[R] Two envelopes problem

[Heinz Tuechler]

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: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
>Behalf Of Duncan Murdoch
>Sent: Tuesday, August 26, 2008 8:15 AM
>To: Jim Lemon
>Cc: r-help at 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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