## Friday, June 28, 2013

### 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.

#### 1 comment:

Alexander R Pruss said...

We can even boost the single-point uniform probability infinitely!

Let f(0.a1a2a3a4a5a6) = 0.a2a4a6...

Then f is a continuum-to-one function: infinitely many, indeed continuum many, things get mapped to one value. Yet f(X) is still uniformly distributed on [0,1).