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.
Define “you” and “ever”. I argue that the “you” who changes there mind tomorrow is not the same observer that decides with 100% probability today, because the one today has information that the one tomorrow doesn’t; namely, actual brain ops, versus memories for tomorrow you.
[∃ 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|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.)
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.
I’m assuming A and B are mutually exclusive, as are C and D, and E and F.
While A and B being mutually exclusive seems reasonable, I don’t think it holds for C and D. And I’m pretty sure that it doesn’t hold at all for E and F.
If I remember visualising 2+2=3 yesterday and 2+2=4 the day before, then E and F are both simultaneously true.
P(A)=.75
P(C)=.75
P(C|A)=.50
These three statements, taken together, are impossible. Consider:
Over the 0.75 probability space where C is true (second statement), A is only true in half that space (third statement). Thus, A is false in the other half of that space; therefore, there is a probability space of at least 0.375 in which A is false. Yet A is only false over a probability space of size 0.25 (first statement).
In your calculations further down, you use the value P(C) = (.75.50+.250) = 0.375; using that value for P(C) instead of 0.75 removes the contradiction.
Similarly, the following set of statements lead to a contradiction, considered together:
The first and third comments are correct. I made some errors in first typing it up that shouldn’t take away from the argument that are now fixed. The third comment is an actual mistake that has also been fixed.
Over the 0.75 probability space where C is true (second statement), A is only true in half that space (third statement).
This is wrong. P(C|A) is read as C given A, which is the chance of C, given that A is true. You’re mixing it up with P(A|C). However, if you switch A and C in your paragraph, it becomes a valid critique, which I’ve fixed, substituting the correct values in. Thanks. (Did I mess anything else up?)
I’m starting to appreciate mathematicians now :)
You need to escape your * symbols so they output correctly.
You’re mixing it up with P(A|C). However, if you switch A and C in your paragraph, it becomes a valid critique, which I’ve fixed, substituting the correct values in. Thanks.
You’re right, I had that backwards.
(Did I mess anything else up?)
Hmmm....
P(F)=.20
P(F)= P(D)*.95+P(C)*.001=0.119125
You have two different values for P(F). Similarly, the value P(E)=0.70 does not match up with P(C), P(D) and the following:
P(memory of X | X happened yesterday)=.95
P(memory of X | X didn’t happen yesterday)=.001
None of which is going to affect your point, which seems to come down to the claim that there exist possible events A, B, C, D, E and F such that P(A|C) = 1.
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.
While A and B being mutually exclusive seems reasonable, I don’t think it holds for C and D. And I’m pretty sure that it doesn’t hold at all for E and F.
If I remember visualising 2+2=3 yesterday and 2+2=4 the day before, then E and F are both simultaneously true.
These three statements, taken together, are impossible. Consider:
Over the 0.75 probability space where C is true (second statement), A is only true in half that space (third statement). Thus, A is false in the other half of that space; therefore, there is a probability space of at least 0.375 in which A is false. Yet A is only false over a probability space of size 0.25 (first statement).
In your calculations further down, you use the value P(C) = (.75.50+.250) = 0.375; using that value for P(C) instead of 0.75 removes the contradiction.
Similarly, the following set of statements lead to a contradiction, considered together:
The first and third comments are correct. I made some errors in first typing it up that shouldn’t take away from the argument that are now fixed. The third comment is an actual mistake that has also been fixed.
This is wrong. P(C|A) is read as C given A, which is the chance of C, given that A is true. You’re mixing it up with P(A|C). However, if you switch A and C in your paragraph, it becomes a valid critique, which I’ve fixed, substituting the correct values in. Thanks. (Did I mess anything else up?)
I’m starting to appreciate mathematicians now :)
You need to escape your * symbols so they output correctly.
You’re right, I had that backwards.
Hmmm....
You have two different values for P(F). Similarly, the value P(E)=0.70 does not match up with P(C), P(D) and the following:
None of which is going to affect your point, which seems to come down to the claim that there exist possible events A, B, C, D, E and F such that P(A|C) = 1.