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From: <hal.domain.name.hidden>

Date: Wed, 2 Jun 1999 22:20:44 -0700

Wei Dai, <weidai.domain.name.hidden>, writes:

*> Bayesian analysis in general does not depend on the Strong SSA, but any
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*> Bayesian analysis where you try to compute P(X | I observe Y) does because
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*> you need the Strong SSA to compute P(I observe Y | X) and P(I observe Y |
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*> not X).
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*>
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*> My example is one of classical Bayesian reasoning, but it is slightly
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*> different from the way you put it, because I don't have direct knowledge
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*> that Mathematica outputs a 9 for the sixth digit of Pi. I do know that I
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*> am reading "N[Pi]=3.14159", and Strong SSA is needed to derive the
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*> probability that I am reading "N[Pi]=3.14159" if Pi doesn't begin with
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*> 3.14159.
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So, are you saying that P(I observe "N[Pi]=3.14159" | PI == 3.14159)

should be considered the probability, given that PI has that value, that

a randomly chosen observer-moment is of me, sitting at a computer, seeing

this value display, out of all observer-moments in all possible universes?

In that case the probability would be extremely low, since only a tiny

fraction of all observer-moments would consist of me doing that. Maybe

it would be something like 1E-100.

However, the probability P(I observe "N[Pi]=3.14159" | PI != 3.14159) would

be even lower, since not only would the randomly chosen observer moment

have to be me running mathematica, but it would also be necessary that

mathematica is wrong. Maybe this probability would be 1E-110.

We would then run the Bayesian formula with these two extremely low

conditional probabilities. From this we would conclude that

P(PI == 3.14159 | I observe "N[Pi]=3.14159") would be high, as we expect.

Is this how you would work this problem, based on the strong SSA?

I'm not sure I am on the right track here...

Hal

Received on Wed Jun 02 1999 - 22:24:20 PDT

Date: Wed, 2 Jun 1999 22:20:44 -0700

Wei Dai, <weidai.domain.name.hidden>, writes:

So, are you saying that P(I observe "N[Pi]=3.14159" | PI == 3.14159)

should be considered the probability, given that PI has that value, that

a randomly chosen observer-moment is of me, sitting at a computer, seeing

this value display, out of all observer-moments in all possible universes?

In that case the probability would be extremely low, since only a tiny

fraction of all observer-moments would consist of me doing that. Maybe

it would be something like 1E-100.

However, the probability P(I observe "N[Pi]=3.14159" | PI != 3.14159) would

be even lower, since not only would the randomly chosen observer moment

have to be me running mathematica, but it would also be necessary that

mathematica is wrong. Maybe this probability would be 1E-110.

We would then run the Bayesian formula with these two extremely low

conditional probabilities. From this we would conclude that

P(PI == 3.14159 | I observe "N[Pi]=3.14159") would be high, as we expect.

Is this how you would work this problem, based on the strong SSA?

I'm not sure I am on the right track here...

Hal

Received on Wed Jun 02 1999 - 22:24:20 PDT

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