Our backyard is full of life

We have a pair of Carolina anoles, in addition to geckos, squirrels and cardinals. Quite delightful!

Even more on infinitesimals and uniform probabilities

Let [0,1) be the set of numbers x such that 0≤x<1. Suppose X is a number uniformly chosen on [0,1) such that every number has equal probability. Let f be a function from [0,1) to [0,1) defined as follows: f(0.a₁a₂a₃...)=0.a₂a₃a₄..., where 0.a₁a₂a₃... is a decimal expansion. In other words, f(x) removes the first digit after the decimal point. Observe that f is a 10-to-1 function. E.g., f(0.01)=f(0.11)=f(0.21)=...=f(0.91)=0.1.

Now, let Y=f(X). Observe that the probability that Y is equal to any particular value is ten times the probability that X is equal to any particular value: P(Y=x)=10P(X=x). For suppose x=0.a₂a₃.... Then: P(Y=0.a₂a₃...)=P(X=0.0a₂a₃...)+P(X=0.1a₂a₃...)+...+P(X=0.9a₂a₃...)=10P(X=x), since all values of X have equal probability.

If the probability of every particular value of X is zero, as classical probability says, there is nothing odd here. We quite properly get P(Y=x)=10P(X=x) as both sides are zero.

But if the probabilities are non-zero (say, because they are infinitesimal), then we have something quite odd. We have two uniformly randomly chosen numbers, X and Y, in [0,1) such that for any x in [0,1) we have P(Y=x)>P(X=x). (The construction is basically that of Williamson.)

Thus, if infinitesimal probabilities of individual outcomes in continuous lotteries are allowed, then it is possible to have two single-winner lotteries such that for every ticket, that ticket is more likely to be the winner on the second lottery. That seems absurd.

Of course, the point is also true for discrete lotteries. Suppose W is chosen from among 0,1,2,3,... with every point having equal probability. Let g(x) be the integer part of x/10 (i.e., we divide x by ten, and drop everything after the decimal point). Then g is a 10-to-1 function. Let Z=g(W). Then for every nonnegative integer n, the probability that Z equals n is 10 times the probability that W equals that integer.

Indifference is dead, again

I've noted that given the Axiom of Choice, the Hausdorff Paradox kills the principle of indifference. But we don't need the Axiom of Choice to kill off Indifference in this way! Hausdorff's proof of his paradox[note 1] also showed, without at this point in the proof using the Axiom of Choice, that:

• There are disjoint countable subsets A, B and C of the (surface of the) sphere and a subgroup G of rotations about the center such that: (a) U=A∪B∪C is invariant under G, and (b) A, B, C and B∪C are all equivalent under rotations from G.

Suppose the only thing we know about some point x is that it lies in U. Now A, B and C are rotationally equivalent under rotations that U is invariant under, so our information about whether x lies in A or in B or in C is exactly on par. By Indifference, hypotheses that are equivalently related to our information get equal probability. Thus P(x∈A)=P(x∈B)=P(x∈C)=1/3 (that they equal 1/3 is due to the fact that the three hypotheses are mutually exclusive and jointly exhaustive). But A and B∪C are also rotationally equivalent under rotations that U is invariant under, so again by Indifference P(x∈A)=P(x∈B∪C)=1/2. And so 1/2=1/3, which is absurd.

Commands and requests

One way to help out a sceptic about something is by arguing that what she is sceptical about is way more widespread than she thinks. Of course, that's a risky strategy, since instead of her scepticism disappearing, she might just widen its scope. Here's one example of this risky strategy.

Authority sceptics are sceptical that another person's commands can generate obligations for us, except in certain unproblematic ways (a command from someone trustworthy might provide epistemic reason to think that the commanded action is independently obligatory; when one has reason to believe that others will follow the command, there might be reasons to coordinate one's activity with theirs; etc.). There is something seemingly magical about generating obligations for another just by commanding.

But isn't it equally magical that we can generate non-obligating reasons for another just by requesting, or obligating reasons for self just by promising? Yet it seems quite absurd to be a request or promise sceptic.

Scoring rules and outcomes of betting behavior

The literature has two ways of measuring the fit between one's credence in a proposition and reality. The "epistemic way" uses a scoring rule to measure the distance between one's credence and the truth (if the proposition is true, then a credence of 0.8 is closer to truth than a credence of 0.6). The "pragmatic way" looks at how well one is going to do if one bets in accordance with one's credence.

A standard condition imposed on scoring rules is propriety. A proper scoring rule is one where you don't expect to improve your by shifting your credence without evidence.

