[R] Two envelopes problem

markleeds at verizon.net markleeds at verizon.net
Tue Aug 26 20:29:54 CEST 2008 

  beautiful. now, i think i got it. the fact that  the calculation works 
in the additive case doesn't make the calculation correct.  the expected 
value  calculation is kind of assuming that the person putting the 
things in the envelopes  chooses what's in the second envelope AFTER 
knowing what's already in the other one. But, if he/she does know and 
still chooses half or double with 50/50 chance, then one can't use the 
original wealth as the reference point/ initial expected value because 
the envelope chooser isn't using it.  I think  it's best for me to think 
about it like that but your example is good to fall back on when I get 
confused again 2 years from now. Thank you very much for your patience 
and explanation.









On Tue, Aug 26, 2008 at  2:06 PM, Duncan Murdoch wrote:

> On 8/26/2008 1:12 PM, markleeds at verizon.net wrote:
>>   Duncan: Just one more thing which Heinz alerted me to. Suppose that 
>> we changed the game so that instead of being double or half of X,
>> we said that one envelope will contain X + 5 and the other contains 
>> X-5. So someone opens it and sees 10 dollars. Now, their remaining 
>> choices
>> are 5 and 15 so the expectation of switching is the same = 10.  So, 
>> in this case, we don't know the distribution of X and yet the game is 
>> fair.
>
> No, you can't do that calculation without knowing the distribution. 
> Take my first example again:  if X is known in advance to be 1,2,3,4, 
> or 5, then an observation of 10 must be X+5, so you'd expect 0 in the 
> other envelope.  The conditional expectation depends on the full 
> distribution, and if you aren't willing to state that, you can't do 
> the calculation properly.
>
> Duncan Murdoch
>
>
>
>> This is
>> why , although I like your example, I still think the issue has 
>> something to do with percentages versus additions.
>>
>> In finance, the cumulative return is ( in non continuous time )  over 
>> some horizon  is a productive of the individual returns over whatever 
>> intervals
>> you want to break the horizon into. In order to make things nicer 
>> statistically ( and for other reasons too like making the assumption 
>> of normaility somkewhat more plausible ) , finance people take the 
>> log of this product in order to to transform the cumulative return 
>> into an additive measure.
>> So, I think there's still something going on with units as far as 
>> adding versus multiplying ? but I'm not sure what and I do still see 
>> what you're saying in
>> your example. Thanks.
>>
>>  Mark
>>
>>
>>
>> On Tue, Aug 26, 2008 at 11:44 AM, Mark Leeds wrote:
>>
>>> 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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>>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
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>>>
>>> ______________________________________________
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