phil6334 rough drafts

* refers to our seminar: Phil6334

I’m putting these notes under “rejected posts” awaiting feedback and corrections.

Contemporary Bayesian epistemology in philosophy appeals to formal probability to capture informal notions of “confirmation”, “support”, “evidence”, and the like, but it seems to create problems for itself by not being scrupulous about identifying the probability space, set of events, etc. and not distinguishing between events and statistical hypotheses. There is usually a presumed reference to games of chance, but even there things can be altered greatly depending on the partition of events. Still, we try to keep to that. The goal is just to give a sense of that research program. (Previous posts on the tacking paradox: Oct. 25, 2013: “Bayesian Confirmation Philosophy and the Tacking Paradox (iv)*” &  Oct 25.

(0) Simple Bayes Boost R: 

H is “confirmed” or supported by x if P(H|x) > P(H) (P(x |H)) > P(x).

H is disconfirmed (or undermined) by x if P(H|x) < P(H), (else x is confirmationally irrelevant to H).

Mayo: The error statistician would already get off at (0): probabilistic affirming the consequent is maximally unreliable, violating the minimal requirement for evidence. That could be altered with context-depedent information about how the data and hypotheses are arrived at, but this is not made explicit.

(a) Paradox of irrelevant conjunctions (‘tacking paradox’)

If x confirms H, then x also confirms (H & J), even if hypothesis J is just “tacked on” to H.[1]

Hawthorne and Fitelson (2004) define:

J is an irrelevant conjunct to H, with respect to evidence x just in case

P(x|H) = P(x|H & J).

(b) Example from earlier: For instance, x might be radioastronomic data in support of:

H: “the GTR deflection of light effect is 1.75” and

J: “the radioactivity of the Fukushima water dumped in the Pacific ocean is within acceptable levels”.

(1) Bayesian (Confirmation) Conjunction: If x Bayesian confirms H, then x Bayesian-confirms:

(H & J), where P(x| H & J ) = P(x|H) for any J consistent with H

(where J is an irrelevant conjunct to H, with respect to evidence x).

If you accept R, (1) goes through.

Mayo: We got off at (0) already.  Frankly I don’t  know why Bayesian epistemologists would allow adding an arbitrary statement or hypothesis not amongst those used in setting out priors. Maybe it’s assumed J is in there somehow (in the background K), but it seems open-ended, and they have not objected.

But let’s just talk about well-defined events in a probability experiment; and limit ourselves to talking about an event providing evidence of another event (e.g., making it more or less expected) in some sense. In one of Fitelson’s examples,   P(black|ace of spade) > P(black), so “black” confirms it’s an ace of spades (presumably in random drawings of card color from an ordinary deck)–despite “ace” being an “irrelevant conjunct” of sorts. Even so, if someone says data x (it’s a stock trader) is evidence it’s an inside trader in a hedge firm, I think it would be assumed that something had been done to probe the added conjuncts.

(2) Using simple B-boost R: (H & J) gets just as much of a boost by x as does H—measuring confirmation as a simple B-boost: R.

C^R(H, x) = C^R((H& J), x) for irrelevant conjunct J.

R: P(H|x)/P(H) (or equivalently, P(x |H)/P(x))

(a) They accept (1) but (2) is found counterintuitive (by many or most Bayesian epistemologists). But if you’ve defined confirmation as a B-boost, why run away from the implications? (A point Maher makes.) It seems they implicitly slide into thinking of what many of us want:

some kind of an assessment of how warranted or well-tested H is (with x and background).

(not merely a ratio which, even if we can get it, won’t mean anything in particular. It might be any old thing, 2, 22, even with H scarcely making x expected.).

(b) The intuitive objection according to Crupi and Tentori (2010) is this (e.g., p. 3): “In order to claim the same amount of positive support from x to a more committal theory “H and J” as from x to H alone, …adding J should contribute by raising further how strongly x is expected assuming H by itself. Otherwise, what would be the specific relevance of J?” (using my letters, emphasis added)

But the point is that it’s given that J is irrelevant. Now if one reports all the relevant information for the inference, one might report something like (H & J) makes x just as expected as H alone. Why not object to the “too was” confirmation (H & J) is getting when nothing has been done to probe J? I think the objection is, or should be, that nothing has been done to show J is the case rather than not. P(x |(H & J)) = P(x |(H & ~J)).

