Re: [R] Two envelopes problem

From: Mark Leeds <markleeds_at_verizon.net>
Date: Tue, 26 Aug 2008 09:51:52 -0400


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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