Re: Cosmology and Boltzmann brains

From: Russell Standish <lists.domain.name.hidden>
Date: Fri, 13 Jun 2008 11:25:56 +1000

On Thu, Jun 12, 2008 at 11:43:26PM +0200, Günther Greindl wrote:
>
> Hi all,
>
> someone on another list alerted me to this post, there is a very
> interesting discussion going on on that blog related to Observer Moments:
>
> http://golem.ph.utexas.edu/category/2008/06/urban_myths_in_contemporary_co.html
>
> Greg Egan has posted too; and has some very interesting things to say.
> Specifically, he says the right things why DA fails:

I'm not sure his application of Bayes is correct. Given the facts of
his hypothetical scenario, and writing e=10^{-4050}

  p(1|A) = e
  p(2|A) = 1-e
  p(1|B) = 1-e
  p(2|B) = e

This is my translation of:

"Now suppose that (somehow) we\u2019re able to extract the following (somewhat fanciful) predictions: theory A implies that in the entire history of the universe, there will be 1050 observers* of class 1 and 105000 observers of class 2, while theory B implies that in the entire history of the universe, there will be 105000 observers of class 1 and 1050 observers of class 2."

Now we further suppose there is no reason to prefer theory A over B,
ie p(A)=p(B).

Then we need to compute the likelihood of theory A given the fact that
we're an observer of class 2, ie:

p(A|2) = p(A & 2) / p(2) = p(2|A) p(A) / p(2) ... (1)

and

p(B|2) = p(B & 2) / p(2) = p(2|B) p(B) / p(2) ... (2)

dividing (1) by (2) gives

p(A|2) / p(B|2) = p(2|A) / p(2|B) = (1-e) / e = 10^{4050}

ie Bayes' theorem most definitely implies theory A is overwhelmingly
supported.

Have I missed something, or is Greg Egan wrong?

In a later posting, he gives absurd example of some extremely
improbably theory A, and applying the above reasoning. Yet the above
reasoning assumes p(A)=p(B), which is not the case in his absurd
example. It may be relevant to the BB argument though. If theory A was
"we are a statistical fluctuation (ie Boltzmann brains)", and theory B
was "evolved by Darwinian evolution", then p(A) << p(B). One cannot
comment on whether one should prefer A or B, since the numerical
values are just pulled out of a hat in any case.

-- 
----------------------------------------------------------------------------
A/Prof Russell Standish                  Phone 0425 253119 (mobile)
Mathematics                         	 
UNSW SYDNEY 2052         	         hpcoder.domain.name.hidden
Australia                                http://www.hpcoders.com.au
----------------------------------------------------------------------------
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups "Everything List" group.
To post to this group, send email to everything-list.domain.name.hidden
To unsubscribe from this group, send email to everything-list-unsubscribe.domain.name.hidden
For more options, visit this group at http://groups.google.com/group/everything-list?hl=en
-~----------~----~----~----~------~----~------~--~---
Received on Thu Jun 12 2008 - 21:26:10 PDT

This archive was generated by hypermail 2.3.0 : Fri Feb 16 2018 - 13:20:14 PST