RE: normalization

From: Marchal <>
Date: Sat Jan 15 10:39:56 2000

This post is my answer to Jacques Mallah. I have send it yesterday
(saturday) to Jacques only, but that was due to an error of manipulation.
(apology to jacques).

But I realise also that my answer (last paragraph) is most probably
false. I think Christopher Maloney is plausibly right concerning the

My 'real' opinion is that such probability computations (which I do
not use in UDA!) are so
much conceptually difficult that it is necessary to approach the whole
problem through some conceptual tools (like Goedel's arithmetisation
and modal logics, for exemple).

I nevertheless send you my whole answer, but please keep in mind this

Note also the link between the SSA (ASSA, or RSSA) and the need to better
defined or represent notion like worlds, observer-moment, memories, etc.

Here is my not quite good answer to Jacques Mallah:

Jacques M. Mallah wrote:

> Actually that doesn't matter. For me a physical system is just a
>mathematical structure which "exists". So let all computations, for
>example, exist.
> But for me the term "implementation" has a special definition
>which I am still trying to formulate precisely, due to the implementation
>problem, so I'm still not sure how you distinguish between a computation
>and an implementation of a computation.

Fair enough.

> Well, first there is the question of what you mean by a 1st person
>expectation and by "yourself".

1st person expectation = expectation by an agent of his own
personal subjective experience.
Exemple: - a child who expect a nightmare.
         - a computationalist practioner who expect to be somewhere
           after a self-duplication.
         - a dog who expect his "master" giving him a bone.

"myself" : well, for a rough third person description, look at my
signature. For a first person description I'm afraid the best
approximation is you (from your first person point of view).

In fine, I modelise the first person by the knower, and I
modelise knowledge by the modal system S4 and S4Grz.

> There is no 1 of the 10 copies which is
>linked more than the others to the original copy. So depending on the
>definition I would either say that you are only at Brussels, or that you
>are at all three places.
> But for practical purposes (such as, if you knew only German,
>deciding which language to learn before being scanned, English or
>Russian), I would of course say that, after the reconstitution, your
>effective probability of seeing Moscow is 90%, so Russian is more
>"likely" to come in handy.


>> If you say that your expectation to find yourself at Maoscow is bigger
>> than the one to find yourself at Washington, then, by definition, I will
>> say that you link the measure with the number of (actual, here) running
>> computations.
> OK, not just link it, but say it is directly proportional
>(but see above). But my point is that if you do that you reject the
>RSSA. For example, suppose that a year later (on 1/1/2001), the copies at
>Moscow are destroyed except for one, while at Washington 8 new copies of
>the version of you that has been there are made. During 2000, what is the
>effective probability that you would be at Moscow? It is 90%; next year,
>it is 10%.
> But according to the RSSA that would not be the case. If you
>are 90% likely to be in Moscow in 2000, each type of version of you could
>only "flow into" its own continuation, so the eff. pr. of being at Moscow
>woulkd remain at 90% for 2001. (Chris Maloney would claim that the
>eff. pr. of being at Moscow would be 10% for both years).
> The ASSA, on the other hand, is naturally compatible with the
>simple rule that the effective probability is proportional to the number
>of copies. So do you, in fact, agree with that rule?

With RSSA? Yes. And you are correct, the "naive" probability remains 90%.
Even without immortality. With mortality the probability at Brussels that
I will die at Moscow in 2001 is 9/10 * 8/10. The multiplication of the
Washingtonian will not change the fact that I have had a bigger probabilty
to appear in Moscow. There will be a bigger set of Moscovian experiences
(including the fear to die with a chance of 8/10). The multiplication of
the one in Washington will not consolate those in Moscow.
With immortality it is even simpler, except that for a real computation
you should take into account the way you kill the Moscovians and you
must bet on the level of substitution.

Look at this: suppose you are annihilated at Brussels again.
And again reconstituted at M and W. Suppose now that at W, after
one day, you are
simply duplicate in you-W1 and you-W2, the day after, you-W2 is killed
and *at the same time* you-W1 is duplicated again, and the day after,
same scenario, all this lasting one year.

With ASSA you could say that at Brussels there is a bigger chance
to find yourself in W, because during all the year there is always two
examplars of you in Washington and only one at Moscow.

With RSSA there is 1/2 chance for W and M. But for the one in Washington
there is 1/2^365 to remain alive (supposing mortality) or healthy
immortality) after one year.

OK? Are we agreeing (at least) on the meaning of ASSA and RSSA ?


PS I am wrong, wrong, wrong ! Isn't it ? What do you think ?

The problem is that an observer-moment is really defined (with comp)
with the set of maximal consistent continuations going through that
moment, and we don't know really how these sets are structured. This
is our very much discussed measure problem.
Received on Sat Jan 15 2000 - 10:39:56 PST

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