[R] Two envelopes problem

[S Ellison]

>>> Duncan Murdoch <murdoch at stats.uwo.ca> 26/08/2008 16:17:34 >>>
>>If this is indeed the case, switch; the expected gain is
>>positive because _you already have the information that you hold the
>> median value of the three possibilities_. The tendency when
presented
> >with the problem is to reason as if this is the case.

>No, you don't know that A is the median.  That doesn't even make
sense, 
>based on the context of the question:  there is no 3-valued random 
>variable here.  

This is my point; apologies if I put it badly. The fact is that you
_don't_ hold the median value and that this is indeed a silly way of
looking at it. My assertion was that this is the way many folk DO look
at it, and that this results in an apparent paradox.

In fact, you inadvertently gave an example when you said
>The unopened envelope can hold only two values, given 
>that yours contains A.
True - for a rather restricted meaning of 'true'. 
As written, it implicitly allows three values; A for the envelope you
hold, and two more (2A and 0.5A) for the alternatives you permit. The
usual (and incorrect) expected gain calculation uses all three; 2A-A for
one choice and 0.5-A for the other. To do that at all, we must be
playing with three possible values for the contents of an envelope. 

This clearly cannot be the case if there are only two possible values,
as we are told in posing the problem. The situation is that you hold
_either_ A _or_ 2A and the other envelope holds (respectively) 2A or A.
We just don't know what A is until we open both envelopes. So for a
given pair of envelopes, it is the choice of coefficient (1 or 2) that
is random.  

If I were to describe this in terms of a random variable I would have
to assume an unknown but - for this run - fixed value A multiplied by a
two-valued random variable with possible values 1 and 2, would I not? We
surely can't have both 0.5 and 2 in our distribution at the same time,
because the original proposition said there were only two possibilities
and they differ by a factor of 2, not 4. 

>You have no basis for putting a probability distribution on those 
>values, because you don't know the distribution of X.
But we do know; or at least we know the distribution of the
coefficient. We have two envelopes of which one is selected at random,
and the ratio of values is 2.  On that basis, assigning a 50:50
probability of ending up with A or 2A on first selection seems
uncontroversial. 

But I'm more than willing to have my intuition corrected - possibly
off-line, of course, since this stopped being R a while back!

Steve E




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