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).
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?
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.
Whaa?
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).
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?
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.