In this paper, the talented researcher Cosma Shalizi argues that to fully accept a subjective Bayesian view, one must also accept an unphysical result that the arrow of time (given by the flow of entropy) should actually go backwards. This is mainly an attempt to argue against the maximum entropy / fully subjective Bayesian view put forward and popularized by E.T. Jaynes.

Over at LessWrong, many contributors are very interested in Bayesian probability theory and also in the subjective Bayesian approach as a basis for formal decision theories and a stepping stone toward strong A.I. Eliezer Yudkowsky is a common contributor there and I was recently reading this post when I came across this comment (several other good comments come shortly after it on the original post's page).

Can anyone comment on the validity of Yudkowsky's rebuttal of Shalizi. Briefly, Yudkowsky's argument is that the physical mechanism by which a reasoning agent updates its beliefs require work and hence has a thermodynamic cost that Shalizi is sweeping under the rug. In another comment, Yudkowsky defends this, saying:

"If you take the perspective of a logically omniscient perfect observer outside the system, the notion of "entropy" is pretty much meaningless, as is "probability" - you never have to use statistical thermodynamics to model anything, you just use the deterministic precise wave equation."

Can any probabilists or statistcal mechanics comment on this? I don't care much about arguments from authority regarding either Shalizi's or Yudkowsky's status, but I would really like to see a summary of the ways that Yudkowsky's three points offer criticism of Shalizi's article.

To conform to FAQ guidelines and make this a concretely answerable question please note that I am asking for a specific, itemized response that takes Yudkowsky's three-step argument and indicates where in the Shalizi article those three steps refute assumptions and/or derivations, or, on the other hand, indicates where in Shalizi's paper the arguments of Yudkowsky are addressed.

I've often heard the Shalizi article touted as iron-clad proof that full-blown subjective Bayesianism can't be defended... but after reading the Shalizi article a few times, it looks like a toy argument to me that could never apply to an observer interacting with whatever is being observed (i.e. all of actual physics). But Shalizi is a great researcher, so I would welcome second opinions because it's highly likely that I don't understand important chunks of this debate.

  • 10
    $\begingroup$ Shalizi likes to be provocative... his argument appears to me to be essentially the same as the creationist argument that evolution violates the second law of thermodynamics because "later" organisms are more complex, in an organized way, than "earlier" organisms, but the second law says entropy is nondecreasing. However, 1) there's nothing in the second law that prevents local decreases in entropy, and 2) the argument implies no-one can learn anything about anything, ever (why should learning via Bayesian updating be any different than any other learning process?) $\endgroup$
    – jbowman
    May 9, 2012 at 1:03
  • 4
    $\begingroup$ I wouldn't get worked up by a debate between Shalizi and Yudkowsky; neither is an authority. (Shalizi writes well, though.) Anyway, don't you think physics.se is a better venue for this question? $\endgroup$
    – Emre
    May 9, 2012 at 5:20
  • 1
    $\begingroup$ Have you read many of Yudkowsky's sequence posts? I think he writes pretty well too. Both of these figures have controversial stances, but Shalizi seems to really have it out for subjective Bayesianism. The reason I asked here is because it ties together deeply with the more purely theoretical stats paper that Shalizi wrote with Andrew Gelman, which also is riddled with philosophical problems (though Gelman is a total pro when it comes to practice). (link) $\endgroup$
    – ely
    May 9, 2012 at 6:07
  • 1
    $\begingroup$ I've been trying to put this down to equations, but I can't seem to do it yet. I think Shazili's biggest problem is his seccond assumption on Section 1, namely, that you can just update the (random) phase point $X$ using Bayes Rule. As Yudkowsky points out, this neglects the fact that when you measure again and update your initial distribution, you have to add up YOUR contribution to the system... $\endgroup$
    – Néstor
    May 12, 2012 at 7:36
  • $\begingroup$ ...and this comes in a lot of forms: trying to control your system (which is unique each time, making the problem maybe essentially stochastic, in whose case the notion of entropy would not make sense...maybe we should talk of entropy rate?). I've been trying to convince myself that this contribution can be modelled as a linear transformation of the phase point-vector $X$: this would explain that the inequality that Shazili uses is not valid, because the resulting entropy would have an extra term (the logarithm of the determinant of the linear transformation). $\endgroup$
    – Néstor
    May 12, 2012 at 7:39

3 Answers 3


In short: 1:0 for Yudkowsky.

