Here’s a quick summary of the paper, to see whether I understand it (and perhaps to distill it down to its essentials.)
Let a ‘hypothesis’ be an algorithm which tells us, among other things, how much utility we get for each of the possible actions we can take.
Assume that we assign a non-zero probability to every hypothesis consistent with our data so far, and that our subjective probability distribution is computable. Assume there is an algorithm F such that F(M) is a hypothesis (consistent with available data) in which whatever we do, we get constant utility u’ satisfying |u’| > M.
Arrange all ‘computations returning natural numbers’ in some (computable) order. Let C_n be the n-th such computation.
Arrange all of the hypotheses with non-zero probabilities in some (computable) order. Now define function f(n) = the index of the hypothesis F(the result of C_n). Somewhat delicate point: f is a computable function (even though some of the C_n do not halt). Therefore, q(n) = 1/(the probability of hypothesis f(n)) is also a computable function.
However, for any computable function g, there must exist (by Busy Beaverish arguments) infinitely many values of n such that g(n) < the result of C_n. This is true of q(n) in particular.
If q(n) < the result of C_n < |the utility of hypothesis f(n)| then the term corresponding to f(n) in our expected utility must have magnitude >= 1. But there are infinitely many such terms, so the expected utility is divergent.
[I hope my various simplifications haven’t broken the proof.
Notes:
I can see a slight gap in the logic where de Blanc states “By definition, phi_G(n) is in S_I”. Surely phi_G(n) may not be a total function, whereas S_I is a set of total functions? How best can we patch things up? Do we need to assume that the agent assigns non-zero probabilities to all partial recursive functions consistent with the data?
I seem to have done away with u_j and simplified the definition of q(n). Is this OK?]
Here’s a quick summary of the paper, to see whether I understand it (and perhaps to distill it down to its essentials.)
Let a ‘hypothesis’ be an algorithm which tells us, among other things, how much utility we get for each of the possible actions we can take.
Assume that we assign a non-zero probability to every hypothesis consistent with our data so far, and that our subjective probability distribution is computable. Assume there is an algorithm F such that F(M) is a hypothesis (consistent with available data) in which whatever we do, we get constant utility u’ satisfying |u’| > M.
Arrange all ‘computations returning natural numbers’ in some (computable) order. Let C_n be the n-th such computation.
Arrange all of the hypotheses with non-zero probabilities in some (computable) order. Now define function f(n) = the index of the hypothesis F(the result of C_n). Somewhat delicate point: f is a computable function (even though some of the C_n do not halt). Therefore, q(n) = 1/(the probability of hypothesis f(n)) is also a computable function.
However, for any computable function g, there must exist (by Busy Beaverish arguments) infinitely many values of n such that g(n) < the result of C_n. This is true of q(n) in particular.
If q(n) < the result of C_n < |the utility of hypothesis f(n)| then the term corresponding to f(n) in our expected utility must have magnitude >= 1. But there are infinitely many such terms, so the expected utility is divergent.
[I hope my various simplifications haven’t broken the proof.
Notes:
I can see a slight gap in the logic where de Blanc states “By definition, phi_G(n) is in S_I”. Surely phi_G(n) may not be a total function, whereas S_I is a set of total functions? How best can we patch things up? Do we need to assume that the agent assigns non-zero probabilities to all partial recursive functions consistent with the data?
I seem to have done away with u_j and simplified the definition of q(n). Is this OK?]