beautiful. now, i think i got it. the fact that the calculation works
in the additive case doesn't make the calculation correct. the expected
value calculation is kind of assuming that the person putting the
things in the envelopes chooses what's in the second envelope AFTER
knowing what's already in the other one. But, if he/she does know and
still chooses half or double with 50/50 chance, then one can't use the
original wealth as the reference point/ initial expected value because
the envelope chooser isn't using it. I think it's best for me to think
about it like that but your example is good to fall back on when I get
confused again 2 years from now. Thank you very much for your patience
and explanation.
On Tue, Aug 26, 2008 at 2:06 PM, Duncan Murdoch wrote:
On 8/26/2008 1:12 PM, [EMAIL PROTECTED] wrote:
Duncan: Just one more thing which Heinz alerted me to. Suppose that
we changed the game so that instead of being double or half of X,
we said that one envelope will contain X + 5 and the other contains
X-5. So someone opens it and sees 10 dollars. Now, their remaining
choices
are 5 and 15 so the expectation of switching is the same = 10. So,
in this case, we don't know the distribution of X and yet the game is
fair.
No, you can't do that calculation without knowing the distribution.
Take my first example again: if X is known in advance to be 1,2,3,4,
or 5, then an observation of 10 must be X+5, so you'd expect 0 in the
other envelope. The conditional expectation depends on the full
distribution, and if you aren't willing to state that, you can't do
the calculation properly.
Duncan Murdoch
This is
why , although I like your example, I still think the issue has
something to do with percentages versus additions.
In finance, the cumulative return is ( in non continuous time ) over
some horizon is a productive of the individual returns over whatever
intervals
you want to break the horizon into. In order to make things nicer
statistically ( and for other reasons too like making the assumption
of normaility somkewhat more plausible ) , finance people take the
log of this product in order to to transform the cumulative return
into an additive measure.
So, I think there's still something going on with units as far as
adding versus multiplying ? but I'm not sure what and I do still see
what you're saying in
your example. Thanks.
Mark
On Tue, Aug 26, 2008 at 11:44 AM, Mark Leeds wrote:
Hi Duncan: I think I get you. Once one takes expectations, there is
an
underlying assumption about the distribution of X and , in this
problem, we
don't have one so taking expectations has no meaning.
If the log utility "fixing" the problem is purely just a
coincidence, then
it's surely an odd one because log(utility) is often used in
economics for
expressing how investors view the notion of accumulating capital
versus the
risk of losing it. I'm not a economist but it's common for them to
use log utility to prove theorems about optimal consumption etc.
Thanks because I think I see it now by your example below.
Mark
-----Original Message-----
From: Duncan Murdoch [mailto:[EMAIL PROTECTED] Sent: Tuesday,
August 26, 2008 11:26 AM
To: Mark Leeds
Cc: r-help@r-project.org
Subject: Re: [R] Two envelopes problem
On 8/26/2008 9:51 AM, Mark Leeds wrote:
Duncan: I think I see what you're saying but the strange thing is
that if
you use the utility function log(x) rather than x, then the
expected
values
are equal.
I think that's more or less a coincidence. If I tell you that the
two envelopes contain X and 2X, and I also tell you that X is
1,2,3,4, or 5, and you open one and observe 10, then you know that
X=5 is the content of the other envelope. The expected utility of
switching is negative using any increasing utility function.
On the other hand, if we know X is one of 6,7,8,9,10, and you
observe a 10, then you know that you got X, so the other envelope
contains 2X = 20, and the expected utility is positive.
As Heinz says, the problem does not give enough information to come
to a decision. The decision *must* depend on the assumed
distribution of X, and the problem statement gives no basis for
choosing one. There are probably some subjective Bayesians who
would assume a particular default prior in a situation like that,
but I wouldn't.
Duncan Murdoch
Somehow, if you are correct and I think you are, then taking the
log , "fixes" the distribution of x which is kind of odd to me. I'm
sorry
to
belabor this non R related discussion and I won't say anything more
about
it
but I worked/talked on this with someone for about a month a few
years
ago
and we gave up so it's interesting for me to see this again.
Mark
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]
On
Behalf Of Duncan Murdoch
Sent: Tuesday, August 26, 2008 8:15 AM
To: Jim Lemon
Cc: r-help@r-project.org; Mario
Subject: Re: [R] Two envelopes problem
On 26/08/2008 7:54 AM, Jim Lemon wrote:
Hi again,
Oops, I meant the expected value of the swap is:
5*0.5 + 20*0.5 = 12.5
Too late, must get to bed.
But that is still wrong. You want a conditional expectation,
conditional on the observed value (10 in this case). The answer
depends on the distribution of the amount X, where the envelopes
contain X and 2X. For example, if you knew that X was at most 5,
you would know you had just observed 2X, and switching would be a
bad idea.
The paradox arises because people want to put a nonsensical Unif(0,
infinity) distribution on X. The Wikipedia article points out that
it can also arise in cases where the distribution on X has infinite
mean: a mathematically valid but still nonsensical possibility.
Duncan Murdoch
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