Mark
My experience was similarly frustrating. Maybe formulating the
problem a bit differently could help to clarify it.
State it like this:
Someone chooses an amount of money x. He puts 2x/3 of it in one
envelope and x/3 in an other. There is no assumption about the
distribution of x.
If you choose one envelope your expectation is x/2 and changing may
lead to a gain or a loss of x/6.
In my view there is no basis for a frequentist conditional
expectation, conditional on the amount in the first envelope. Of
course, after opening the first envelope and finding a, you know for
sure that x can only be 3a or 3a/2, but to me there seems to be no
basis to assign probabilities to these two alternatives.
I am aware of the long lasting discussion and of course this will not end it.
Heinz
At 14:51 26.08.2008, 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. 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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