Newsgroups: bit.listserv.edstat-l
Date:         Sun, 11 Oct 1992 23:22:54 EDT
Reply-To:     William Kahn <71020.1025@COMPUSERVE.COM>
From:         William Kahn <71020.1025@COMPUSERVE.COM>
Subject:      envelope

There are two envelopes, one with twice as much money as the other.  You
open one and find 100$.  What is the chance the other has $50?.

Answer: 1/3.  Bayesian proof (with apologies to Jerry Dallal :-) )

Let L be the amount in the lower envelope (L>=0).
Let p(x) be our prior on L.
Let U be the amount in the upper envelope (U>=0).
Let q(x) be our prior on U.
Let A(x) be our prior on amount in envelope A (randomly selected).
Then A(x)=.5p(x)+.5q(x).
Because U=2L, by transformation of random variables, q(x)=.5p(x/2).
So, A(x)=.5p(x)+.25p(x/2).
Let AB(x,y) be the joint distribution of the amounts in envelopes A and B.
Clearly AB has mass concentrated on two straight lines in the first
    quadrant, one line with slope 1/2, the other with slope 2.  The
    probability distribution along each line, as measured along each line,
    is the same.

      |              *
      |             *
      |            *
      |           *
      |          *
      |         *
Amount|        *
in B  |       *
      |      *                          *
      |     *                      *
      |    *                  *
      |   *              *
      |  *          *
      | *      *
      |*  *
      |_____________________________________________________
             Amount in Envelope A

Joint distribution apriori is concentrated identically on the two lines
above, slope 2 and 1/2.

Let d be the distance along one of the lines from the origin.
Let h(d) be the probability density.

Let AU(x) be the contribution to A(x) from the upper line.
Let AL(x) be the contribution to A(x) from the lower line.
A(x)=AU(x)+AL(x)

Let F=sqrt(1*1+2*2)=sqrt(5).

AL(x)=(F/2)*h(Fx/2) and AU(x)=F*h(Fx).
  Note use of Jacobian of transformation for each of the lines.

Then A(x)=(F/2)*h(Fx/2) + F*h(Fx).
So, .5p(x)+.25p(x/2) = (F/2)*h(Fx/2) + F*h(Fx).
So, h(x)=(1/2F)*p(x/F).
We now know the joint apriori distribution AB given the marginal apriori.

For any x=X, A(X) is the sum of two probabilities, AU(x) and AL(x).
That is, we can find 100$ because it is the lower envelope or the upper
envelope.  To determine the probability that it is the upper envelope we
need P = AU(x)/(AU(x)+AL(x))

Plug in our formulas for AU(x), AL(x) and h(x) to get

P = 2p(X)/(2*p(X)+p(X/2)).

This is the general probability that the observed value (100$) has come
from the upper envelope given any prior on the lower amount L.  If we
assume a vague prior for L, then p(x)=p(x/2) so P=2/3.

The envelope problem solved (at least for simple continuous priors).

Note that the essence of the whole game is the Jacobian of the
transformation.  If we are told the ratio twixt envelopes is 3 then the
Jacobians change (F becomes sqrt(10)) and we end up with P=3/4.  The
solution is properly symmetric in its arguments.

If envelope A has 100$ then we have 2/3 chance it is the larger one.  At
the same time if envelope B has 100$ then it has a 2/3 chance to be the
larger one.

The conditioning on a fixed amount makes it more likely to contain the
larger amount.  If the problem is stated as one envelope has 20$ more
than the other, the current solution technique still works: both lines
have slope=1 and P=1/2.

This is not a paradox evolving out of improper priors.  The essense
remains even if you have a uniform prior out to a finite value.

The key is that because of the ratio condition there are "more ways" the
upper envelope can give 100$ then can the lower envelope.  This has been
alluded to by previous authors who solved the integer version.

Is this
A proof?  (Ask an editor)
A convincing argument?  (Anyone else convinced?)
Scholarly research?  (Need to want to bother to push it past an editor)
Correct?  (Ahh-what is truth?)
Worth giving on an oral exam? (Sadistic satisfaction for all)
Fun for a class discussion? (Like wow--for sure man)
Explainable to spouses, children, and college kids? (Not in current form)

Bill Kahn
W. L. Gore and Associates
71020.1025@compuserve.com
Comments: Gated by NETNEWS@AUVM.AMERICAN.EDU