SIA fears (expected) infinity

Clearly what we need to do first is to turn this into a Sleeping Beauty variant.

Let C = 6/pi^2, and suppose that we choose to wake Sleeping Beauty up k times with probability C/k^2. Then Sleeping Beauty is put in the awkward position that the expected number of times she wakes up is infinite. When asked "what is the probability that you were only woken up once?" the SSA, of course, suggests that Sleeping Beauty should answer C = 6/pi^2, while the SIA sort of gives up and maybe suggests 0 as a possible answer (If you object to 0 as a probability, recall that we're dealing with an infinite sample space, which you should also refuse to believe in).

I argue that this is a legitimate answer to give. Why? Recall that in the standard Sleeping Beauty problem, there is a way to specifically elicit SIA-based probabilities for Sleeping Beauty. We ask "Would you rather receive $1 if the coin came up heads (and $0 otherwise), or $1 if the coin came up tails?" By answering "heads", Sleeping Beauty earns $1 twice over the course of the experiment, for a total of $2; by answering tails, she earns $1. We can vary the payoffs to confirm that her best strategy is to act as though the probability the coin came up heads is 1/3.

Now let's consider the infinite case. We ask "Would you rather receive a googolplex dollars if you are to be woken up only once, or $1 if you are to be woken up more than once?" This is actually a bit awkward because money has nonlinear utility; but it's trivial to see that Sleeping Beauty maximizes her expected winnings by choosing the second option. So she is acting as though the probability she's only woken up once is less than 1/googolplex, and similarly for any other large number. The only probability consistent with this is 0.

Some neat points here.

This reminds me of some of the discussion in this paper by Weatherson, which makes a number of infinity-related claims against indifference principles for assigning the probability that you are a certain member of a class of subjectively indistinguishable observers.

It seems to me that is we have infinite population, which include all possible observers, then SIA merges with SSA. For example, in presumptuous philosopher, it would mean that there are two regions of the multiverse: one with trillion observers, and another with trillion of trillions, and it will be not surprising to be located in the larger one. 

SIA in PP becomes absurd only for a finite universe (and no other universes), where only one of two regions exists. But the absurdity is in the definition: it is absurd to think that the universe could be provably finite, as there should be some force above the universe which limits its size. 

Stuart, thanks for this.

If we are using Solomonoff induction, won't the expected population be infinite?

Very crudely, finite worlds of size n will have probability about 2^-K(n) in the Solomonoff prior, where K is the Kolmogorov complexity of the binary representation of n. This works out at about 1/n x 1/log n x 1/log log n x... for most values of n, taking base 2 logarithms, and repeating the logs until we hit a small constant. The probability is higher for the simple "non-random" values of n like a googleplex or 3^^^3.

Then if the expected population in worlds of size n is proportional to n (again this is very crude), we get an expected population proportional to:

Sigma {n=1 to infinity} n x 2^-K(n)

which is at least

Sigma {n=1 to infinity} 1/log n x 1/log log n terms

and that is a divergent sum. So SIA predicts that for any size n, we are almost certainly in a world of size bigger than n and we can't normalize the distribution! Problem.

There might be a better story with a less crude treatment, but frankly I'm doubtful. As I understand it, using SIA is equivalent (in your anthropic decision theory) to having an additive utility function which grows in proportion to the population size (or at least to the population of people whose decisions are linked to yours). And so the utility function is unbounded. And unbounded utility functions are a known problem with Solomonoff induction, at least according to Peter de Blanc (http://arxiv.org/abs/0712.4318). So I think the crude treatment is revealing a real problem here.

If you sum up C/j^2, that's finite, with the value given by Riemann's zeta function. Zeta of 2 is around 1.645.

It's C/j that goes up to infinity, logarithmically.