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jimrandomh comments on Pascal's Mugging as an epistemic problem - Less Wrong Discussion

3 [deleted] 04 October 2010 05:52PM

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Comment author: jimrandomh 06 October 2010 07:54:27PM *  0 points [-]

This is similar to the formulation I gave here, but I don't think your version works. You could construct a series of different sets of knowledge X(n) that differ only in that they have different numbers n plugged in, and a bounding function B(n) such that

for all n P(E|X(n))U(E|X(n)) < B(n), but
lim[n->inf] P(E|X(n))U(E|X(n)) = inf

Basically, the mugger gets around your bound by crafting a state of knowledge X for you.

I'm pretty sure the formulation given in my linked comment also protects against Pascal's Reformed Mugger.

Comment author: RichardKennaway 07 October 2010 10:52:17AM 0 points [-]

Basically, the mugger gets around your bound by crafting a state of knowledge X for you.

This is giving too much power to the hypothetical mugger. If he can make me believe (I should have called X prior belief rather than prior knowledge) anything he chooses, then I don't have anything. My entire state of mind is what it is only at his whim. Did you intend something less than this?

One could strengthen the axiom by requiring a bound on P(E|X) U(E|X) uniform in both E and X. However, if utiilty is unbounded, this implies that there is an amount so great that I can never believe it is attainable, even if it is. A decision theory that a priori rules out belief in something that could be true is also flawed.

Comment author: jimrandomh 07 October 2010 11:49:04AM 0 points [-]

He doesn't get to make you believe anything he chooses; making you believe statements of the form "The mugger said X(n)" is entirely sufficient.

Comment author: RichardKennaway 07 October 2010 12:20:13PM 0 points [-]

There would have to be statements X(n) such that the maximum over E of P(E|The mugger said X(n)) U(E|The mugger said X(n)) is unbounded in n. I don't see why there should be, even if the maximum over E of P(E|X) U(E|X) is unbounded in X.

Comment author: jimrandomh 07 October 2010 12:54:21PM 0 points [-]

There would have to be statements X(n) such that the maximum over E of P(E|The mugger said X(n)) U(E|The mugger said X(n)) is unbounded in n.

Yes, and that is precisely what I said causes vulnerability to Pascal's Mugging and should therefore be forbidden. Does your version of the anti-mugging axiom ensure that no such X exists, and can you prove it mathematically?

Comment author: RichardKennaway 07 October 2010 09:03:00PM *  0 points [-]

It does not ensure that no such X exists, but I think this scenario is outside the scope of your suggestion, which is expressed in terms of P(X) and U(X), rather than conditional probabilities and utilities.

What do you think of the other potential defect in a decision theory resulting from too strong an anti-mugging axiom: the inability to believe in the possibility of a sufficiently large amount of utility, regardless of any evidence?

Comment author: jimrandomh 07 October 2010 09:57:10PM 0 points [-]

Oh, so that's where the confusion is coming from; the probabilities and utilities in my formulation are conditional, I just chose the notation poorly. Since X is a function of type number=>evidence-set, P(X(n)) means the probability of something (which I never assigned a variable name) given X(n), and U(X(n)) is the utility of that same thing given X. Giving that something a name, as in your notation, these would be P(E|X) and U(E|X).

Being unable to believe in sufficiently large amounts of utility regardless of any evidence would be very bad; we need to be careful not to phrase our anti-mugging defenses in ways that would do that. This is a problem with globally bounded utility functions, for example. I'm pretty sure that requiring all parameterized statements to produce expected utility that does not diverge to infinity as the parameter increases, does not cause any such problems.