Kindly comments on Original Research on Less Wrong - Less Wrong

21 Post author: lukeprog 29 October 2012 10:50PM

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Comment author: Kindly 31 October 2012 02:50:39PM *  0 points [-]

I'm not saying that frequentists believe this. I'm saying that the frequentist math (which computes Pr[E|H0]) is equivalent to computing Pr[H0|E] with respect to a prior distribution under which Pr[H0]=Pr[E]. Furthermore, this is a reasonable thing to look at, because from that point of view the way statistical significances combine actually makes sense.

Comment author: Eliezer_Yudkowsky 01 November 2012 04:41:59AM 1 point [-]

Pr[H0]=Pr[E]

Whaa?

Comment author: Kindly 01 November 2012 04:55:36AM 0 points [-]

Well, we have, in general, Pr[H0|E] = Pr[E|H0] * Pr[H0]/Pr[E]. Frequentists compute Pr[E|H0] instead of Pr[H0|E], but this turns out not to matter if Pr[H0]/Pr[E] cancels, which happens when the above equality holds.

From a certain point of view, this is just mathematical sleight of hand, of course. Also, the "E" is actually some class of outcomes that are grouped together (e.g. all outcomes in which 8 or more coins, out of 10, came up heads). But if we combine sequences of experimental results in the correct way, then this means that the frequentist and Bayesian result differ only by a constant factor (precisely the factor which we assumed, above, to be 1).

Comment author: Eliezer_Yudkowsky 01 November 2012 07:09:31AM 4 points [-]

Why the heck would the probability of seeing the evidence, conditional on the mix of all hypotheses being considered, exactly equal the prior probability of the null hypothesis?

Comment author: Kindly 01 November 2012 01:42:59PM *  1 point [-]

It wouldn't. Probably a better way to explain it would have been to factor their ratio out as a constant.

Anyway, I've totally messed up explaining this, so I will fold for now and direct you to a completely different argument made elsewhere in the comments which is more worthy of being considered.