I think the two ways come to the same thing, at least in the case of a single proposition. Any appropriate (yeah, some more precision is needed) betting scenario gives rise to a proper scoring rule, where your score for a credence is minus the expected utility on the assumption that you bet according to your credence in the scenario. And, conversely, any proper scoring rule can be generated in this way from an appropriate betting scenario (or at least a limit of them—this is where the details get a bit sketchy).

Cannonball and regress

From my first year of graduate school, I've been pushing a cannonball argument against Hume's idea that a regress of causes provides a complete explanation. The argument isn't very complex, but it has some complexity (it talks of complete states and all that), and it has occurred to me that it can be simplified.

At noon the cannonball is at rest and precisely then a cannon is fired. At every time after noon, the cannonball is moving. (Maybe the whole thing takes place in space.) We now have this dialogue:

• You: Why is the cannonball ball moving at one minute after noon?
• Me: Because it's moving at half a minute after noon and there is inertia.
• You: But why is the cannonball ball moving at half a minute after noon?
• Me: Because it's moving at a quarter of a minute after noon and there is inertia.
• You: But why is the cannonball ball moving at a quarter of a minute after noon?
• Me: Because it's moving at an eighth of a minute after noon and there is inertia. Don't you see the pattern?
• You: I do see the pattern, but why is it moving at any of these times: a minute after noon, half a minute after noon, a quarter of a minute after noon, an eighth of a minute after noon and so on? Why is the cannonball moving at all at any time after noon?
• Me: Don't you see, I've explained each item in the chain, and so I've explained the chain!

You should object that as long as I haven't mentioned the cannon being fired, I haven't explained why it is that the cannonball is moving at all. If my chainwise explanation was a good explanation, then a complete explanation of all the motion of the of the cannonball could be given without mentioning the cannon. And that's absurd. So an infinite chain does not give an explanation of itself.

Now suppose that there was no such time as noon and no cannon. Would the infinite chain then explain itself? No! For the chain is no more explanatory if there is
no such time as noon and no cannon. Taking away the real explanation does not turn the chain into an explanation.

Deflation of truth and doxastic evolutionary teleology

Our beliefs are states of a faculty that aims to represent the truth by means of them. This is a part of what makes these mental states be beliefs (as opposed to, say, desires). Now on evolutionary theories of teleology, this will have to be explained in something like this way: There is a faculty that we have that (a) in fact tends to hold true beliefs, and (b) that it tended to hold true beliefs conferred a survival advantage on our ancestors that led to us having it. This is what makes the faculty count as aimed at truth.

But if one has a deflationary view of truth, one doesn't get to say this. On a deflationary theory of truth, truth is wholly characterized by something like instances of the Tarski schema, such as:

• "Snow is white" is true if and only if snow is white.

A standard deflationary account of (b) would be an infinite sentence:

• (b') a survival advantage was conferred on our ancestors by the fact that the faculty tended to hold a belief provided that: (i) it was the belief that snow is white, and snow is white, or (ii) it was the belief that snow is not white, and snow is not white, or (iii) it was the belief that tigers are dangerous, and tigers are dangerous, or (iv) it was the belief that tigers are not dangerous, and tigers are not dangerous, or (v) it was the belief that there is life on other planets, and there is life on other planets, or (vi) it was the belief that there is no life on other planets, and there is no life on other planets, or ....

But (b') is simply false. It wasn't the quantified infinite disjunction that conferred a survival advantage on our ancestors. It was at most a quantified finite sub-disjunction. In particular, while (i)-(iv) may have been a part of what conferred a survival advantage, (v) and (vi) were not.

The anti-deflationist does better. For she sees truth as a genuine property and can say (b) as is, if she takes truth to be a fundamental property, or can replace (b) by some natural analysis. But the deflationist does not see truth as a genuine property and hence cannot use (b), and thus cannot give an evolutionary account of the truth-directedness of our faculties.

So what? Well, I think it is undeniable that our doxastic faculties are truth-directed. It is, I think, particularly crucial for naturalists to be able to say this, since it will be needed for an account of intentionality. But a naturalist (in the contemporary, non-Aristotelian sense) can only give an evolutionary account of teleology. Thus, a naturalist cannot be a deflationist about truth.

But neither can a naturalist be an anti-deflationist about truth. For if a naturalist were an anti-deflationist about truth, she would have to hold that truth is to be analyzed in terms of the sorts of properties science speaks of. But science speaks only of first-order properties, and truth is not a first order property (here's one quick argument based on an idea by Leon Porter: if it were, then the liar paradox could be formulated in entirely first-order scientific terms, yielding a contradiction in the language of science).

