Re: Bayesian boxes and Independence of Scales

From: <GSLevy.domain.name.hidden>
Date: Thu, 13 May 1999 00:00:26 EDT

In a message dated 99-05-12 07:45:08 EDT, Gilles Henri wrote:

<<
May be but how do you justify to take a geometrical average? And what if
you assume m and m^2 in the boxes instead of m and 2m?

I think the problem is that m is not restricted and can take any value from
zero to infinity (even with discrete values). So actually the probability
of finding any particular value of m is strictly zero, which does not make
sense from the beginning. >>

and Jacques Mallah wrote

<< If you want to assume that m was initially drawn from a
distribution that is uniform for log m on (-00,00), so far so good. But
that's completely different from what you did above with the geometric
average.
Suppose you make that assumption. Then the value of m will most
likely be either extremely small or extremely large. That's the source of
the apparent paradox: 1.25 x = x is OK if x = 0 or x=00. >>

Gilles and Jacques, we are going in opposite direction in our reasoning and
this is the reason we don't meet. You are using conventional logic, and I am
using anthropic logic.

Assumption 1)
I start out by assuming that m is what it is, however improbable it is From
an outside observer, the probability that m is what it is approaches zero. In
fact the range of choice for m is the whole set of real numbers.However,
this is the m that we observe. (ANTHROPY) and, given this m, the probability
of me seeing this particular m is one.

Assumption 2)
It is common sense that it does not make any sense to switch. (I am willing
to play some money gambling games with you if you disagree on this one). The
expected value of the money in the other box is m. Because of the PERFECT
SYMMETRY of the situation, there is NO REASON for the expected value in the
other box to be other than m. Lack of causes for the boxes to be different
translates into perfect symmetry and the equivalence in the expected value
between the two boxes. The laws of physics are the same in all reference
systems (after choosing box A or after choosing box B)

Conclusion: The distribution of m depends on the problem and must be such
that the expected value for the other box is equal to m. For the problem
0.5m and 2m the distribution MAY BE logarithmic (sufficient condition) but
there may be other distributions. For the problem sqrt(m) and m**2, the
distribution is such that
fn( sqrt(m)) + fn( m**2) = m. I am too lazy to work out all the possible
functions that fn could be.
In other words the problem itself defines the distribution.

<< In any case it is easily proved that it's false, since if you were
immortal you would expect to be very old.>>

It all depends how you measure my age. In fact, my lifeline extends
uninterrupted probably four billions years (or possibly more), since it first
appeared on earth, and we may even be cousins. And by the way, your lifeline
also extends that much. Joyeux Anniversaire! Happy Birthday! :-)

George
Received on Wed May 12 1999 - 21:02:48 PDT

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