Cosma Shalizi considers a probability distribution subjected to some measurements. He updates the probabilities accordingly (here it is not important if it is the Bayensian inference or anything else).

No surprising at all, the entropy of the probability distribution decreases.

However, he makes a wrong conclusion that it says something about the arrow of time:

These assumptions reverse the arrow of time, i.e., they make entropy non-increasing.

As it was pointed out in comments, what matters to thermodynamics, is the entropy of a closed system. That is, according to the second law of thermodynamics, entropy of a closed system cannot decrease. It says nothing about the entropy of a subsystem (or an open system); otherwise you couldn't use your fridge.

And once we measure sth (i.e. interact and gather information) it is not a closed system anymore. Either we cannot use the second law, or - we need to consider a closed system made of the measured system and the observer (i.e. ourselves).

In particular, when we measure the exact state of a particle (while before we knew its distribution), indeed we lower its entropy. However, to store the information we need to increase our entropy by at least the same amount (typically there is huge overhead).

So Eliezer Yudkowsky makes a good point:

1) Measurements use work (or at least erasure in preparation for the next measurement uses work).

Actually, the remark about work is not the most important here. While the thermodynamics is about relating (or trading) entropy to energy, you can get around (i.e. we don't need to resort to Landauer's principle, of which Shalizi is skeptical). To gather some new information you need to erase the previous information.

To be consistent with classical mechanics (and quantum as well), you cannot make a function arbitrarily mapping anything to all zeros (with no side effects). You can make a function mapping your memory to all zero, but at the same time dumping the information somewhere, which effectively increases the entropy of the environment.

(The above originates from Hamiltonian dynamics - i.e. preservation of the phase space in the classical case, and unitarity of evolution in the quantum case.)

PS: A trick for today - "reducing entropy":

  • Flip an unbiased coin, but don't look at the result ($H = 1$ bit).
  • Open your eyes. Now you know its state, so its entropy is $H = 0$ bits.
  • 1
    $\begingroup$ is this tl;dr version correct-ish: "Shalizi's paper is just a specialized restatement of Maxwell's demon"? $\endgroup$ May 16, 2012 at 19:47
  • $\begingroup$ @ArtemKaznatcheev Basically yes. But more in taste closed vs open systems. But for the ones who don't like reading there is the first line ;). $\endgroup$ May 16, 2012 at 19:51
  • $\begingroup$ I like this answer, but I am having a hard time reconciling with a discussion on another thread. Look at this link and find the thread/answer started by the user "pragmatist". If you add a paragraph or two addressing that argument (or explaining why that argument is valid / disagrees with your answer above), I will be happy to accept. $\endgroup$
    – ely
    May 19, 2012 at 18:20
  • $\begingroup$ @EMS Well, "Could you comment a discussion?" is not the best suited for SE (and in general, there are many arguments). Moreover in fact I justified critique of Shalizi's paper. Including also critique of a critique of a critique of a paper is asking for too much. Could you be more specific, i.e. point exact points? However: "When we do statistical mechanics, we are not usually interested in the entropy of the system plus the observer" - false (open vs closed systems), "the system evolution will not be unitary" - true, but even classically you cannot decrease total entropy. $\endgroup$ May 19, 2012 at 18:46
  • $\begingroup$ @EMS The erasure principle is deeper than stat. mech. - as I said, if it not satisfy it both refutes quantum and classical mechanics. And once more: you cannot apply rules for closed systems to open systems - so most of arguments by pragmatist either are not scientific (i.e. in what to believe or not) or ignoring physics. $\endgroup$ May 19, 2012 at 18:58

Shalizi's flaw is very basic and derives from assumption I, that the time evolution is invertible (reversible).

The time evolution of INDIVIDUAL states is reversible. The time evolution of a distribution over ALL OF PHASE SPACE is most certainly not reversible, unless the system is in equilibrium. The paper treats time-evolution of distributions over all of phase space, not that of individual states, and so the assumption of invertibility is totally unphysical. In the equilibrium case, the results are trivial.

