Re: Bayesian boxes and expectation value

From: <GSLevy.domain.name.hidden>
Date: Thu, 20 May 1999 02:57:25 EDT

In a message dated 99-05-19 18:25:27 EDT, Jacques Mallah writes:

<< Regardless, once we fix some quantity as the one we are
 considering, we have to use the usual formulas.

> I still maintain that the scale issue in the Bayesian problem comes in the
> back door and must be dealt with. In the Bayesian case, the problem itself
> defines the scale to be used. And selecting a logarithmic scale (for
example)
> guarantees that the expectation value of the box not chosen is exactly
equal
> to m, which agrees with common sense.

        I hope you now see the error in your above paragraph. The
log of the expectation value of the amount in the box is not equal to the
expectation value of the log of the amount.>>


MMM.... there is nothing better than an experiment to resolve the issue one
way or another.

I first thought of making presenting you with a Bayesian experiment selecting
m from a set of natural numbers ranging from 0 to M. The problem with this
approach is that 2m will always be even while m will have a 50% chance of
being even. So it will be easy for the chooser (you) to decide on average
which one is the m box and which one is the 2m box.

I could force the both boxes to be odd by redesigning the experiment as
follows: m is always picked from the set of odd natural numbers in the range
from 1 to M ( where M is odd). And the second box is filled with 2m+1 dollars
to insure that it is also odd. Now both boxes are odd and you have no way of
knowing which box contains more money than the other one. One more thing: m
is always in whole dollars.

Now let me present you with the following four tests, each test using a
different range M which is unknown to you.

Test 1. You open one of two boxes and find one dollar. What do you do? Pick
the other box? How much are you expecting to win if you do? (Answer is
obvious)

Test 2. You open one box and find three dollars. What do you do? How much are
you expecting to win if you do?

Test 3. You open one box and find 87 dollars. What do you do? How much are
you expecting to win if you do?

Test 4. You open one box and find 782364981238562094058281094092365745841
dollars. What do you do? How much are you expecting to win if you do?


Responding to Test 1 was easy because you knew that m ranges from 1 to M. The
probability of finding 2m+1 in the other box was one.
Responding to Test 2 was more difficult, but given the low value (3) found in
the box, it may be better to switch. However I do not know how to quantify
this.
Test 3 and 4 have much higher values..., however using the same reasonning,
it may be argued that it is better to switch. Again I do not know how to
quantify this.

So far it appears that you may be right.. It may be preferrable to switch on
the basis of information gleaned from the fact that our distribution has a
lower limit at 1.

So let's pick a distribution with no definite lower limit. Let's pick the
positive set of real numbers but EXCLUDE zero. Any number above zero will do.
There is no explicit "last digit" visible to the chooser (you) since real
numbers generally require an infinite number of digits after the decimal
point.

Now let's repeat the four tests:

Test 1a: You pick a box and find 0.0000001412349886934509831341231..... What
do you do and how much is your expected return?

Test 2a: You pick a box and find 3.873480139834690210093.... What do you do
and how much is your expected return?

Test 3a: You pick a box and find 87.2230984930987319030379021.... What do you
do?

Test 4a: You pick a box and find
782364981238562094058281094092365745841.3459024948240482874559425.....
What do you do?

In the last four tests the lack of "reference" such as a boundary lets you
float along what I feel to be a logarithmic scale. There does not appear to
be any way to take a "FIX" as they say in the field of navigation. There does
not appear to be any rational for switching. Or is there? One may argue that
any distribution that my mind is capable of generating is centered around 1
and on that basis there may be an extremely small advantage in switching when
the number is below one and not switching when the number is above one. I
could certainly counter this strategy by willingly biasing myself to generate
distributions centered above one. And you knowing what I know will adapt your
strategy too. In the limit the end result will be no advantage in switching.
Certainly the simple minded expectation value of 0.5(2m + 0.5m) = 1.25m for
switching does not make any sense. If you believe that it does make sense, I
think I just have found a way to make a quick buck.

George :-)
Received on Wed May 19 1999 - 23:59:42 PDT