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Wes_W comments on Absolute Authority - Less Wrong
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Comments (72)
You could consider a proposition to be infinite evidence for itself, I guess. That seems like maybe a kinda defensible interpretation of P(A|A) = 1. I don't think it gets you anything useful, though.
[∃ B: P(A|B) ∈ (0,1)] → [P(A) ∈ (0,1)]. Better?
If, having made them, your own probability assessments are meaningless and unusable, who cares what values you assign? Set P(A) = 328+5i and P(B) = octopus, for all it matters.
Additionally, I'm not sure it matters when the mind-changing actually occurs. At the instant of assignment, your mind as it is right that moment should already have a value for P(A|B) - how you would counterfactually update on the evidence is already a fact about you. If you would, counterfactually assuming your current mind operates without interference until it can see and process that evidence, update to some credence other than 1, it is already at that moment incorrect to assign a credence of 1. Whether that chain of events does in fact end up happening won't retroactively make you right or wrong; it was already the right or wrong choice when you made it.
Or, if you get mind-hacked, your choice might be totally moot. But this is generally a poor excuse to deliberately make bad choices.
Yes, it makes it clearer what you're doing wrong. I'll do what I should have done earlier, and formalize my argument:
Let's call "2+2=4" A, "2+2=3" B, "I can visualize 2+2=4" C, "I can visualize 2+2=3" D, "I can remember visualizing 2+2=4" E, "I can remember visualizing 2+2=3" F.
So, my claim is that P(A|C) is 1, likewise P(B|D). (Remember, I don't think it's like this in real life, I'm trying to show that the argument put forward to prove that is not sufficient.)
What is the Bayes formula for tomorrow's assessment?
Not, P(A|C,D), which (if <1) would indeed disprove P(A|C)=1.
But, instead, P(A|E,D). This can be less than 1 while P(A|C)=1. I'll just make up some arbitrary numbers as priors to show that.
I'm assuming A and B are mutually exclusive, as are C and D.
P(A)=.75
P(B)=.25 (just assume that it's either 2 or 3)
P(C)=.375
P(D)=.125
P(memory of X | X happened yesterday)=.95
P(memory of X | X didn't happen yesterday)=.001
P(E)=P(C)*.95+P(~C)*.001=0.356875
P(F)= P(D)*.95+P(~D)*.001=0.119625
P(C|A)=.50
P(C|B)=0
P(D|A)=0
P(D|B)=.50
P(A|C) = P(C|A)P(A)/P(C)=(.50*.75)/(.75*.50+.25*0)=1
P(A|C,D) is undefined, because C and D are mutually exclusive (which corresponds to not being able to visualize both 2+2=3 and 2+2=4 at the same time)
P(F,D)=P(D)*.95=0.11875
P(A|E,D)= P(E,D|A)P(A)/P(E,D)=0 (Because D|A is zero).
Using my numbers, you need to derive a mathematical contradiction if there are, truly "technical reasons" for this being impossible.
The mistake you (and EY) are making is that you're not comparing P(A) to P(A|B) for some A,B, but P(A|B) to P(A|C) for some A,B,C.
Added: I made two minor errors in definitions that have been corrected. E and F are not exclusive, and C and D shouldn't be defined as "current", but rather as having happened, which can only be confirmed definately if they are current. However, they have the evidential power whenever they happened, it's just if they didn't happen now, they're devalued because of fragile memory.
Added: Fixed numerical error and F where it was supposed to be E. (And errors with evaluating E and F. I really should not have assumed any values that I could have calculated from values I already assumed. I have less degrees of freedom than I thought.)
blink
Well huh. I suppose I ought to concede that point.
There are probabilities of 0 and (implicitly) 1 in the problem setup. I'm not confident it's valid to start with that; I worry it just pushes the problem back a step. But clearly, it is at least possible for probabilities of 1 to propagate to other propositions which did not start at 1. I'll have to think about it for a while.