# Re: Request for a glossary of acronyms

From: Jesse Mazer <lasermazer.domain.name.hidden>
Date: Wed, 04 Feb 2004 18:19:46 -0500

Saibal Mitra wrote:
>
>This means that the relative measure is completely fixed by the absolute
>measure. Also the relative measure is no longer defined when probabilities
>are not conserved (e.g. when the observer may not survive an experiment as
>in quantum suicide). I don't see why you need a theory of consciousness.

The theory of consciousness is needed because I think the conditional
probability of observer-moment A experiencing observer-moment B next should
be based on something like the "similarity" of the two, along with the
absolute probability of B. This would provide reason to expect that my next
moment will probably have most of the same memories, personality, etc. as my

As for probabilities not being conserved, what do you mean by that? I am
assuming that the sum of all the conditional probabilities between A and all
possible "next" observer-moments is 1, which is based on the quantum
immortality idea that my experience will never completely end, that I will
always have some kind of next experience (although there is some small
probability it will be very different from my current one).

Finally, as for your statement that "the relative measure is completely
fixed by the absolute measure" I think you're wrong on that, or maybe you
were misunderstanding the condition I was describing in that post. Imagine
the multiverse contained only three distinct possible observer-moments, A,
B, and C. Let's represent the absolute probability of A as P(A), and the
conditional probability of A's next experience being B as P(B|A). In that
case, the condition I was describing would amount to the following:

P(A|A)*P(A) + P(A|B)*P(B) + P(A|C)*P(C) = P(A)
P(B|A)*P(A) + P(B|B)*P(B) + P(B|C)*P(C) = P(B)
P(C|A)*P(A) + P(C|B)*P(B) + P(C|C)*P(C) = P(C)

And of course, since these are supposed to be probabilities we should also
have the condition P(A) + P(B) + P(C) = 1, P(A|A) + P(B|A) + P(C|A) = 1 (A
must have *some* next experience with probability 1), P(A|B) + P(B|B) +
P(C|B) = 1 (same goes for B), P(A|C) + P(B|C) + P(C|C) = 1 (same goes for
C). These last 3 conditions allow you to reduce the number of unknown
conditional probabilities (for example, P(A|A) can be replaced by (1 -
P(B|A) - P(C|A)), but you're still left with only three equations and six
distinct conditional probabilities which are unknown, so knowing the values
of the absolute probabilities should not uniquely determine the conditional
probabilities.

>Let P(S) denote the probability that an observer finds itself in state S.
>Now S has to contain everything that the observer knows, including who he
>is
>and all previous observations he remembers making. The ''conditional''
>probability that ''this'' observer will finds himself in state S' given
>that
>he was in state S an hour ago is simply P(S')/P(S).

This won't work--plugging into the first equation above, you'd get
(P(A)/P(A)) * P(A) + (P(B)/P(A)) * P(B) + P(P(C)/P(A)) * P(C), which is not
equal to P(A). It would work if you instead used 1/N * (P(S)/P(S')), where N
is the total number of distinct possible observer-moments, but obviously
that won't work if the number of distinct possible observer-moments is
infinite. And as I said, this condition should not *uniquely* imply a
certain set of conditional probabilities given the absolute probabilities,
so even with a finite N this wouldn't be the only way to satisfy the
condition.

Jesse

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Received on Wed Feb 04 2004 - 18:25:03 PST

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