mcoram comments on Alternative to Bayesian Score - Less Wrong

6 Post author: Coscott 27 July 2013 07:26PM

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Comment author: mcoram 11 February 2014 11:53:54PM *  0 points [-]

There's no math error.

Why is it consistent that assigning a probability of 99% to one half of a binary proposition that turns out false is much better than assigning a probability of 1% to the opposite half that turns out true?

I think there's some confusion. Coscott said these three facts:

Let f(x) be the output if the question is true, and let g(x) be the output if the question is false.

f(x)=g(1-x)

f(x)=log(x)

In consequence, g(x)=log(1-x). So if x=0.99 and the question is false, the output is g(x)=log(1-x)=log(0.01). Or if x=0.01 and the question is true, the output is f(x)=log(x)=log(0.01). So the symmetry that you desire is true.

Comment author: Decius 12 February 2014 04:55:30AM 0 points [-]

But that doesn't output 1 for estimates of 100%, 0 for estimates of 50%, and -inf (or even -1) to estimates of 0%, or even something that can be normalized to either of those triples.

Comment author: mcoram 12 February 2014 05:03:48AM 0 points [-]

Here's the "normalized" version: f(x)=1+log2(x), g(x)=1+log2(1-x) (i.e. scale f and g by 1/log(2) and add 1).

Now f(1)=1, f(.5)=0, f(0)=-Inf ; g(1)=-Inf, g(.5)=0, g(0)=1.

Ok?

Comment author: Decius 13 February 2014 12:56:20AM 0 points [-]

Huh. I thought that wasn't a Bayesian score (not maximized by estimating correctly), but doing the math the maximum is at the right point for 1/4, 1/100, 3/4, and 99/100, and 1/2.

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