I posted a while back about an interesting experiment in group intelligence I’d like someone to perform. I thought of another interesting experiment I’d like to see performed, so it’s apparently starting to become a habit.

The experiment is as follows:

Put two people in a room. Tell them to talk to eachother. After some fixed period of time (say, 10 minutes), take them out of the room and give each of them a questionnaire. The questionnaire has N questions on of it, each of which is a binary choice. These questions could be things like “Do you mostly agree with this complex political statement?”, “You are given a choice between these two scenarios, which do you pick?” etc. Things which are very much about your value and behaviours rather than simple statements of objective facts.

Each of these questions must be filled out twice. Once with your answer, once with what you think your partner is most likely to answer.

Basic questions it would be interesting to answer:

• What questions are people particularly good at bad or predicting?
• Is the prediction rate asymmetric? If one person successfully predicts the other well, is the other likely to be worse or at prediction?
• Are predictions better when the answers are different or the same?

More advanced questions that would be interesting to answer:

• How do prediction rates differ when we vary the length of the conversation?
• How do prediction rates differ with different primings for the conversation? e.g. rather than instructing “Talk to eachother”, say “Talk to eachother about your family”
• How do prediction rates differ if instead of talking to eachother in person you talk via an instant messaging program? Or via a phone?

It occurred to me yesterday that there is an extremely common and natural example of a total ordering (modulo indifference. i.e. it’s really a total pre-ordering) of distributions which doesn’t satisfy The VNM Axioms.

(Note: We are talking about distributions rather than random variables in this, so we will follow the VNM notation of using pA + (1-p)B to mean “choose A with probability p or B with probability (1 – p)”. This has the effect of a similar combination on the distributions. It’s not addition of random variables).

Specifically, ordering distributions by their lower median value (assuming a total ordering over outcomes). More generally, by any percentile value.

This does not satisfy continuity: To see that it does not satisfy continuity, suppose our outcomes are 1, 2 and 3. Let A, B, C be lotteries choosing each of these with probability 1. Then A < B < C. But the lower median of \(pA + (1-p)B\) is 1 if \(p \geq \frac{1}{2}\) or 3 if \(p < \frac{1}{2}\). The median of \(B\) is always 2, so we are never indifferent between them. You could argue that the problem is that at \(p = \frac{1}{2}\) we should take an averaging of the two values and the median should be 2 there, but that doesn't save you. Instead consider where we have 5 values and A, B, C choose 1, 2 and 5. Then at \(p = \frac{1}{2}\) the median will be 3, so we're still not indifferent to it at any point. It also does not satisfy independence. Here's an example: Suppose we have two possible outcomes, 0 and 1. Consider distributions \(A = [0.35, 0.65]\), \(B = [0.4, 0.6]\), \(C = [0.9, 0.1]\). Let \(p = 0.75\). Then the median of \(A\) and \(B\) are both 1, while the median of \(C\) is 0. Further the median of \(pA + (1-p)C\) is 1, because the probability of it being 0 is \(0.75 * 0.35 + 0.25 * 0.9 \approx 0.49 < 0.5\). The median of \(pB + (1-p)C\) however is 0 because the probability of it being 0 is \(0.525\). So \(A \preceq B\) but \(pA + (1-p)C \succ pB + (1-p)C\) as desired. (Disclaimer: I didn't work this out by inspection, I totally just ran a computer program to find examples for me because I was pretty sure they must exist)

I wrote previously about a dominance relation amongst \(\mathbb{N}\) valued random variables. I’ve realised a nice characterisation of the relationship, which is what this post is about.

We’ll change the setting slightly to \(\mathbb{R}\) valued random variables, as proving this theorem works more nicely in this context, but the result will still hold for \(\mathbb{N}\) as a subset of it.

Definition: Let \(X, Y\) be real valued random variables. Say \(x \preceq y\) if \(\forall t. P(X \geq t) \leq P(Y \geq t)\).

