Re: [R] Two envelopes problem

From: Mark Leeds <markleeds_at_verizon.net>
Date: Tue, 26 Aug 2008 11:44:19 -0400


Hi Duncan: I think I get you. Once one takes expectations, there is an underlying assumption about the distribution of X and , in this problem, we don't have one so taking expectations has no meaning.

If the log utility "fixing" the problem is purely just a coincidence, then it's surely an odd one because log(utility) is often used in economics for expressing how investors view the notion of accumulating capital versus the risk of losing it. I'm not a economist but it's common for them to use log utility to prove theorems about optimal consumption etc.

Thanks because I think I see it now by your example below.

                                           Mark





-----Original Message-----
From: Duncan Murdoch [mailto:murdoch_at_stats.uwo.ca] Sent: Tuesday, August 26, 2008 11:26 AM
To: Mark Leeds
Cc: r-help_at_r-project.org
Subject: Re: [R] Two envelopes problem

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: 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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