On 8/26/2008 9:51 AM, Mark Leeds wrote:
Duncan: I think I see what you're saying but the strange thing is that if
you use the utility function log(x) rather than x, then the expected values
are equal.
I think that's more or less a coincidence. If I tell you that the two
envelopes contain X and 2X, and I also tell you that X is 1,2,3,4, or 5,
and you open one and observe 10, then you know that X=5 is the content
of the other envelope. The expected utility of switching is negative
using any increasing utility function.
On the other hand, if we know X is one of 6,7,8,9,10, and you observe a
10, then you know that you got X, so the other envelope contains 2X =
20, and the expected utility is positive.
As Heinz says, the problem does not give enough information to come to a
decision. The decision *must* depend on the assumed distribution of X,
and the problem statement gives no basis for choosing one. There are
probably some subjective Bayesians who would assume a particular default
prior in a situation like that, but I wouldn't.
Duncan Murdoch
Somehow, if you are correct and I think you are, then taking the
log , "fixes" the distribution of x which is kind of odd to me. I'm sorry to
belabor this non R related discussion and I won't say anything more about it
but I worked/talked on this with someone for about a month a few years ago
and we gave up so it's interesting for me to see this again.
Mark
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Duncan Murdoch
Sent: Tuesday, August 26, 2008 8:15 AM
To: Jim Lemon
Cc: r-help@r-project.org; Mario
Subject: Re: [R] Two envelopes problem
On 26/08/2008 7:54 AM, Jim Lemon wrote:
Hi again,
Oops, I meant the expected value of the swap is:
5*0.5 + 20*0.5 = 12.5
Too late, must get to bed.
But that is still wrong. You want a conditional expectation,
conditional on the observed value (10 in this case). The answer depends
on the distribution of the amount X, where the envelopes contain X and
2X. For example, if you knew that X was at most 5, you would know you
had just observed 2X, and switching would be a bad idea.
The paradox arises because people want to put a nonsensical Unif(0,
infinity) distribution on X. The Wikipedia article points out that it
can also arise in cases where the distribution on X has infinite mean:
a mathematically valid but still nonsensical possibility.
Duncan Murdoch
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