[R] Two envelopes problem

[John C Frain]

A very important point is missing here.  If there is x in one envelope
and 2x in the other the expected gain is 3x/2.  If the idea is to
switch after observing the second envelope the expected gain is again
3x/2.  In the case being put x will be either 5 or 10.  But x is a
parameter and in this case does not have a probability distribution.
Then one can not take an expectation with respect to x.

John Frain

2008/8/26 S Ellison <S.Ellison at lgc.co.uk>:
>
>
>>>> 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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