Gray_Area comments on What is Evidence? - Less Wrong

60 Post author: Eliezer_Yudkowsky 22 September 2007 06:43AM

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Comment author: Gray_Area 22 September 2007 09:15:37AM 3 points [-]

Why not just say e is evidence for X if P(X) is not equal to P(X|e)?

Incidentally, I don't really see the difference between probabilistic dependence (as above) and entanglement. Entanglement is dependence in the quantum setting.

Comment author: Larks 06 October 2010 07:14:49AM 7 points [-]

Trivially, because P(X|e) could be less than P(X)

Comment author: Will_Sawin 09 July 2011 08:57:56PM 3 points [-]

Quantum wave amplitudes behave in some ways like probabilities and in other ways unlike probabilities. Because of this, some concepts have analogues, while others don't.

But no concepts are exactly equivalent. For example, evidence isn't integrally linked to complex numbers, while entanglement is.

Comment author: ec429 14 September 2011 06:09:45PM 0 points [-]

Nonetheless, it is instructive (imho) to consider how (assigned) probability is a property of the observer, and not an inherent property of the system. If a qubit is (|0> + |1>)/sqrt(2), and I measure it and observe 0, then I'm entangled with it so relative to me it's now |0>. But what's really happened is that I became (|observed 0> + |observed 1>)/sqrt(2), or rather, that the whole system became (|0,observed 0> + |1,observed 1>)/sqrt(2). This is closely analogous to the Law of Conservation of Probability; if you take Expectations conditional on the observation, then take Expectation of the whole thing, you get the original expectation back. This is because observing the system doesn't change the system, it just changes you. This is obvious in Bayesian probability in the classical-mechanics world; the only reason it doesn't seem obvious in the quantum realm is that we've been told over and over that "observing a quantum system changes it".

Quite honestly, I don't see how a Bayesian can possibly be a Copenhagenist. Quantum probability is Bayesian probability, because quantum entanglement is just the territory updating itself on an observation, in the same way that Bayesian 'evidence entanglement' is updating one's map on an observation.

Comment author: Will_Sawin 16 September 2011 03:49:48AM 2 points [-]

Classical probability preserves amplitude, quantum preserves |amplitude|^2.

They're different things, and they could, potentially, be even more different.

Comment author: ec429 16 September 2011 03:58:19AM 1 point [-]

Um, but isn't that just a convention? Why should we treat the "amplitude" of a classical probability as being the probability?

Does the problem have something to do with the extra directionality quantum probabilities have by virtue of the amplitude being in C? (so that |0> and (-1*|0>) can cancel each other out)

Comment author: Will_Sawin 21 September 2011 04:46:46AM 1 point [-]

Classical probability transformations preserve amplitude and quantum ones preserve |amplitude|^2. That's not a whole reason, but it's part of one.

Yes, that's part of the difference. Quantum transformations are linear in a two-dimensional wave amplitude but preserve a 1-dimensional |amplitude|^2. Classical transformations are linear in one-dimensional probability and preserve 1-dimensional probability.

Comment author: ec429 21 September 2011 12:46:19PM 0 points [-]

Ah, I get it now, thanks!

(Copenhagen is still wrong though ;)

Comment author: potato 18 July 2011 11:16:56PM 6 points [-]

"This should not be confused with the technical sense of "entanglement" used in physics - here I'm just talking about "entanglement" in the sense of two things that end up in correlated states because of the links of cause and effect between them."

That's literally in the third paragraph.

I think you mean, if P(x)<P(x|e) then e is evidence for x. That is a good definition for evidence, but it doesn't function on the same level as Yudkowsky's above. Yudkowsky is explaining not just what function evidence has in truth finding, he is also explaining how evidence is built into a physical system, e.g., camera, human, or other entanglement device. The Bayesian def of evidence you gave tells us what evidence is, but it doesn't tell us how evidence works, which Yudkowsky's does.

Comment author: ibbyitis 30 August 2016 04:25:50PM 0 points [-]

X : precence of flower A in certain area e : there are bees on that area then you would possibly have that P(X) < P(X|e), given that bees help doing pollinization. Then should we phrase "probability of having flower A in an area is greater if we have bees, therefore e is evidence for X (bees are evidence for flower A)" and what if X is "having presents brought by santa claus", and e is "we are in USA instead of cambodia" (which increases the probability of having presents because that date is more commonly celebrated with presents in USA).

Comment author: alex_zag_al 16 September 2012 01:44:18AM *  0 points [-]

That definition does not always coincide with what is described in the article; something can be evidence even if P(X|e) = P(X).

Imagine that two cards from a shuffled deck are placed face-down on a table, one on the left and one on the right. Omega has promised to put a monument on the moon iff they are the same color.

Omega looks at the left card, and then the right, and then disappears in a puff of smoke.

What he does when he's out of sight is entangled with the identity of the card on the right. Change the card to one of a different color and, all else being equal, Omega's action changes.

But, if you flip over the card on the right and see that it's red, that doesn't change the degree to which you expect to see the monument when you look through your telescope. P(monument|right card is red) = P(monument) = 25/51

It does change your conditional beliefs, though, such as what the world would be like if the left card turned out to also be red: P(monument|left is red & right is red) > P(monument|left is red)

Comment author: moshez 31 December 2012 11:40:50AM 0 points [-]

Of course e can be evidence even if P(X|e)=P(X) -- it just cannot be evidence for X. It can be evidence for Y if P(Y|e)>P(Y), and this is exactly the case you describe. If Y is "there is a monument and left is red or there is no monument and left is black", then e is (infinite, if Omega is truthful with probability 1) evidence for Y, even though it is 0 evidence for X.

Similarly, you watching your shoelace untied is zero evidence for my shoelaces...