Interpret Bayesian probability as frequentist probability?

Question

It is usually said that the Bayesian probability is a subjective concept, quantifying one's degree of belief in something, while the frequentist probability is the the fraction of certain outcomes when observation is conducted many times (either in space or in time). My simple questions are: 1. If we consider statistics as a science, instead of a philosophy or even a psychology, isn't it necessary that a statistical prediction (or the outcome of a statistical calculation) be tested in the real world (at least in principle), as physical sciences usually do? 2. If statistical calculations need to be tested (at least in principle), is the method of test the same as physical sciences, namely, through experiments and observations? My thinking is that if the answer to both questions are yes, then the Bayesian probability can be interpreted as frequentist, and ultimately the two are not so different. _You may stop reading here and give your comments._ For example, suppose a geologist says that "there is a 0.9 probability that dinosaur was extinguished by a meteorite impact some 70 million years ago", a Bayesian may interpret the number 0.9 as the "degree of confidence" of this geologist. To me, this "degree of confidence" is too "void" to understand. I would prefer to understand it as something like the following: Given the current knowledge of the world, especially of the paleontological and geological discoveries, if there is a God who can create the universe for many times, then among the Earths (within these universes) that have been created and share the same features as our present Earth, dinosaurs are killed by meteorite impact in around 90% of them. Well, of course, apparently there is not a God who is willing to do such kind of "experiment" for us (we may observe many extraterrestrial worlds instead, though that's also extremely difficult if not impossible in the imaginable future). But without such kind of "thought experiment", I can hardly understand what does the number 0.9 mean, and, how to prove that it is correct. If the number 0.9 is not testable, even in principle or in imagination, or it is not meant to be tested, then I think it should not be spoken out in the first place. But if you agree that "degree of confidence" also needs to be tested, then isn't it equivalent to the frequentist understanding of probability? Note that in this example, the prior knowledge of paleontological and geological discoveries simply helps selecting out a subsample of the (hypothetical) universes in the "frequentist" approach to probability. I mean, if I were a frequentist (I am not sure whether I am, since I don't quite understand yet), I would not mind making use of prior knowledge, since using prior knowledge is not in contradictory with my "belief" in the frequentist approach.

Answer 1

The "subjective" initial apriori probability in Bayesian statistics is like the null hypothesis in frequentist statistics.

With Bayes Theorem, you get sort of an average of the initial hypothesis & the alternative hypothesis.

Bayes moves away more slowly from Ho, while the frequestist approach rejects Ho and replaces it with Ha.

Bayes sort of averages them to get (Ho+Ha)/2

Frequentist probability is inductive logic. Bayesian probability is deductive logic P(H/E)P(E)=P(E/H)P(H) is a logical tautology.

Take the Monty Hall paradox. You're shown 3 doors, one has a prize behind it, the other 2 have goats. You choose door A. Monty opens door B showing a goat. Should you switch to door C? The frequentist statistical conclusion, such as computer generated with random numbers, says yes, switch as you'll get the prize 2 out of 3 times.

But Bayes Theorem says it doesn't matter. P(C/~B)P(~B)=P(~B/C)P(C)

P(~B)=2/3. P(C) apriori = 1/3 before Monty opens B

P(C/~B)= probability C has the prize if B doesn't

P(~B/C)=1 If C has the prize, B doesn't for certain

Plug them all together:

(1/2)(2/3) = (1)P(C/~B)

1/3 = P(C/~B)

the probability hasn't changed

the frequentist statistics have a hidden assumption, that Monty deliberately selects a door without the prize

The Bayes Theorem assumes Monty selects a door at random