[jason1990]

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Bayesians define probability as subjective belief, frequentist by repeated experiment. Both are equally rigorous mathematically, but both have severe technical problems.

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All statisticians, whether they call themselves Bayesians or Frequentists, use the same formal mathematics, which is simply measure theoretic probability theory. I don't know of any severe technical problems with that branch of mathematics. Perhaps you are talking about philosophical problems.

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A Bayesian would say the probability that you have the higher amount in your envelope is 50% before you open it, but since the amount you see will change your subjective estimate, it will be different from 50% after you look. That is consistent and leads to the commonsense result, but it's hard to deal with probabilities that change based on unobservable mental states.

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The probabilities change based on concrete observable information. To the Bayesian, the sum total of money in both envelopes is a random variable with whatever distribution he decided to use to model his beliefs. If he knew the total amount, the decision of whether or not to switch would be trivial. Once he sees the amount in one envelope, it is possible for that to give him information which would affect the probability distribution of that sum total. It is not difficult to deal with this change in probability, since it is given by the formulas for conditional probability. The only objection that can be made to this is to claim that the Bayesian had no right in the first place to declare the sum total to be a random variable with such-and-such a distribution.

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A frequentist would note that if you repeat the experiment many times, you end up with the larger amount 50% of the time

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What, exactly, is the frequentist repeating? Is he resealing the envelopes, shuffling them up, and picking them again? Then the contents are fixed, say n and 2n, and it is his choice which is random. Or is he being supplied with new envelopes every time? In this case, it is both the contents and his choice which may be regarded as random, and the frequentist cannot proceed with his analysis until he either determines or assumes something about the distribution of the random quantities inside those envelopes. However, once something is known about that, even the frequentist would agree that it is entirely possible for his observation of the first amount to persuade him to switch.

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Before we open our envelope, call the amount inside X.

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Do you mean for X to denote a number or a random variable? Even if you regard the amounts in the envelopes as fixed (say n and 2n), the amount in your particular chosen envelope is random. It is the random variable which is n with probability 0.5 and 2n with probability 0.5. So it seems that you are suggesting X is a random variable.

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We know that the other envelope has a 50% chance of holding 0.5*X and a 50% chance of holding 2*X.

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This sentence clearly indicates that the amount in the other envelope is being regarded as a random variable. So I am now even more convinced that you meant for X to be random. Moreover, if I assume that you meant for X to be a fixed number, then you just told me that there is a 50% chance the sum total of both envelopes is 1.5X and a 50% chance it is 3X. Hence, if X is a fixed number, then you are regarding the overall contents of the envelopes as random. However, you have not stated your assumed distribution on that random quantity, nor have you explained how that distribution implies that 1.5X and 3X are equally likely. So I seem to have no choice but to conclude that X is meant to be a random variable.

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Its expected holding is 1.25*X.

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Now wait just a minute. Expected value is a number, not a random variable. So it looks like you're regarding the contents of the chosen envelope as a fixed number. So, as above, you must be regarding the sum total of both envelopes as a random quantity. But, as stated above, you did not tell us your assumed distribution on that random quantity, nor did you explain how that distribution implies that 1.5X and 3X are equally likely. Perhaps you implicitly assumed that the sum total of both envelopes is an unknown quantity which is equally likely to be any positive number. If that's the case, then you have committed exactly the Bayesian fallacy I talked about in my other post.

Or perhaps you really did mean for X to be a random variable, and what you called "expected holding" was really supposed to be the conditional expectation. In that case, when you said "the other envelope has a 50% chance of holding 0.5*X", you were really talking about conditional probability, saying that P(Y=0.5X|X)=0.5, where Y is the contents of the other envelope. But if the sum of the envelopes is not random, then this statement is false, since P(Y=0.5X|X) is the random variable which is 1 if X=2n and 0 if X=n. If this is what you meant, then the problem lies in the fact that you used unconditional probabilities to compute a conditional expectation. In this case, the fallacy in what you wrote has nothing to do with Bayesians vs. Frequentists. It has to do with the problems that arise when we do mathematics informally, without proper regard for rigor. I suppose this is another lesson the Paradox can teach us. I hadn't thought about that.

I would very much like to see you write this down rigorously -- i.e., formally define your probability space, your probability measure(s), and your random variables. You claim to be demonstrating fundamental flaws in probability theory that stem from the Bayesian/Frequentist debate. But without a rigorous presentation, I fail to see how you've demonstrated anything other than (a) the uniform distribution fallacy, or (b) the fact that being inconsistent about what is random and what is fixed, and about what is conditional and what is not, can lead to the wrong answer.