Re: momentary and persistent minds

From: Wei Dai <weidai.domain.name.hidden>
Date: Thu, 4 Jun 1998 16:58:47 -0700

On Fri, May 29, 1998 at 04:03:00AM +0000, Nick Bostrom wrote:
> It seems we can interpret "I will observe X" as meaning: "There is a
> future brain-state B2, similar in certain respects to the brain-state
> B1 which instanciates this present cognition C1, such that B2
> instanciates C2, and C2 includes an observation of X.".

That is not going to give you nice results. For example if there
is no wavefunction collapse, all possible brain-states exist in the future
and "I will observe X" would have probability 1 for all X under your
interpretation.

I think I've found a way to define these kinds of probabilities that is
consistent (it satisfies the axioms of probability theory) and still
somewhat intuitive. Suppose we have a set I of observer-instants, a
mapping G from observer-instants to their current perceptions and
memories, a theory of identity, i.e., a function F that maps each
observer-instant to the set of observer-instants that are identified with
its future, and a probability measure Q on the observer-instants.

Define:
P(I currently observe X)=Q({i|X is in G(i)})
P(I did observe X)=Q({i|X is in G(j) for some j such that i is in F(j)})
P(I will observe X)=Q({i|there exist some j in F(i) such that X is in G(j)})
P(I will observe X and I currently observe Y)=Q({i|there exist some j in
F(i) such that X is in G(j)} intersect {i|Y is in G(i)}
(and similarly for other kinds of conjunctions)

The basic idea is that "I will observe X" is interpreted as "A randomly
distributed observer-instant has the property that some future version of
it observes X", which seems obvious now that I think about it.

For an example, consider the setup in the single-coin copying paradox. We
have one observer-instant (a) at time 0 with measure 2/6, two
observer-instants (b,c) at time 1 with measure 1/6 each, three
observer-instants (d,e,f) at time 2 with measure 1/6 each. Also,
F(a)={b,c,d,e,f}, F(b)={d,e}, and F(c)={f}.
G(a)={clock reading 0}
G(b)={clock reading 1, coin landing heads, memories of being a}
G(c)={clock reading 1, coin landing tails, memories of being a}
G(d)={clock reading 2, heads-up coin, memories of being a and b}
G(e)={clock reading 2, heads-up coin, memories of being a and b}
G(f)={clock reading 2, tails-up coin, memories of being a and c}

According to the above definitions, P(I will see a heads-up coin and clock
reading 2|I currently see clock reading 0) = 1/2, whereas P(I see a
heads-up coin|I see clock reading 2) = 2/3. In general under this
definition P(I currently observe X|I observed Y) <> P(I will observe X|I
currently observe Y) whenever there is some possibility of copying or
death between observations X and Y.
Received on Thu Jun 04 1998 - 17:02:57 PDT