Newsgroups: bit.listserv.edstat-l
Date:         Wed, 7 Oct 1992 17:46:22 CDT
Reply-To:     thisted@GALTON.UCHICAGO.EDU
From:         thisted@GALTON.UCHICAGO.EDU
Subject:      Take this envelope...please!

Dick Darlington writes of a nice "fully Bayesian" analysis of the envelope
problem.  And Mary Ann Metzger gives a nice "general solution".

For my money (!), the envelope problem is a problem in part because it
appears to be well specified, but in fact is not.  One must make several
assumptions, and most of us are so eager to reach a conclusion that we
don't recognize that we have done so.  This situation is not so unusual
when it comes to statistical consulting--the client (often) has some data,
wants us to answer a particular (ill-posed) question, and it is tempting to
provide a stock, off-the-shelf answer that doesn't take much account of the
process by which the data were generated.

Both Bayesians and frequentists deal with problems of inference by
modelling what is known and what is observed about the data-generation
process.  The model should be a genuine attempt to describe the
data-generation process in so far as it is known.  Most of the discussions
of this problem have implicitly assumed a model.  It helps if one makes it
painfully clear what one is assuming.  For instance, Mary Ann Metzger
concludes that if the envelope contains an odd number of dollars, then the
optimal strategy is to switch.  The unstated assumption is that the
envelope contains only paper currency.  No one said that the envelope
couldn't contain coins -- we all just assumed it.  [And *I* just subtlely
misstated Mary Ann's argument, since she what she actually said was, "If Y
is odd, then switch.  An odd integer must always be the smaller amount."
She did not say that Y could not be measured in cents.]

With this in mind, let me tell you a story.

I observe two envelopes, both sealed.  My benefactor tells me that one
envelope contains twice the amount contained in the other.  Neither
envelope jingles.  I feel them both; they are even of identical thickness
(I have very sensitive fingertips). I weigh the envelopes, and they weigh
exactly the same.  I hold them up to the light, but unfortunately they are
opaque.  My benefactor tells me to choose an envelope, and she will let me
keep what I find inside.  I select one of the envelopes and open it to find
a crisp $100 bill.

My benefactor now offers to let me switch envelopes.  "Absolutely not!" I
cry.

Reasoning: The envelopes are of equal thickness.  The envelope that I
opened contained a $100 bill.  Therefore, the other envelope must also
contain exactly *one* bill, either a $50 or a $200.  If it contains the
$200 bill, I know that it is counterfeit, and if it contains a $50, I will
have lost half my fortune.

In effect, I am conditioning on additional information, namely the valid
denominations of US paper currency, and I am adopting a prior that severely
discounts the possibility that the other envelope might contain a 100 pound
sterling note (worth almost 2 x $100).

Dennis Lindley always maintains that it is incorrect to use Bayes's Theorem
in the form

      P(A|D) = P(D|A) x P(A)/P(D),

but rather

      P(A|D,H) = P(D|A,H) x P(A|H)/P(D|H)

where H represents everything (else) that one knows, in effect, the context
of the problem.  The form of the model comes in part from our attempts to
refine what we know of the context.

And, by the way, after I smugly tucked the $100 bill into my wallet, my
benefactor smiled enigmatically and then opened up the other envelope.
From it, she extracted a certified check in the amount of two hundred
dollars.


Ron Thisted
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