ThisSpaceAvailable comments on Understanding and justifying Solomonoff induction - Less Wrong

1 Post author: gedymin 15 January 2014 01:16AM

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Comment author: ThisSpaceAvailable 20 January 2014 11:40:01PM 0 points [-]

So, the precise statement would be that whatever weight you assign each program, if f(n) := sum of the weights of all programs of length n F(n) := sum of f(k) for all k < n then limit as n goes to infinity of F(n) must be finite (if the weights are normalized, then the limit must be 1) and thus limit as n goes to infinity of f(n) must be zero. In fact, by the ratio test, the limit of f(n)/f(n+1) must be greater than or equal to one. Since there are 2^n programs of size n, that means that if a(n) := average weight of programs of size n then limit as n goes to infinity of a(n)/a(n+1) must be greater than or equal to 2

We are therefore highly constrained as far as the average weight for programs of size n, but less so for the maximum weight. For instance, we could have a weighting system such that for all n, there is some program of size n with weight (.999^n)/10000. We could even have a weighting system such that for all k, there is some n > k such that there is a program of length n and weight (.9999^log(log (log n) ) )/10

Do I have that correct?

PS I believe that you meant "monotonic", rather than "monotonous" :)

Comment author: cousin_it 21 January 2014 09:30:52AM *  1 point [-]

In fact, by the ratio test, the limit of f(n)/f(n+1) must be greater than or equal to one.

The limit is >=1 if it exists, but it doesn't have to exist. And the lim inf can be arbitrarily close to zero.

limit as n goes to infinity of a(n)/a(n+1) must be greater than or equal to 2

Ditto.