So naturalism is in trouble.

More details on Banach-Tarski and Axiom of Choice

Earlier I claimed that even without the Axiom of Choice, the proof of the Banach-Tarski Paradox yields paradoxical results, so that the Axiom of Choice isn't the source of paradoxicality.

Here is something more precise. Without the Axiom of Choice, one can prove the following.

Almost-BT. Suppose B is a punctured ball, i.e., a solid ball without a center point, and v is a translation sufficiently large that vB does not intersect B. Then there are rigid motions (combinations of rotations and translations) α and β such that B can be partitioned into an infinite collection U of countable subsets with the property that each member A of the partition U can be further partitioned into four subsets A₁, A₂, A₃ and A₄, such that (a) A₁ and αA₂ are a partition of A and (b) vA₃ and βA₄ are a partition of vA. (This follows from the method of proof of Theorem 4.5 in the Wagon book on Banach-Tarski.)

In other words, one can partition B into a bunch of sets, each of which can then be divided into four pieces and reassembled into two copies of itself oriented the same way. Moreover, all the reassembly can be done using the same rigid motions. When we add the Axiom of Choice to the mix, we can basically take B_i to be the union of "the" sets A_i ("the" is in quotation marks because the sets A_i aren't unique given A; if they were, we wouldn't need Choice), and then B=B₁∪B₂∪B₃∪B₄=B₁∪αB₂ and vB=vB₃∪βB₄, and we have Banach-Tarski for punctured balls. (It's a bit of extra work to get Banach-Tarski for non-punctured balls, but that extra work doesn't need Choice.) But even without the Axiom of Choice, Almost-BT seems pretty counterintuitive.

Blowguns and their darts

In case this heated discussion has made you desire a blowgun, here are my instructions for blowgun and dart making. :-) Stay safe! (Here's a good safety tip!)

Spoof arguments against the Axiom of Choice

As a first year graduate student, I wrote this pseudonymous spoof, inspired by the Sokal affair. It's rather immature in places, but enjoy!

The Banach-Tarski Paradox and the Axiom of Choice

The Banach-Tarski theorem (BTT) says that, given the Axiom of Choice, a continuous ball can be decomposed into a finite number of pieces that can be rearranged to form two balls of equal size. That's weird, and is taken by some to be an argument against the Axiom of Choice.

I don't think we should take it as such an argument. Sure, BTT is paradoxical. But when one looks at the proof, one notes that the proof makes use of paradoxical results that do not depend on the Axiom of Choice. For instance, a lemma in standard proofs of BTT is the surprising fact that you can take any circle that's missing a countable number of points, decompose that circle into two disjoint (messy!) pieces, and reassemble the pieces, without overlap, to get a complete circle. That lemma is about as weird as BTT, but it doesn't use the Axiom of Choice at all.[note 1] Moreover, the proof uses paradoxical decompositions of various countable sets, and these are, well, paradoxical, but do not involve the Axiom of Choice. Using the Axiom of Choice lets you put all the paradoxicality together into a neat package, but when I think about the proof of the result, I just don't see Choice as the source of paradoxicality. In fact, once one sees all the other ingredients of the proof of BTT, the Axiom of Choice step seems quite intuitive.

Another way to put the point is this: Once one reflects enough on all the pieces of the proof of BTT that do not use Choice and accepts them, BTT no longer seems very surprising. When you cut things up into strangely scattered pieces, it's not that surprising that you can put them back together in various ways.

Popper functions and null sets

Let's go back to the problem that I keep on thinking about: How to distinguish possibilities that are classically of null probability. For instance, given a uniform choice of a point on some nice set (say, a ball) in Euclidean space, we want to say something like P({x,y})>P({x}), when x and y are distinct: it's more likely that one would hit one of two points than that one would hit a particular point. A series of my blog posts (and at least one article) showed that infinitesimals aren't the way. What about conditional probabilities?

For various reasons, instead of taking unconditional probabilities to be fundamental and defining conditional probabilities in terms of them, one may want to take conditional probabilities as fundamental. The standard method is to use Popper functions (I'll assume the linked axioms below). One might hope to use Popper functions to do things like make sense of the difference between the probability of two points in a continuous case (say, where a point is uniformly chosen in some nice subset of a Euclidean space) and the probability of a single point. For instance, one might hope that P({x,y}|{x,y})>P({x}|{x,y}) whenever x and y are distinct.