The arrow of time comes from this fact, actually, that time evolution of distributions are not reversible (the reason gradients run down and gases spread out). The irreversibility is known to emerge out of 'collision terms'

If you take this into account, his argument falls apart. Information entropy = thermodynamic entropy, still, for now. :D

  • 1
    $\begingroup$ Because at a fundamental level QM is deterministic--the Schrodinger equation precisely describes how a system evolves over time and there's no uncertainty about that--and it is linear, it would seem that reversibility in the evolution of individual states would immediately imply reversibility in any distribution of such states. I would therefore like to see your mathematical justification of your assertion to the contrary, because it will show more clearly what you are now only implicitly assuming about the dynamical equations. $\endgroup$
    – whuber
    Mar 4, 2013 at 23:13
  • $\begingroup$ For an equilibrium distribution, things are trivial, time evolution is reversible. For a dissipative system, where phase space volume is not constant, many states of the initial distribution might be mapped to a single state of the final distribution, or vice versa (no longer reversible). This is clear in the case of, eg, free expansion of an ideal gas. The motion of each individual particle is clearly reversible, but the expansion itself is not, since it involves a change in phase space volume. The gas never 'unexpands'. If your still not happy, I can work out some math for you. $\endgroup$
    – Ethan
    Mar 5, 2013 at 6:58
  • $\begingroup$ Since you're accusing Shalizi of being wrong about this, offering some kind of objective mathematical support would be a good idea. But be careful not to stray too far from the focus of this site, which is about analyzing data, not physics! Nevertheless, the free expansion example does not seem dispositive to me, because in a (hypothetically) compact universe there appears to be no such thing: the gas expands into somewhere else. $\endgroup$
    – whuber
    Mar 5, 2013 at 15:45
  • $\begingroup$ Right sometimes I forget which stackexchange I'm on. Maybe I'll get something started over there. But for the gas, the entropy change is TdS = dU + pdV but dU is zero is we're adiabatic so dS = pdV/T. By ideal gas law dS = nRdV/V so going from v1 to v2 changes the entropy by ln(v2/v1). Basically all spontaneous macroscopic processes (ie. reproducible) are irreversible. But perhaps getting this from basic principles isn't trivial (Boltzmann spent his life on it) $\endgroup$
    – Ethan
    Mar 5, 2013 at 16:57

The linked paper explicitly assumes that

The evolution operator T is invertible.

But if you use QM in the conventional way, then this assumption doesn't hold. Suppose you have a state X1 which can evolve into either X2 or X3 with equal probability. You would say that state X1 evolves into the weighted set [1/2 X2 + 1/2 X3]. Shalizi proves that this set has no more entropy than X1 did.

But we, as observers or as part of that system, only get to look at one of the branches, either X2 or X3. Picking which of those two branches we get to look at adds one bit of new entropy, and this selection is not invertible. This is where the increase in entropy with time comes from. What Shalizi has done, is to use math in which all entropy originates in quantum branching, then forget that quantum branching happens.

  • 2
    $\begingroup$ The paper (as the second law) deals with closed systems. Quantum mechanics is completely reversible on a closed system (i.e. all operators are unitary). The only non-reversible operation in quantum mechanics is measurement; if you measure a closed system then it is no longer closed from the thermodynamics perspective. If your observer is inside the system, and measures a sub-system, then the observer+sub-system evolve unitarily together, and thus the operation is invertible (this trick is informally called the "Church of the larger Hilbert Space"). Thus, your argument from "QM" is wrong. $\endgroup$ May 16, 2012 at 19:25
  • 1
    $\begingroup$ This is only if you believe the Copenhagen interpretation though (or others that separate 'measurement' from the unitary processes). Many Worlds holds that measurement is just the usual unitary laws and hence is perfectly reversible; it's just an artifact of the universe's initial state that it's probabilistically unlikely to see its reversal (I may not be explaining it very well, I'm not a physicist). At any rate, I'm not convinced that this answer should be downvoted due to this criticism. $\endgroup$
    – ely
    May 16, 2012 at 19:29
  • 1
    $\begingroup$ @EMS It doesn't matter what interpretation you use, QM of a closed system is reversible. But in the bigger context of the original question, the details of the answerer being wrong about QM is irrelevant: Shalizi already addresses this point in section II.A in a more general sense; even a correct form of this answer does not go beyond the shortcoming Shalizi himself points out. $\endgroup$ May 16, 2012 at 19:34
  • 1
    $\begingroup$ As mentioned in another thread discussing this, this answer appears to just be the flipside of the other answer given: if you insist on the closed system requirement, then you must find your source of entropy (i.e. Shalizi's "closed system" has to include the person with the bit of entropy for 'happening to proceed down one (unknown) branch of the two branches'. That is, it seems like this answer is also saying that Shalizi's paper is just a restatement of Maxwell's Demon. Again, I may be misunderstanding it due to lack of formal physics training. $\endgroup$
    – ely
    May 16, 2012 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.