(c) Switch from R to LR: What Fitelson (Hawthorne and Fitelson 2004) do is employ, as a measure of the B-boost, what some call the likelihood ratio (LR).

C^LR(H, x) = P(x | H)/P(x | ~H).

(3) Let x confirm H, then

(*) C^LR(H, x) > C^LR((H& J), x)

For J an irrelevant conjunct to H.

So even though x confirms (H & J) it doesn’t get as much as H does, at least if one uses LR. (It does get as much using R).

They see (*) as solving the irrelevant conjunct problem.

(4) Now let x disconfirm Q, and x confirm ~Q, then

(*) C^LR(~Q, x) > C^LR((~Q & J), x)

For J an irrelevant conjunct to Q: P(x|Q) = P(x|J & Q).

Crupi and Tentori (2010) notice an untoward consequence of using LR confirmation in the case of disconfirmation (substituting their Q for H above): If x disconfirms Q, then (Q & J) isn’t as badly disconfirmed as Q is, for J an irrelevant conjunct to Q. But this just follows from (*), doesn’t it? That is, from (*), we’d get (**) [possibly an equality goes somewhere.)

(**) C^LR(Q, x) < C^LR((Q & J), x).

This says that if x disconfirms Q, (Q & J) isn’t as badly disconfirmed as Q is. This they find counterintuitive.

But if (**) is counterintuitive, then so is (*).

(5) Why (**) makes sense if you wish to use LR:

The numerators in the LR calculations are the same:

P(x|Q & J) = P(x|Q) and P(x|H & J) = P(x|H) since in both cases J is an irrelevant conjunct.

But P(x|~(Q & J)) < P(x|~Q)

Since x disconfirms Q, x is more probable given ~Q than it is given (~Q v ~J). This explains why

(**) C^LR(Q, x) < C^LR((Q & J), x)

So if (**) is counterintuitive then so is (*).

(a) Example Q: unready for college.

If x = high scores on a battery of college readiness tests, then x disconfirms Q and confirms ~Q.

What should J be? Suppose having one’s favorite number be an even number (rather than an odd number) is found irrelevant to scores.

(i) P(x|~(Q & J)) = P(high scores| either college ready or ~J)

(ii) P(x|~Q ) = P(high scores| college ready)

(ii) might be ~1 (as in the earlier discussion), while (i) considerable less.

The high scores can occur even among those whose favorite number is odd.This explains why

(**) C^LR(Q, x) < C^LR((Q & J), x)

In the case where x confirms H, it’s reversed

P(x |~(H & J)) > P(x |~H)

(b) Using one of Fitelson’s examples, but for ~Q:

e.g., Q: not-spades    x: black      J: ace

P(x |~Q) = 1.

P(x | Q)= 1/3

P(x |~(Q & J)) = 25/49

i.e., P(black|spade or not ace)=25/49

Note: C^LR [(Q & J), x) = P(x |(Q & J))/P(x |~(Q & J))

Please share corrections, questions.

Previous slides are:

http://errorstatistics.com/2014/02/09/phil-6334-day-3-feb-6-2014/

http://errorstatistics.com/2014/01/31/phil-6334-day-2-slides/

REFERENCES:

Chalmers (1999). What Is This Thing Called Science, 3rd ed. Indianapolis; Cambridge: Hacking.

Crupi & Tentori (2010). Irrelevant Conjunction: Statement and Solution of a New Paradox, Phil Sci, 77, 1–13.

Hawthorne & Fitelson (2004). Re-Solving Irrelevant Conjunction with Probabilistic Independence, Phil Sci 71: 505–514.

Maher (2004). Bayesianism and Irrelevant Conjunction, Phil Sci 71: 515–520.

Musgrave (2010) “Critical Rationalism, Explanation, and Severe Tests,” in Error and Inference (D.Mayo & A. Spanos eds). CUP: 88-112.