For a non-expert, what's different between Bayesian and frequentist approaches?

Re:For a non-expert, what's different between Bayesian and f... by raverbashing on Friday February 12, 2016 @08:19AM (#11086951)

Yes, what I mean is that, if you got a "yes" from the detector, it either means that:

- The sun hasn't gone nova and the detector is lying (with 1/36 chance)

- The sun has gone nova and the detector is not lying (negligible)

Now, given the result, one can calculate the probability of which of the cases

And the betting aspect of it makes sense as well

[ Parent | on HN ]

Re:For a non-expert, what's different between Bayesian and f... by bonoboTP on Friday February 12, 2016 @11:11AM (#11088215)

The "1/36 chance" and the "negligible" aren't the corresponding/analogous probabilities. 1/36 is P(detector "yes" | not nova). The "negligible" is P(nova).

What we want to calculate is P(nova | detector "yes"). Using Bayes' rule it's

P(nova | detector "yes") = P(detector "yes" | nova) P(nova) / (P(detector "yes" | nova) P(nova) + P(detector "yes" | not nova) P(not nova)) = 35/36 * P_nova / (35/36 * P_nova + 1/36 * (1-P_nova))

P(nova | detector "yes") > 0.5 if P_nova > 1/36.

This is also quite intuitive. Unless going nova is more likely than malfunctioning, you should bet on malfunctioning. This is very much like David Hume's quote:

"no testimony is sufficient to establish a miracle, unless the testimony be of such a kind, that its falsehood would be more miraculous, than the fact, which it endeavors to establish." (test: would you say the same amount of "holy crap! WTF!!" if it turned out that the testimony was false, compared to if it turned out that the miraculous thing actually happened).

So: no detector "yes" is sufficient to establish the sun going nova, unless the detector saying "yes" be such, that it lying (1/36) would be more miraculous, than the sun going nova.

It's fascinating how precisely this can be formulated and how clear it is with probability theory.

[ Parent | on HN ]