# Re: Bayesian boxes and Independence of Scales

From: Jacques M Mallah <jqm1584.domain.name.hidden>
Date: Fri, 14 May 1999 15:47:10 -0400

On Thu, 13 May 1999 GSLevy.domain.name.hidden wrote:
> 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.

So you believe in 'a new kind of logic', allowing your the conflict
between your beliefs and logic to continue without forcing you to abandon
the beliefs. How unoriginal. Every third crackpot does that.

> 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.

P(m|m) = 1. Hey, that's not unconventional yet. But wait, you've
thrown in the concept of an outside observer, apparently defined as one
who doesn't know m.

> 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.

Again, totally obvious and conventional, mentioned in the original
post that started this.

> 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.

Ok, now you seem to think that the expected value for the other
box is exp((log(2m)+log(m/2))/2) given that the first box contains m and
given a 50% chance that the second box contains 2m. Ok, that's
unconventional logic all right! Weird conclusions from unrelated
assumptions.

> << 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.

I see you're still in wackyland! I take it that you are your own
I think even your allies would part company with you here. At
least it would seem to imply that in 100 years, you would most likely have
no memory of being your current self who's reading this, right? If not it
would be very strange to find yourself on the early transition to
conventional immortality!
Higgo James not only believes he's immortal, he also believes that
he's less than a Plank time old. I thought that was strange, but you're
up there with him.
Anyway, 4*10^9 years is nothing. If you were really immortal, you
should expect to be a lot older than that.

- - - - - - -
Jacques Mallah (jqm1584.domain.name.hidden)