>
> On Mon, 15 Nov 1999, Russell Standish wrote:
> > > Given the measure distribution of observation-moments, as a
> > > function on observables (such as Y1 and X),
> > > p(Y1|X) = p(Y1 and X) / p(X)
> > > Not so hard, was it?
> > > [Note that here X was the observation of being Jack Mallah, and
> > > Y1 was basically the observation of being old. See previous posts on
> > > this thread if you want exact details of Y1; nothing else about it is
> > > relevent here I think.]
> >
> > ASSA doesn't give p(Y1 and X) either.
>
> Obviously, and as I've repeatedly said, some prescription for the
> measure distribution is also needed. That is true even to just get p(X).
>
> > > Huh? Why should p(not Y1, and X) = 0 ? Especially since my
> > > current observations are (not Y1, and X)!!!
> >
> > Your current observations are [sic] p(Y3|X), where Y3 = Jacques Mallah's
> > is observed to be young. Y3 is not equivalent to (not Y1). Just because
> > you see yourself young does not preclude seeing yourself old at a
> > later date!
>
> Here your misunderstanding is clearly exposed. The way I've
> defined p(A), it is the effective probability of an observation-moment
> with the property 'A'.
Oh dear - and I thought we were debating whether the RSSA is
consistent with Bayesian statistics. Now you revert to the ASSA, which
I quite accept is consistent.
> Definitions of identity, of 'me' or 'not me', are irrelevant to
> finding p(A). By definition, if my current observation is A, and A and B
> are such that it is not possible for the same observation-moment to have
> both, then I observe (not B).
> If you want to talk about the probability that, using some
> definition of identity that ties together many observation moments, "I"
> will eventually observe Y1 - that will depend on the definition of
> identity. It is NOT what I have been talking about, nor do I wish to talk
> about it until you understand the much more basic concept of the measure
> of an observer-moment.
I have no problem with the concept of observer moment. It appears you
have a problem with the concept of connecting up a set of such
observer moments into an observer. One cannot discuss QTI or RSSA
without doing this.
In light of our previous discussions, p(X) was defined as the
probability of being observer "Jacques Mallah", not the probability of
being observer "Jacques Mallah" at a particular observer moment.
There is an obvious normalisation problem with the usual model of
branching histories in MWI (I see from your signature you at least
accept that!). Since the total number of histories (belonging to say a
particular observer) is some exponentially growing function of time,
and extends indefinitely into the future, the total measure of an
observer is unnormalisable, without some renormalisation applied at
each "timestep" (which seems rather arbitrary - unless you've got some
better ideas). Your measure argument, which is a variation of the
Leslie-Carter Doomsday argument, implicitly relies on a normalised
measure distribution of observer moments. I seem to remember this
normalisation problem was discussed earlier this year, but I'm not
sure (without rereading large tracts of the archives)
Now, with RSSA, this normalisation problem is not an issue, as only
the relative measures between successive time steps is important, not
the overall measure.
There is a more important reason why the ASSA is
unbelievable. Basically, the ASSA implies that the first person view
of the world is identical with the third person (the observer moment I
am experiencing now is selected from precisely the same distribution
as other people's observer moments). There are many examples that show
the opposite (eg Tegmark's suicide experiment, Marchal's
"KILL-THE-USER" instruction) that are basically "Schroedinger's cat as
observer" variants.
None of this is in defence of QTI! It merely is to show that your
measure argument fails - unless you happen to be a Copenhagener :)
As I have mentioned before, in order for QTI to work, there must also be no
possible "cul-de-sac" branches. In Bruno's model logic, I believe this
would be expressed as
\forall \alpha \in W, \models_\alpha^W \neg \Box (\Box\top \wedge \Box\bot)
or in slightly more English notation (\top == true, \bot == false,
\wedge == and)
for every world alpha in the model W, there cannot be a successor
world that can only access a terminal world
where \Box p is trivially true in a terminal world, regardless of the
truth table of p, but \Box true and \Box false cannot both be true at
the same time.
Am I right Bruno? I'm testing out the modal logic I've been studying.
In anycase, I haven't got a clue as to how one might start proving
this, or working out under what conditions it might hold.
>
> - - - - - - -
> Jacques Mallah (jqm1584.domain.name.hidden)
> Graduate Student / Many Worlder / Devil's Advocate
> "I know what no one else knows" - 'Runaway Train', Soul Asylum
> My URL: http://pages.nyu.edu/~jqm1584/
>
>
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Dr. Russell Standish Director
High Performance Computing Support Unit,
University of NSW Phone 9385 6967
Sydney 2052 Fax 9385 6965
Australia R.Standish.domain.name.hidden
Room 2075, Red Centre
http://parallel.hpc.unsw.edu.au/rks
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Received on Sun Dec 05 1999 - 20:57:32 PST