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From: Russell Standish <R.Standish.domain.name.hidden>

Date: Tue, 23 Oct 2001 09:11:00 +1000 (EST)

juergen.domain.name.hidden wrote:

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

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

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

*>
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*>
*

----------------------------------------------------------------------------

Dr. Russell Standish Director

High Performance Computing Support Unit, Phone 9385 6967, 8308 3119 (mobile)

UNSW SYDNEY 2052 Fax 9385 6965, 0425 253119 (")

Australia R.Standish.domain.name.hidden

Room 2075, Red Centre http://parallel.hpc.unsw.edu.au/rks

International prefix +612, Interstate prefix 02

----------------------------------------------------------------------------

Received on Mon Oct 22 2001 - 16:27:21 PDT

Date: Tue, 23 Oct 2001 09:11:00 +1000 (EST)

juergen.domain.name.hidden wrote:

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.

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

----------------------------------------------------------------------------

Dr. Russell Standish Director

High Performance Computing Support Unit, Phone 9385 6967, 8308 3119 (mobile)

UNSW SYDNEY 2052 Fax 9385 6965, 0425 253119 (")

Australia R.Standish.domain.name.hidden

Room 2075, Red Centre http://parallel.hpc.unsw.edu.au/rks

International prefix +612, Interstate prefix 02

----------------------------------------------------------------------------

Received on Mon Oct 22 2001 - 16:27:21 PDT

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