# Re: zombie wives

From: Jacques M. Mallah <jqm1584.domain.name.hidden>
Date: Mon, 30 Aug 1999 22:10:48 -0400 (EDT)

On Sun, 29 Aug 1999, Christopher Maloney wrote:
> > > P(A) = Sum over i [ P(A|Bi) P(Bi) ]
>
> A and Bi are labels used to indicate states of an SAS. [e.g. i=1,2,3]
> Note that the set of Bi are mutually exclusive and exhaustive. That means
> P(B1 AND B2 AND B3) = 0, and P(B1 OR B2 OR B3) = 1. But the set of all Bi
> don't label "all the things an observer could see".

Of course. The word exhaustive mislead me, but I realized soon
after the mail what it must have meant. The above is fine for considering
a single observer-moment.

> "Jacques M. Mallah" wrote:
> > I guess your 'identity function' is coming into play. Otherwise
> > the above makes no sense ... unless you are not using an MWI coin flip.
> > Without the MWI, either the H branch exists or the T branch but not both.
> > In any case you have changed the game when bringing time into the picture.
> > Am I now to understand that what Tegmark meant to say is that
> > P(A,t2) = Sum over i [ P(A,t2|Bi,t1) P(Bi,t1) ]
> > which not the same at all as what you quoted him as saying.
>
> Actually, I didn't change the game, Tegmark did. I've re-read his paper,
> section III-B-2, "Making Predictions", and here's what he says:
>
> Suppose that a SAS at location L0 has perceived Y up until that
> instant, and that this SAS is interested in predicting the
> perception X that it will have a subjective time interval delta-t
> into the future, at location L1.
> ... for any mutually exclusive and collectively exhaustive set of
> possibilities Bi, the probability of an event A is given by
> P(A) = Sum over i { P(A|Bi)P(Bi) }. Using this twice, we find that
> the probability of X given Y is
> P(X|Y) = Sum over L0 { Sum over L1 { P(X|L1)P(L1|L0)P(L0|Y) } }
>
> He uses L0 and L1 to label "locations" in the entire ensemble of all
> universes and all SAS-observer-moments. L0 == (i0, j0, t0), where
> i0 labels the structure, j0 labels the SAS, and t0 labels the subjective
> time.

Some identity function is already clearly needed if P(X|Y) is to
mean what he said.
Clearly the L's do not label observables, except for t, while the
X and Y do.
So what does j0 mean? Possibly a label for an implementation of a
computation, but hard to say since his identity function is quite
different from the concept of identity I endorse.

> Tegmark clearly, in this line of reasoning, assumes some sort of
> (unspecified) identity function, and makes that explicit in his discussion
> of the second term above, P(L1|L0). This is supposed to be the probability
> that the SAS "will be", or will find herself, at L1, given that she is now
> at L0. He says:
>
> Since a SAS is by definition only in a single mathematical structure i
> and since t1 = t0 + delta-t, the second factor will clearly be of the
> form:
> P(L1|L0) = P(i1, j1, t1 | i0, j0, t0)
> = kd(i0, i1) Dd(t1 - t0 - delta-t) P(j1 | i0, j0)
>
> Where "kd" is the Kronecker delta function, and "d" is the Dirac delta
> function. So P here is a probability density function of six independent
> variables. We know we're interested in subjective times t0 and t1, so
> P(L1|L0) vanishes for every value of (t1 - t0) except delta-t.

OK. The first factor, kd(i0,i1) indicates that 'jumping' to
another 'world' is not possible, presumably regardless of whether the
structure 'i' you started on dissappears. So that's fine with me,
probably not fine with some of you.
The second factor is part of the definition of L, and probably
should be a Kronecker delta since subjective times are digital.
The third factor I do not understand. Apparently one can leap
from j0 to j1. (Supposedly, that is.) But when he says
> instant, and that this SAS is interested in predicting the
> perception X that it will have a subjective time interval delta-t

it seems that this third factor *should be* delta_j0,j1 since j
labels the SAS, and the SAS is said to be the one to have both of the
perceptions in question. Then j would be the identity function.
Due to copying, an extended computation can give rise to multiple
perceptions simultaneously, so if that is what he meant, he must have not
taken copying into account.

