## Tuesday, July 30, 2013

### Uniform probabilities and Borel paradox

Question 1: I uniformly choose a random number X from the interval [0,1] (all points from 0 to 1, inclusive). What is the conditional probability that X is 1/9, given that X is either 1/9 or 4/9? I.e., if I receive the information that either 1/9 or 4/9 was picked, how confident should I be that 1/9 was picked?

Answer? Surely, the right answer is: 1/2. Both points are equally likely.

Question 2: I shoot a dart at a circular target of radius 1, with uniform distribution over the target, and I measure the distance R between where the dart hits and the center of the target. What is the conditional probability that R is 1/3, given that R is 1/3 or 2/3?

Answer? We expect this to be less than one half, because the circle of radius 2/3 is bigger. More precisely, the circle of radius R has circumference 2πR. The conditional probability of being on some circle, given that one is on one of two circles, is presumably proportional to the circumference of the relevant circle. Thus: P(R=1/3|R=1/3 or R=2/3)=2π(1/3)(2π(1/3)+2π(2/3))=1/3.
But now let Y=R2. Observe that Y is uniformly distributed over the interval [0,1]. Here's why. Y is in the interval [a,b] (where ba) provided that R is in the interval [√a,√b]. But the probability of R being in that interval is equal to the probability that the dart lands between √a and √b units away from the center of the target. The region where this happens has area π(√b)2−π(√a)2. The total area of the circle is π(1)2. So the fraction of the area of the circle where Y is in [a,b] is equal to (π(√b)2−π(√a)2)/π=ba. Thus, the probability that Y is in [a,b] is equal to ba, which is exactly what we have in the case of a uniform distribution.[note 1]

Let's now go back to Question 1. The only thing I stipulated was that X is uniformly distributed over [0,1]. Well, we've seen that Y is uniformly distributed over [0,1]. So, what we said about X's conditional probability should hold for Y. Thus, the conditional probability of Y being 1/9 given that it's 1/9 or 4/9 should be 1/2. But let's see: P(Y=1/9|Y=1/9 or Y=4/9)=P(R=√(1/9)|R=√(1/9) or Y=√(4/9))=P(R=1/3|R=1/3 or R=2/3), by definition of Y. But we've already worked out the latter conditional probability as the answer to Question 2: it's 1/3.

(This is of course a version of the Borel paradox.)

So what is going on?

Well, we're conditioning on events of zero probability. That's fishy. One thing we could learn from this story is that saying that some measurement is uniformly distributed in the sense in which that's normally understood does not convey all the relevant information about that measurement. For to compute conditional probabilities on null sets, one needs more information on how the "uniform" measurement was generated. For if it was generated as the square of the distance from the center of our target of unit radius, the conditional probabilities will be different than if it is generated in a more truly uniform manner.

It is tempting to say that true uniformity of a number in [0,1] requires that the Popper function associated with the process be invariant under isometries, in this case translations. That's fine in one dimension, but in three dimensions such strong isometric invariance cannot hold.

A different move is simply to admit that the notion of conditional probability just doesn't make sense when we're conditioning on sets of measure zero. Sure, we can sometimes talk about what credences it would be rational to give in some condition, where that condition has null probability. But that is a matter of rationality, not of formal probability theory.

I've previously made this point with infinitesimals.

#### 1 comment:

Alexander R Pruss said...

One can also run this in reverse, challenging the intuition behind the answer to Question 2. Here's a way to generate a random point in the unit circle. Start by independently generating uniformly distributed numbers Y and Z in [0,1] in the stronger, intuitive sense of "uniformly". Then let our random point be (sqrt(Y), 2pi Z) in polar coordinates. It's easy to check that this is uniform in the probabilistic sense. Then the conditional probability of being on the R=1/3 circle given being on the R=1/3 or R=2/3 circle will be equal, as R = sqrt(Y).