Bubbles Bad; Ripples Good

… Data aequatione quotcunque fluentes quantitates involvente fluxiones invenire et vice versa …

Category: classical analysis

Arrow’s Impossibility Theorem

Partially prompted by Terry’s buzz, I decided to take a look at Arrow’s Impossibility Theorem. The name I have heard before, since I participated in CollegeBowl as an undergraduate, and questions about Arrow’s theorem are perennial favourites. The theorem’s most famous interpretation is in voting theory:

Some definitions

  1. Given a set of electors E and a finite set of candidates C, a preference \pi assigns to each elector e \in E an ordering of the set C. In particular, we can write \pi_e(c_1) > \pi_e(c_2) for the statement “the elector e prefers candidate c_1 to candidate c_2“. The set of all possible preferences is denoted \Pi.
  2. A voting system v assigns to each preference \pi\in\Pi an ordering of the set C.
  3. Given a preference \pi and two candidates c_1,c_2, a bloc biased toward c_1 is defined as the subset b(\pi,c_1,c_2) := \{ e\in E | \pi_e(c_1) > \pi_e(c_2) \}
  4. The voting system is said to be
    1. unanimous if, whenever all electors prefer candidate c_1 to c_2, the voting system will return as such. In other words, “\pi_e(c_1) > \pi_e(c_2) \forall e\in E \implies v(\pi,c_1) > v(\pi,c_2)“.
    2. independent if the voting results comparing candidates c_1 and c_2 only depend on the individual preferences between them. In particular, whether v(\pi,c_1) > v(\pi,c_2) only depends on b(\pi,c_1,c_2). An independent system is said to be monotonic if, in addition, a strictly larger biased bloc will give the same voting result: if v(\pi,c_1) > v(\pi,c_2) and b(\pi,c_1,c_2) \subset b(\pi',c_1,c_2), then v(\pi',c_1) > v(\pi',c_2) necessarily.
    3. dictator-free if there isn’t one elector e_0\in E whose vote always coincides with the end-result. In other words, we define a dictator to be an elector e_0 such that v(\pi,c_1) > v(\pi,c_2) \iff \pi_{e_0}(c_1) > \pi_{e_0}(c_2) for any \pi\in \Pi, c_1,c_2\in C.
  5. A voting system is said to be fair if it is unanimous, independent and monotonic, and has no dictators.

And the theorem states

Arrow’s Impossibility Theorem
In an election consisting of a finite set of electors E with at least three candidates C, there can be no fair voting system.

As we shall see, finiteness of the set of electors and the lower-bound on the number of candidates are crucial. In the case where there are only two candidates, the simple-majority test is a fair voting system. (Finiteness is more subtle.) It is also easy to see that if we allow dictators, i.e. force the voting results to align with the preference of a particular predetermined individual, then unanimity, independence, and monotonicity are all trivially satisfied.

What’s wrong with the simple majority test in more than three candidates? The problem is that it is not, by definition, a proper voting system: it can create loops! Imagine we have three electors e_1, e_2, e_3 and three candidates c_1,c_2,c_3. The simple majority test says that v(\pi,c_1) > v(\pi,c_2) if and only if two or more of the electors prefer c_1 to c_2. But this causes a problem in the following scenario:

e_1: c_1 > c_2 > c_3
e_2: c_2 > c_3 > c_1
e_3: c_3 > c_1 > c_2

then the voting result will have v(c_1) > v(c_2), v(c_2) > v(c_3), and v(c_3) > v(c_1), a circular situation which implies that the “result” is not an ordering of the candidates! (An ordering of the set requires the comparisons to be transitive.) So the simple-majority test is, in fact, not a valid voting system.

From this first example, we see already that, in the situation of more than 2 candidates, designing a voting system is a non-trivial problem. Making them fair, as we shall see, will be impossible. Read the rest of this entry »


Compactness. Part 1: Degrees of freedom

This series of posts has its genesis in a couple questions posted on MathOverflow.net. It is taking me a bit longer to write this than expected: I am not entirely sure what level of an audience I should address the post to. Another complication is that it is hard to get ideas organized when I am only writing about this during downtime where I hit a snag in research. The main idea that I want to get to is “a failure of sequential compactness of a domain represents the presence of some sort of infinity.” In this first post I will motivate the definition of compactness in mathematics.

We start with the pigeonhole principle: “If one tries to stuff pigeons into pigeonholes, and there are more pigeons than there are pigeonholes, then at least one hole is shared by multiple pigeons.” This can be easily generalized the the case where there are infinitely many pigeons, which is the version we will use

Pigeonhole principle. If one tries to stuff infinitely many pigeons into a finite number of pigeonholes, then at least one hole is occupied by infinitely many pigeons.

Now imagine being the pigeon sorter: I hand you a pigeon, you put it into a pigeonhole. However you sort the pigeons, you are making a (infinite) sequence of assignments of pigeons to pigeonholes. Mathematically, you represent a sequence of points (choices) in a finite set (the list of all holes). In this example, we will say that a point in the set of choices is a cluster point relative to a sequence of choices if that particular point is visited infinitely many times. A mathematical way of formulating the pigeonhole principle then is:

Let S be a finite set. Let (a_n) be a sequence of points in S. Then there exist some point p \in S such that p is a cluster point of (a_n).

The key here is the word finite: if you have an infinitely long row of pigeonholes (say running to the right of you), and I keep handing you pigeons, as long as you keep stepping to the right after stuffing a pigeon into a hole, you can keep a “one pigeon per hole (or fewer)” policy. Read the rest of this entry »

A little Hilbert space problem

First let us consider the following question on a finite dimensional vector space. Let (V, \langle\rangle) be a k-dimensional Hermitian-product space. Let (e_i)_{1\leq i \leq k} be an orthonormal basis for V. Let T:V\to V be the linear operator defined by T(e_i) = e_{i+1} when i < k, and T(e_k) = 0. Does there exist any non-trivial vector v\in V such that \langle v,v\rangle = 1 and \langle v, T^jv\rangle = 0?

The answer, in this case, is no. Present v = \sum v_i e_i where v_i are complex numbers. Let a be the smallest index such that v_a \neq 0 and v_i = 0, i < a. Similarly let b be the largest non-vanishing index. If a = b, then v = v_a e_a is a multiple of a standard basis element, and so is trivial. So assume a < b. Now, by the requirement \langle v, T^{b-a}v\rangle = 0, we see that v_a v_b = 0, which contradicts our assumption that a,b are the minimum and maximum non-vanishing indices. In this proof, we used crucially that V is finite dimensional, so that a largest element b can exist.

Now, onto the real question

Take the complex Hilbert space \ell^2(\mathbb{N}), i.e. the set of all complex sequences (a_i)_{0\leq i < \infty} satisfying \sum_{i\in\mathbb{N}} |a_i|^2 < \infty. Let e = (1,0,0,\ldots), and let latex T$ be the right shift operator: (Ta)_{i+1} = a_i and (Ta)_0 = 0. Then T^ke is an orthonormal basis of \ell^2, and we have \langle e, T^ke\rangle = \delta_0^k. Does there exist non-trivial elements of \ell^2 for which \langle v, T^kv\rangle = \delta_0^k hold?

The answer is yes by the way. Read the rest of this entry »