# Re: normalization

From: Jacques M. Mallah <jqm1584.domain.name.hidden>
Date: Tue, 1 Feb 2000 17:22:14 -0500 (EST)

On Tue, 18 Jan 2000 GSLevy.domain.name.hidden wrote:
> jqm1584.domain.name.hidden writes:
> > On Tue, 18 Jan 2000 GSLevy.domain.name.hidden wrote:
> > > jqm1584.domain.name.hidden writes:
> > > > The RSSA is not another way of viewing the world; it is a
> > > > category error.
> > >
> > > I use the RSSA as the basis for calculating what I call the relative
> > > probability, in this group the first person probability, or, equivalently,
> > > the probability conditional on the life of the observer. The ASSA is by
> > > extension, the assumption for calculating the 3rd person probability.
> > >
> > > Let us perform a thought experiment.
> > > Imagine that you are the scientist in the Schroedinger cat experiment.
> >
> > Scratch that. Right now let's stick to the example with Bruno and
> > the 3 cities, because it's better for the current point.
> > Suppose Bruno, in 1999, wants to know if he is more likely to be
> > in Washington or in Moscow during 2001.
> > First of all, that is not a well defined question, because
> > "Bruno" must be defined. Suppose we define it to mean the set of all
> > Bruno-like observations, where by "Bruno-like" we can assume we know what
> > qualifies.
> > But then the question becomes meaningless, because it is 100%
> > certain that he will be in *both* cities. A 3rd person would have to
> > agree with that, he is in *both* cities.
> > So let's ask a meaningful question. Among the set of Bruno-like
> > observations in 2001, what is the effective probability of such an
> > observation being in Moscow?
> > This is just a conditional effective probability so we use the
> > same rule we always use:
> > p(Moscow|Bruno in 2001) =
> > M(Moscow, Bru. 2001) / [M(Moscow, Bru. 2001) + M(Washington, Bru. 2001)]
> > where M is the measure.
> > So in this case the conditional effective probability of him
> > seeing Moscow at that time is 10%, and in *1999* he knows he should brush
> > up on his English because his future 'selves' will be affected by that.
> >
>
> Fine, you have computed the third person probability.

Bullshit. There is no such thing. As I said, a 3rd person would
say that he is in both cities for certain. What I computed was the
effective probability for Bruno's.

> Unfortunately, your
> example does not have the option of having an independent observer, and
> therefore does not illustrate the concept I am trying to communicate.

How can an example not have an option? Evidently your concept of
"independent observer" is closely tied to QM physics, and the fact that I
am discussing a different situation messes it up.

> with an observer who is not threatened with death and a subject who is. It is
> the only way to bring out the concept of relative probability or 1st and 3rd
> person probability.

I have another view, namely, that the example you mentioned
(Schrodinger's Cat) is conducive to letting you make your errors, and it
is the "only way" because your views don't work in a more general
situation.
But because you used the magic word, OK, I'll illustrate the same
methods I used above, using the Cat.

As far as the human experimenter can tell, the experiment has a
50% chance to kill the Cat.
Suppose the Cat, before the experiment, wants to know if he is
more likely to be alive or dead after the experiment.
First of all, that is not a well defined question, because
"the Cat" must be defined. Suppose we define it to mean the set of all
Cat-like observations, where by "Cat-like" we can assume we know what
qualifies.
Now let's ask a more precise question. Among the set of Cat-like
observations after the experiment, what is the effective probability of
such an observation being of a live Cat?
This is just a conditional effective probability so we use the
same rule we always use:
p(alive|Cat sees after experiment) = M(alive, Cat after) / M(alive, Cat after)
where M is the measure.
So in this case the conditional effective probability of him
seeing that he is alive, given that he makes the observation after the
experiment, is 100%. This is the quantity that is analagous to the 10%
effective probability Bruno had for being in Moscow (in 2001), according
to the ASSA. (While the RSSA claimed 90% for Bruno, but would agree with
100% for the Cat.)
OK. Obviously this was a rather trivial example of a conditional
effective probability. Let's ask one more question. Assume that the Cat
was 10 years old at the time of the experiment, and would have lived to 20
if he had not been subjected to the experiment. Over his lifetime, what
is the effective probability for him to be older than 10?
p(10+|Cat) = M(10+|Cat) / [M(10+|Cat) + M(<10|Cat)] = .5 / (.5 + 1) = 1/3

- - - - - - -
Jacques Mallah (jqm1584.domain.name.hidden)
Physicist / Many Worlder / Devil's Advocate
"I know what no one else knows" - 'Runaway Train', Soul Asylum
My URL: http://pages.nyu.edu/~jqm1584/
Received on Tue Feb 01 2000 - 14:25:33 PST

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