> The problem in the notation is exactly because of this unspecified
> identity function. The way I used it, I intended that P(H,t3|H,t1)
> should mean "the probability that she'll remember having seen heads,
> and that she now sees the clock at t3, given that when she saw the
> clock at t1, she remembered having seen heads".
> But in a strict reading of the notation, one would assume it to mean
> "the probability that she remembers having seen heads and the time
> is now t3, given that she remembers having seen heads and the time is
> now t1," which of course is 0 (the time cannot be both t3 and t1).

Right. That's where he changed the game.

> So you have to assume some sort of identity function for this sort
> of reasoning to work at all. Which is exactly what I was trying to
> say before, viz: In option 4, this sort of probabalistic reasoning is
> not possible.
>
> But I say that if we throw this out, then all is lost. ALL we are are
> able to do, in physics, is make predictions about future observations,
> based on past experiences. And all of these predictions are necessarily
> subjective, for each one of us.

Wrong! First, we are not aware of past experiences whatsoever.
We know only what our memory banks tell us now. That's all we need. We
can base beliefs about other times on that, but there's no way to prove
such theories. More to the point, we (or at least our brains!) can make
decisions, of which stating predictions is a special case, based on our
beliefs about the future plus our desires (~utility function).
Second, an observer is not the sort of thing that can *do*
anything. A computer can do things, and a computer gives rise to
observations, and those two facts are closely related in my opinion. But
one can not identify a computer as an observer because a *single* computer
can give rise to multiple different conscious observations simultaneously.
If one were to admit zombies as a possibilty (which I don't; they are a
useful concept though) the difference would be even more obvious.
So I don't see how an identity function, or lack thereof, could
make any difference. Computers will continue to function regardless of
such ideology.

> So let me re-assert what I mean by P(H,t3|H,t1), as
> "the probability that she'll remember having seen heads,
> and that she now sees the clock at t3, given that when she saw the
> clock at t1, she remembered having seen heads".
> And let me ask you, how can this not be 1? (Or at least as near as
> dammit?)

It is not a meaningful question, except in the sense of both
observations being derived from the same implementation as I discussed in
a previous post.

> If you like, forget about the copy machine and use a death
> machine instead: if the coin lands heads, then she'll have a 50/50
> chance of being killed. Should she expect a 1/2 chance of remembering
> having seen heads at t1, but a 1/3 chance of remembering having seen

Yes, but only if she's alive then.

> I suppose you would say yes, because of the absolute SSA. But then you'd
> have to conclude, I guess, that the reason we see things obey physical
> laws now is that the chance of our dying is relatively low.

What??? I have no idea what you mean by that. Of course, our
chance of dying is relatively low only because the particular laws we see
are relatively favorable to our continued existance. If all laws were
equally favorable predictions would be a lot harder to make.

> Anyway, the reason, I guess, I adhere to the QTI is that the concept of
> my own personal death just doesn't make any sense to me. My own death
> is by definition outside the realm of my own experience. This ties in
> directly with the discussion above, and with the strong sense I have that
> my identity continues through subjective time.

The boundary of a line segment is by definition outside the realm
of a line. Therefore all line segments continue indefinitely.

> []
> Yes, I think you're right about [memory vs. measure]. The expectation
> value for my memory should be right around the median.

Finally, at least one thing you learned.

> > Life will continue but with decreasing measure. Still it seems
> > that you can make a refutable prediction: namely, that the universe we are
> > in is not optimised for us to be here, but is optimised to give you a long
> > lifetime. Basically you are saying that what the measure ratio (say,
> > between two universes) will be in the future affects the measure ratio in
> > the present. For example a universe in which lives decay polynomially
> > would be favored over one in which they decay exponentially.
>
> I have made a refutable prediction: that we necessarily will find ourselves
> in a universe that will support eternal life.

That's not refutable. Thanks to the MWI, almost any universe
will. The prediction I suggested is much stronger and seems implied by

From: Christopher Maloney <dude.domain.name.hidden>
>But this doesn't make any sense! Define "old"! No matter what age you
>are, you could always ask why you don't find yourself to be "younger" or
>"older". When I'm 10,000 years old, I think I'll still consider myself
>young, so I'd still be able to reason on the basis that I don't find
>myself to be very old.

We have been over this *many* times. A) "Old" is when your finite
brain is too old to even know how old it is. B) In any case, "old" is
much older than the age that would be expected if QTI is false: In QTI
one should expect one's age *relative to* the other people you see to be
large, certainly *****not***** less than or of order 1! If you were
10,000 and everyone else was <100 at least you would have reason to supect
the rules might be different for you than for third parties.

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