From: John C Frain <frainj_at_gmail.com>

Date: Tue, 26 Aug 2008 19:57:47 +0100

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Tue 26 Aug 2008 - 19:00:22 GMT

Date: Tue, 26 Aug 2008 19:57:47 +0100

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@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
**>
**>
**>
**>
**> *******************************************************************
**> This email and any attachments are confidential. Any u...{{dropped:20}}
*

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