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Old 07-01-2006, 11:42 PM
AaronBrown AaronBrown is offline
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Default Re: The envelope problem, and a possible solution

[ QUOTE ]
I still don't understand what these two "approaches" are. Here's one approach: we assume that
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[1] if A and B are events, then P(A | B) = P(A and B)/P(B),
[2] if A and B are independent events, then P(A | B) = P(A),
[3] if A and B are mutually exclusive events, then P(A or B) = P(A) + P(B),
[4] if A is an event, then 0 &lt;= P(A) &lt;= 1.[/list]As far as I can tell, if we assume these four things, then it cannot be the case that

P(Y = 2X | X = k) = 0.5

for all k.

What is the other approach? Which of these assumptions does it reject? Does it involve one of these other "alternative" probability theories you mentioned? Because as far as I can tell, the work of Shannon and Arrow-Debreu was within the context of Kolmogorov's formalism, and Savage's main departure from the formalism was his use of "measures" which were only finitely additive. But the above assumptions do not require countable additivity, nor do they even require sigma-algebras, only algebras. You also suggest that this other approach, whatever it is, is used every day in statistics. So are you saying that there are mainstream statisticians who reject one or more of the above assumptions on a daily basis?

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I did not mean to suggest that any of the five mathematicians above believe in switching envelopes. I mentioned them only to demonstrate we do not have consensus over the best mathematical way to handle probability. It's not that these camps think they have found mathematical errors in the others, it's that they think they have mathematics that correspond better to the natural idea of probability, which predates mathematical formalism.

I accept all four of your assumptions, and don't know anyone who does not. But how do they lead to the conclusion?

The only problem I know of with P(Y = 2X | X = k) = 0.5 for all k is that the expected value of X must be infinite. I understand there are technical complexities with dealing with distributions that allow this statement to be true, but that's not the same as proving it false. Even if you outlaw them entirely, that's only a formal resolution of the paradox, not a refutation of the force of its argument.

The two common approaches to resolution are to deny that the probabilities are equal, as you do, and to forbid computing the expected value of an unknown quantity, as Kolmogorov would.
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