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From: Christopher Maloney <dude.domain.name.hidden>

Date: Sun, 29 Aug 1999 23:15:18 -0400

"Jacques M. Mallah" wrote:

*>
*

*> On Wed, 18 Aug 1999, Christopher Maloney wrote:
*

*> >
*

*> > Consciousness "flowing" is perhaps a bad image. I mean nothing more than
*

*> > that certain observer moments are related by some sort of identity function
*

*> > to other observer moments.
*

*>
*

*> I have described my view. I see no random identity functions.
*

*> The only thing that even resembles an identity function in my view is that
*

*> an extended computation may occur which gives rise to several observer
*

*> moments, but in that case all copies are equally related to the original.
*

*> Your so called identity function sounds like hidden variables,
*

*> related to the many-minds type.
*

*>
*

*> [re: Tegmark's formula]
*

*> > It's very simple, although I guess I haven't written it out in gory
*

*> > detail before. Tegmark's formula "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) ]
*

*>
*

*> What is the meaning of this formula? If P(Bi) is the effective
*

*> probability that an observer would see Bi, then Bi is a label for all the
*

*> things an observer could see. 'A' is then a particular characteristic
*

*> that an observer-moment can have. But then P(A|Bi) is either zero or one,
*

*> which is not what I think you meant. Explain.
*

A and Bi are labels used to indicate states of an SAS. For example, let's

define

A == remembers having just seen the Ace of Spades,

B1 == remembers having just seen an Ace,

B2 == remembers having just seen a card that is not an Ace.

B3 == does not remember having just seen a card.

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".

In the above example, P(A|B1) = 1/4, P(A|B2) = 0, P(A|B3) = 0. With an

appropriate restriction of our domain, we might guess that P(B3) = 1/2,

then if we assume a random draw from a standard deck of cards,

P(B1) = 1/26, P(B2) = 12/26. Then the formula says

P(A) = P(A|B1)P(B1) + P(A|B2)P(B2) + P(A|B3)P(B3)

= 1/4 * 1/26 + 0 + 0 = 1/104.

*>
*

*> > I want to compute P(H, t3). Let B1 be (H, t1) and B2 be (T, t1), then
*

*> >
*

*> > P(H, t3) = P(H,t3 | H,t1) P(H,t1) + P(H,t3 | T,t1) P(T,t1)
*

*> >
*

*> > We assume in this option (4) that P(H,t1) = P(T,t1) = 1/2.
*

*>
*

*> 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.

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.

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).

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.

And that's why I think this thought experiment is CENTRAL to the whole

topic that we are talking about. Granted, we don't have copy machines.

But death is in some sense very real (we see it all around us), and it

has the converse affect of a copy machine. Yet we still propose to

make subjective predictions about the future based on each of our

past observations.

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?) 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

heads at t3?

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.

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.

*> >
*

*> > But you said it yourself, if I remember a random bit, "it cuts the total
*

*> > measure of each type of [me] in half." I don't know what you mean by
*

*> > "type". I'd say it cuts the measure of me in half. That's exactly my
*

*> > point.
*

*>
*

*> The bit is either 0 or 1. If you don't know the value, let your
*

*> measure be M_d. If you know it's 0, you have measure M_0; if you know
*

*> it's 1, M_1. So M_d = M_0 + M_1. Suppose there are two guys, Dummy and
*

*> Ben. Dummy is ignorant and has M_d. Ben knows the bit and the versions
*

*> of him that saw 1 have M_1, the versions that saw 0 have M_0. A randomly
*

*> chosen observer is equally likely to be Dummy or to be one of the Bens.
*

Yes, I think you're right about this. The expectation value for my memory

should be right around the median.

*>
*

*> > [JM wrote]
*

*> > > The Omega Point CRAP is disproven because the universe is open.
*

*> > > (CRAP=causally retroactive anthropic principle)
*

*> >
*

*> > Cute! But I thing the jury is still out. I probably shouldn't use the
*

*> > term "Omega-Point", since that refers to a specific theory of the infinite
*

*> > future evolution of the universe. I sometimes use it to refer in general
*

*> > to the infinite future in which life continues.
*

*>
*

*> 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.

Date: Sun, 29 Aug 1999 23:15:18 -0400

"Jacques M. Mallah" wrote:

A and Bi are labels used to indicate states of an SAS. For example, let's

define

A == remembers having just seen the Ace of Spades,

B1 == remembers having just seen an Ace,

B2 == remembers having just seen a card that is not an Ace.

B3 == does not remember having just seen a card.

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".

In the above example, P(A|B1) = 1/4, P(A|B2) = 0, P(A|B3) = 0. With an

appropriate restriction of our domain, we might guess that P(B3) = 1/2,

then if we assume a random draw from a standard deck of cards,

P(B1) = 1/26, P(B2) = 12/26. Then the formula says

P(A) = P(A|B1)P(B1) + P(A|B2)P(B2) + P(A|B3)P(B3)

= 1/4 * 1/26 + 0 + 0 = 1/104.

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.

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.

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).

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.

And that's why I think this thought experiment is CENTRAL to the whole

topic that we are talking about. Granted, we don't have copy machines.

But death is in some sense very real (we see it all around us), and it

has the converse affect of a copy machine. Yet we still propose to

make subjective predictions about the future based on each of our

past observations.

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?) 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

heads at t3?

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.

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.

Yes, I think you're right about this. The expectation value for my memory

should be right around the median.

I have made a refutable prediction: that we necessarily will find ourselves

in a universe that will support eternal life.

-- Chris Maloney http://www.chrismaloney.com "Donuts are so sweet and tasty." -- Homer SimpsonReceived on Sun Aug 29 1999 - 20:33:26 PDT

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