Theorem: \(X \preceq Y\) iff for every monotone function \(h : \mathbb{R} \to \mathbb{R}\), \(E(h(X)) \leq E(h(y))\).

Proof of this will need a version of a lemma from the last post:

Lemma: Let \(X\) be a real-valued random variable. Then \(E(X) = \int\limits_{-\infty}^\infty P(X \geq t) dt\).

I’m going to skip proving this for now. I have proofs, but they’re a little fiddly and I got lost in the details when trying to write this up pre-lunch (it’s easy to prove given some stronger continuity assumptions if you use integration by parts). So IOU one proof.

Proof of theorem:

If \(X \preceq Y\) and \(h\) is monotone, then \(E(h(X)) = \int\limits_{-\infty}^\infty P(h(X) \geq t) dt\) and similarly for \(y\).

But \(h\) is monotone, so \(H_t = \{x : h(x) \geq t\}\) is either \([y, \infty)\) or \((y, \infty)\) for some \(y\). The latter can be written as \(\bigcup [y_n, \infty)\) for some sequence \(y_n\). We know that \(P(X \geq y) \leq P(Y \geq y)\), so we know that \(P(X \in H_t) \leq P(Y \in H_t)\) due to this characterisation. But these probabilities are respectively \(P(h(X) \geq t)\) and \(P(h(Y) \geq t)\).

So \[\begin{align*}
E(h(X)) & = \int\limits_{-\infty}^\infty P(h(X) \geq t) dt \\
& \leq \int\limits_{-\infty}^\infty P(h(Y) \geq t) dt \\
& = E(Y) \\
\end{align*}\]

as desired.

Now for the converse:

Let \(t \in \mathbb{R}\) with \(P(X \geq t) > P(Y \geq t)\). Let \(h(x) = 0\) if \(x < t\) and \(h(x) = 1\) if \(x \geq t\). Then \(h\) is monotone increasing and \(E(h(X)) = P(X \geq t) > P(Y \geq t) = E(h(Y))\) as desired.

QED

As you might have noticed, I have Flattr attached to my blog posts. It’s a system which gives me entire pennies in income and is mostly there as a vanity project.

Terence kindly Flattred my post on bribing MPs, which got me to thinking about another strategy for crowd-funding subverting the course of the democratic process by making our elected officials actually pay attention to the people they’re representing: We adopt the Flattr model. Let’s call it Bribr.

Here’s how I imagine it would work:

Every time a bill, debate or other significant event happened in parliament, we would have an electronic representation of which MPs took which sides. You could then choose to cast your vote of approval with a particular position. Every MP who took that position gets one bribry point from you (a lot of stuff goes through parliament. I imagine you could filter by some significant metrics in order to not get overwhelmed).

At the end of the month you are asked how happy you were with parliament this month. Happiness is a quantity measured in pounds sterling. That quantity is then divided by MPs according to how much you bribred each of them: If you only engaged in bribry over one issue and 10 MPs voted on that issue in accordance with your favour, they’d each get 10% of your monthly allocation.

So each MP ends the month with a certain amount of money, spread over bribs from many different people. Rather than giving them money directly there are a bunch of reward tiers – ideally you want a hundred or so different reward possibilities just to keep things varied – and the MP gets a reward appropriate to the amount of money they have assigned to them, with any not spent rolling over to the next month. Any MP who ends up with a truly ridiculous amount of money (say, multi thousands of pounds) gets one of the most lavish rewards plus a note saying “Hey. We really love what you’re doing. Is there anything you’d like that’s worth about this much?”

Moral status? I’m actually more comfortable with this then I am with the other one. You’re essentially just rewarding MPs for doing their jobs well. I’d still have to hold my nose a bit to run something like this, but I don’t think I’d have a problem using it. Legal status? Not sure. I don’t know enough about the bribery statutes to say one way or another. It’s clearly much closer to bribery than the other version, because people are being directly rewarded for their votes. Again, would have to talk to an actual lawyer.