January 09, 2007

More On the Self-Undermining Argument

At the end of the last post I performed a slightly fanciful calculation to determine what should happen if there are two epistemic peers who disagree on the epistemology of disagreement: One thinks that you should not adjust your credences because of your epistemic peers' views, the other thinks that you should adjust your credence so that it's between your original credence and your peer's. (Call the second view the E view, for Elga, as in the previous post. Christensen and Feldman hold related views.)

My calculation was that the second expert should wind up with a credence of 0.586 in the E view, but I'm thinking that I made a mistake; it should be 2/3.

The question is this: According to the E view, how far do you need to go in adjusting your credences? After you've adjusted your credence to reflect your peer's view, must you then look around again, note that your peer still has a different credence, and adjust again? Or is the ideal credence an average of your peers' credences and the credence that you would have if not for the adjustment you'd performed?

I think it has to be the latter, for two reasons. The first is an absurdity result; if you have to keep adjusting your credence to reflect your peer's, then a stubborn peer could force everyone else's credence toward their own. If you start out with a credence of 1 in p, and my credence is 0, after the first adjustment yours is 0.5, after the second it's 0.25, and you get closer and closer to my credence of 0.

The second reason is more straightforward. After you've adjusted your credence and I haven't adjusted mine, you're basing your credence on more evidence than I am. Your credence is based on the opinions of all epistemic peers, while mine is only based on my own opinion. Since I'm ignoring some relevant evidence, I'm no longer your peer, and you don't have to adjust your view to account for mine. (But if you went back to your original credence, you'd be ignoring the same evidence, and I'd be your peer again, so you'd have to adjust back.) So, more precisely, your credence should be an average of the credence that all the peers would have without adjusting their credences.

[It's entirely possible that this is all addressed in the literature somewhere.]

In the previous calculation I said the following: Suppose that the original believer in E winds up with credence c in E. According to the methods endorsed by the E view, their credence in E should be c/2: midway between their credence and the credence (0) of the other peer. But now it seems that, according to the E view, their credence should still be 0.5: Midway between the other peer's credence and the credence (1) that they would have in E without the adjustment for the other peer's credence.

The weighted sum calculation relies on the idea that someone with a credence of c in E should give a credence to another proposition, P, as follows: If according to the methods of E they should give credence x to P, and according to the methods of not-E say they should give credence y to P, then their credence in P should be cx + (1 - c) * y. Now, according to the methods of not-E, they should trust their own judgment of the arguments for E and give credence 1 to E; according to the methods of E they should give credence 0.5 to E. Note that these are not conditional probabilitles (thanks to David Christensen here). According to the weighted sum, the credence in E, c, should be c * 0.5 + (1 - c) *1. Solving c = c/2 + 1 - c, we get c = 2/3.

Incidentally, no matter how many experts there are opposing E this calculation can never push credence in E below 0.5. Suppose there are infinitely many peers opposed to E, so that the methods of E dictate that your credence in E should be 0. Since the methods of not-E dictate that your credence in E should be 1, the weighted sum equation is c = c * 0 + (1 - c) * 1; solving we get c = 1/2.

Thanks also to Mike Almeida in Brian Weatherson's comments; I still owe him an answer to his last comment. [UPDATE: Not anymore.]

UPDATE: This answer should generalize pretty easily; the other answer should generalize less easily.

Suppose that there are two experts, A and B; A begins with a credence of x in EW, B begins with a credence of y in EW. (That is, these are the credences each has when they don't take the other's credence into account.) Then according to the methods of EW, each should have a credence in EW of (x + y)/2; according to the methods of not-EW, each should have their original credence in EW. So if B's credence in EW is c, the weighted sum for B's credence in EW is c * (x+y)/2 + (1 - c) * y. Solving c = c * (x+y)/2 + (1-c)*y, we get y = c + cy - c * (x+y)/2, or c = 2y/(2 + 2y - (x+y)) = 2y/(y - x + 2). Say x = 0, y = 1/2, then c = 2/5; say x = 1/4, y = 3/4, then c = 3/5. If x = 3/4, y = 1/4, then c = 1/3. Of course if y = 0 then c = 0.

Working out the other answer seems like it'll involve simultaneous linear equations, which I'm not up for at the moment.

Posted by Matt Weiner at January 9, 2007 08:49 PM