[EMAIL PROTECTED] wrote:
> 
> 
> 
> > From [EMAIL PROTECTED]:
> > [EMAIL PROTECTED] wrote:
> > > M measure:
> > > M(empty string)=1
> > > M(x) = M(x0)+M(x1) nonnegative for all finite x.
> > 
> > This sounds more like a probability distribution than a measure. In
> > the set of all descriptions, we only consider infinite length
> > bitstrings. Finite length bitstrings are not members. However, we can
> > identify finite length bitstrings with subsets of descriptions. The
> > empty string corresponds to the full set of all descriptions, so the
> > first line M(empty string)=1 implies that the measure is normalisable
> > (ie a probability distribution).
> 
> 
> Please check out definitions of measure and distribution! 
> Normalisability is not the critical issue.
> 
> Clearly: Sum_x M(x) is infinite. So M is not a probability
> distribution. M(x) is just measure of all strings starting with x:
> M(x) = M(x0)+M(x1) = M(x00)+M(x01)+M(x10)+M(x11) = ....

For a measure to be normalisable, the sum over a *disjoint* set of
subsets must be finite. If the set of subsets is not disjoint (ie the
intersection are not empty) then the sum may well be infinite.

Bringing this to the case of finite strings. Each finite string is
actually a subset of the set of infinite strings, each containing the
same finite prefix. So the string 101010 is actually a subset of 10101
and so on. The Sum_x M(x), where I assume x ranges over all strings
will of course be infinite. However, since the set of finite strings
is not disjoint, this doesn't imply M(x) is unnormalisable.

Now when you realise that every finite string x is a subset of the empty
string, it becomes clear that M(x) is normalised to precisely 1.

> 
> Neglecting finite universes means loss of generality though.
> Hence measures mu(x) in the ATOE paper do not neglect finite x: 
> 
> mu(empty string)=1
> mu(x) = P(x)+mu(x0)+mu(x1)  (all nonnegative).
> 
> And here P is a probability distribution indeed! 
> P(x)>0 possible only for x with finite description. 
> 

The P(x)>0 case above actually breaks the countably subadditive
property, so mu(x) cannot be called a measure... I'm not entirely sure
what you're getting at here.

                                        Cheers

> Juergen Schmidhuber
> 
> http://www.idsia.ch/~juergen/
> http://www.idsia.ch/~juergen/everything/html.html
> http://www.idsia.ch/~juergen/toesv2/
> 
> 
> 



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