[R] Two envelopes problem

Mark Leeds markleeds at verizon.net
Tue Aug 26 15:51:52 CEST 2008

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.


-----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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