This won't work, however. Instead of working with propositions, I will work with sets—the definitions of Popper functions neatly adapt. Let Ω be a solid three-dimensional ball. Assume that P(A|B) is defined for all A and B in some algebra F of subsets of Ω.

Say that a set A in F is P-trivial provided that P(C|A)=1 for all C (including for the empty set). The empty set is trivial, of course. In order for us to have any hope of saying things like P({x,y}|{x,y})>P({x}|{x,y}), we better have sets with two points be non-trivial. Now, it's not hard to show that a finite union of trivial sets is trivial and that the subsets of a trivial set are trivial, so in order for all the finite sets to be non-trivial, it's necessary and sufficient that singletons be non-trivial.

Moreover, we want rotational symmetry. Say that F and P are rotationally symmetric provided that for any rotation r around the origin and A and B in F, rA and rB are in F, and P(rA|rB)=P(A|B).

Theorem. If P is rotationally symmetric and F includes all countable subsets of Ω, then for any sphere of positive radius around the origin lying in Ω there is at least one P-trivial countably infinite set on the surface of
that sphere. Thus, all finite sets not containing the origin are trivial.

If we are to be extending something like classical probabilities, we do want countable subsets of Ω to be in F. The triviality of finite sets follows from the fact that some singleton on every sphere of non-zero radius about the origin is trivial (since any subset of a trivial set is trivial) and if one singleton on that sphere is trivial, then by rotational invariance, they all are, while finite unions of trivial sets are trivial. So all we need to prove is the existence of that trivial countably infinite set.

The proof is easily based on standard ideas from the proof of the Banach-Tarski Paradox.

Proof. Let SO(3) be the rotation group around the origin. Famously, there is a subgroup G that is isomorphic to the free group F₂ on two elements. Choose a point ω in Ω such that ρ(ω)=ω only if ρ is the identity rotation. (This is a counting argument: G is a countable group, and each non-identity member of G has two fixed points on any given sphere around the center, so for any fixed sphere there will be only countably many fixed points of non-identity members of G on it, and hence there will be a point on the sphere that isn't a fixed point of any non-identity member.) Let H={ρ(ω):ρ∈G}. Then H is a countable subset of Ω.

For a reductio ad absurdum, suppose H is non-trivial. Then P(−|H) is a finitely additive probability measure on H. Moreover, H=ρH for any ρ∈G, so by rotational invariance P(ρK|H)=P(ρK|ρH)=P(K|H) for any ρ∈G, and so P(−|H) is a finitely additive G-invariant measure on H. Using the bijection between G and H given by f(ρ)=ρ(ω) (we use the choice of ω to see that f is one-to-one), we can then get a finitely additive G-left-invariant measure on G. But G is isomorphic to F₂ and hence is not amenable and hence has no such invariant measure. (One could also neatly demonstrate a paradoxical decomposition here.) That's a contradiction, so H must be trivial. QED

Even though this argument uses ideas from the proof of Banach-Tarski, and famously the latter uses the Axiom of Choice, this argument does not use the Axiom of Choice.

One can perhaps get out of this by having a more onerous requirement on the sets A and B that are on either side of the bar in "P(A|B)" than that they are fit in a single algebra F. We want any countable set to be an acceptable A, but perhaps we don't want to allow every countable set as an acceptable B. I don't know what natural requirement could be put here.

Yet another account of omnipotence

The following account of omnipotence runs into the McEar objection:

1. x is omnipotent iff x can do anything whose doing is consistent with the nature of x.

For suppose McEar has the essential property of doing nothing other than scratching his ear, and suppose he can scratch his ear. Then (1) counts McEar as omnipotent. That's no good.

The Pearce-Pruss account of omnipotence escapes this. But so does this minor twist on (1):

2. x is omnipotent iff x can do anything whose doing is consistent with the nature of a perfect being.

There are things consistent with the nature of a perfect being that McEar can't do, say create a pebble.

Perhaps, though, there is a circularity problem. For a perfect being has all perfections. And one of the perfections is omnipotence. However, I do not know that this is fatal. Compare:

3. a fully self-knowledgeable person is one who knows all her mental attributes.

This seems a perfectly reasonable definition, even if one of the mental attributes of such a person is being fully self-knowledgeable.

A funny uniform distribution?

Let X and Y be independent random variables uniformly distributed over [0,1]. Let Z=max(X²,Y²). Then it's easy to check that Z is also uniformly distributed over [0,1].

