Jim,

I'll argue that solomonoff probabilities are in fact like Pi, that is,
computable in the limit.

I still do not understand why you think these combinations are necessary. It
is not necessary to make some sort of ordering of the sum to get it to
converge: ordering only matters for infinite sums which include negative
numbers. (Perhaps that's where you're getting the idea?)

Here's my proof, rewritten from an earlier post, using the properties of
infinite sums of non-negative numbers.

(preliminaries)

Define the computation as follows: we start with a string S which we want to
know the Solomonoff probability of. We are given a time-limit T. We start
with P=0, where P is a real number with precision 2*log_4(T) or more. We use
some binary encoding for programs which (unlike normal programming
languages) does not contain syntactically invalid programs, but still will
(of course) contain infinite loops and so on. We run program "0" and "1" for
T/4 each, "00", "01", "10" and "11" for T/16 each, and in general run each
program of length N for floor[T/(4^N)] until T/(4^N) is less than 1. Each
time we run a program and the result is S, we add 1/(4^N) to P.

(assertion)

P converges to some value as T is increased.

(proof)

If every single program were to output S, then T would converge to 1/4 + 1/4
+ 1/16 + 1/16 + 1/16 + 1/16 + ... that is, 2*(1/(4^1)) + 4*(1/(4^2)) +
8*(1/(4^3)) + ... which comes to 1/2 + 1/4 + 1/8 + 1/16 + .. i.e. 1/(2^1) +
1/(2^2) + 1/(2^3) + ... ; it is well-known that this sequence converges to
1. Thus 1 is an upper bound for P: it could only get that high if every
single program were to output S.

0 is a lower bound, since we start there and never subtract anything. In
fact, every time we add more to P, we have a new lower bound: P will never
go below a number once it reaches it. The sum can only increase. Infinite
sums with this property must either converge to a finite number, or go to
infinity. However, we already know that 1 is an upper bound for P; so, it
cannot go to infinity. Hence, it must converge.

--Abram

On Mon, Jul 26, 2010 at 9:14 AM, Jim Bromer <[email protected]> wrote:

> As far as I can tell right now, my theories that Solomonoff Induction is
> trans-infinite were wrong.  Now that I realize that the mathematics do not
> support these conjectures, I have to acknowledge that I would not be able to
> prove or even offer a sketch of a proof of my theories.  Although I did not
> use rigourous mathematics while I have tried to make an assessment of the
> Solomonoff method, the first principle of rigourous mathematics is to
> acknowledge that the mathematics does not support your supposition when they
> don't.
>
> Solomonoff Induction isn't a mathematical theory because the desired
> results are not computable.  As I mentioned before, there are a great many
> functions that are not computable but which are useful and important because
> they tend toward a limit which can be seen in with a reasonable number of
> calculations using the methods available.  Pi is one such function.  (I am
> presuming that pi would require an infinite expansion which seems right.)
>
> I have explained, and I think it is a correct explanation, that there is no
> way that you could make an apriori computation of all possible
> combinations taken from infinite values.  So you could not even come up with
> a theoretical construct that could take account of that level of
> complexity.  It is true that you could come up with a theoretical
> computational method that could take account of any finite number of values,
> and that is what we are talking about when we talk about the infinite, but
> in this context it only points to a theoretical paradox.  Your theoretical
> solution could not take the final step of computing a probability for a
> string until it had run through the infinite combinations and this is
> impossible.  The same problem does not occur for pi because the function
> that produces it tends toward a limit.
>
> The reason I thought Solomonoff Induction was trans-infinite was because I
> thought it used the Bayesian notion to compute the probability using all
> possible programs that produced a particular substring following a given
> prefix.  Now I understand that the desired function is the computation of
> only the probability of a particular string (not all possible substrings
> that are identical to the string) following a given prefix.  I want to study
> the method a little further during the next few weeks, but I just wanted to
> make it clear that, as far as now understand the program, that I do not
> think that it is trans-infinite.
>
> Jim Bromer
>    *agi* | Archives <https://www.listbox.com/member/archive/303/=now>
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-- 
Abram Demski
http://lo-tho.blogspot.com/
http://groups.google.com/group/one-logic



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