked post was mistaken on a point of mathematics: specifically, I said that it is not the case that P(A|B) – P(A)= P(A v B|B) – P(A v B) + P(A v ~B|B) – P(A v ~B). However, Guy Srinivasan pointed out that my supposed disproof contained a mistake. Redoing my calculations, I find that the equation in question is in fact an identity. I regret the error.]P(Av~B|B) does not equal P(A). P((Av~B) & B) equals P(A).

Edit: it doesn’t, of course. P(Av~B|B) = P(A|B) and the other thing I said is just silly.

Whoops! You’re right. I’ll edit the comment to reflect this correction.

Just for general convenience: P((A+~B)B) = P(AB).

I don’t see how that follows. Is weight being assigned as a result of Popper’s reputation?

Was that supposed to be (A & B) v (A & ~B)? What it has is always true. Also, what is =||=?

If I can understand this correctly, they’re saying that induction is false because the only accuracy in it is due to the hidden Bayes-structure. This is true of everything.

DanielLC,

Hi, I am the author.

The =||= just means bientailment. It’s short for,

A |= (A v B) & (A v ~B) and (A v B) & (A v ~B) |= A

Where |= means entailment or logical consequence. =||= is analogous to a biconditional.

The point is that each side of a bientailment is logically equivalent, but the breakdown allows us to see how B alters the probability of different logical consequences of A.

Thank you, I think I understand most of it now. I don’t see (at this hour) where the absolute values come from, but that doesn’t seem to matter much. Let’s focus on this line:

The conjunction of those two does contradict itself, and if you actually write out the probability of the contradiction—using the standard product rule p(CD)=p(D)p(C|D) rather than multiplying their separate probabilities p(D)p(C) together—you’ll see that it always equals zero.

But each separate claim (A, and B~A), can increase in probability provided that they each take probability from somewhere else, namely from ~B~A. I see no problem with regarding this as an increase for our subjective confidence in A and a separate increase for B~A. Again, each grows by replacing an option (or doubt) which no longer exists for us. Some of that doubt-in-A simply changed into a different form of doubt-in-A, but some of it changed into confidence. The total doubt therefore goes down even though one part increases.