But now suppose we think that uniform random variables have equal infinitesimal probabilities of hitting every point. Thus, P(X=a)=P(Y=a)=α for every a, where α is some infinitesimal. What, then, is P(Z=a)? Well Z=a if and only if one of three mutually exclusive possibilities occurs:

1. X<a^1/2 and Y=a^1/2
2. X=a^1/2 and Y<a^1/2
3. X=Y=a^1/2.

Now, P(X<a^1/2)=a^1/2−O(α) (the last term is due to end-point effects: P(X<1)=1−α) and P(Y=a^1/2)=α. Thus, P((1))=αa^1/2+O(α²). By the same token, P((2))=αa^1/2+O(α²). And P(X=Y=a^1/2)=α² by independence. Thus, P(Z=a)=P((1))+P((2))+P((3))=2αa^1/2+O(α²).

In other words, Z is a uniformly distributed random variable by standard probabilistic criteria, but the probability of Z hitting different points is different: P(Z=a) is basically an infinitesimal multiple of √a.

What is happening here is that if one attempts to attach infinitesimal probabilities to the individual outcomes of bona fide classical probabilities, the infinitesimal individual outcome probabilities float free from the distribution. You can have the same individual outcome probabilities and different distributions or, as in this post, different (nonuniform) individual outcome probabilities and the same (uniform!) distribution.

Uniform distributions

Consider two random variables, X and Y, whose probability densities p_X and p_Y are shown in the following graph, with p_X(x)=1, in blue, and p_Y(x)=2x, in red.

Looking at the graph, it is tempting to say things like this: X is a uniform distribution and has equal probability of having any value between 0 and 1, while values closer to 0 are much less likely than values close to 1 for Y. We might even look at the graph and say things like: P(X=0.1)=P(X=0.2) while P(Y=0.2)>P(Y=0.1).

Of course, with these continuous distributions, classical probability theory assigns equal zero probability to every value: P(X=a)=P(Y=a)=0 for all a. But this seems wrong, and so we may want to bring in infinitesimals to remedy this, assigning to P(Y=0.2) an infinitesimal twice as big as the one we assign to P(Y=0.1), while P(X=0.2)=P(X=0.1).

Or we might attempt to express the pointwise non-uniformity of Y by using conditional probability P(Y=0.2|Y=0.1 or Y=0.2)=2/3 and P(Y=0.1|Y=0.1 or Y=0.2)=1/3, while P(X=0.2|X=0.1 or X=0.2)=1/2=P(X=0.1|X=0.1 or X=0.2).

In other words, it is tempting to say: X is pointwise uniform while Y is not.

Such pointwise thinking is problematic, however. For I could have generated Y by taking our uniformly distributed random variable X and setting Y=X^1/2. (It's an easy exercise to see that if X is uniform then the probability density of X^1/2 is given by p(x)=2x.) Suppose that I am right in what I said about the uniformity of pointwise and conditional probabilities for X. Then P(Y=0.1)=P(X=0.01)=P(X=0.04)=P(Y=0.2). And P(Y=0.2|Y=0.1 or Y=0.2)=P(X=0.04|X=0.01 or X=0.04)=1/2=P(X=0.01|X=0.01 or X=0.04)=P(Y=0.1|Y=0.1 or Y=0.2), since Y=0.1 if and only if X=0.01 and Y=0.2 if and only if X=0.04.

So in fact, Y could have the nonuniform distribution of the red line in the graph and yet be just as pointwise uniform as X.

Lesson 1: It is a mistake to describe a uniform distribution on a continuous set as one "where every outcome is equally likely". For even if one finds a way of making nontrivial sense of this, by infinitesimals or conditional probabilities say (and I think similar arguments will work for any other plausible characterization), a nonuniform distribution can satisfy this constraint just as happily.

Lesson 2: One cannot characterize continuous distributions by facts about pointwise probabilities. It is tempting to characterize the uniform distribution by P(X=a)=P(X=b) (infinitesimal version, but similarly for conditional probabilities) and the nonuniform one by P(Y=a)=(a/b)P(Y=b). But in fact both could have the same pointwise properties. I find this lesson deeply puzzling. Intuitively, it seems that chances of aggregate outcomes (like the chance that X is between 0.1 and 0.2) should come out of pointwise chances. But no.

The converse characterization would also be problematic: pointwise facts can't be derived from the distribution facts. For imagine a random variable Z which is such that Z=X unless X=1/2, and Z=1/4 if X=1/2 (cf. this paper). This variable has the same distribution as X, but it has obviously different pointwise probability facts.