Suppose you do two independent experiments, A and B, each of which uniformly generates a number in the interval I = [0, 1).
Here are some properties we would like to have on our probability assignment P:
There is a value α such that P(A = x)=P(B = x)=α for all x ∈ I and P((A, B)=z)=α2 for all z ∈ I2.
For every subset U of I2 consisting of a finite union of straight lines, P((A, B)∈U) is well-defined.
For any measurable U ⊆ I2, if P((A, B)∈U|A = x)=y for all x ∈ I, then P((A, B)∈U)=y.
For any measurable U ⊆ I2, if P((A, B)∈U|B = x)=y for all x ∈ I, then P((A, B)∈U)=y.
The assignment P satisfies the axioms of finitely additive probability with values in some real field.
Here is an interesting consequence. Let U consist of two line segments, one from (0, 0) to (1, 1/2) and the other from (0, 1/2) to (1, 1). Then every vertical line in I2 intersects U in exactly two points. This is measurable by (2). It follows from (1) and (5) that P((A, B)∈U|A = x)=2α for all x ∈ I. Thus, P((A, B)∈U)=2α by (3). On the other hand, every horizontal line in I2 meets U in exactly one point, so P((A, B)∈U|B = x)=α by (1) and P((A, B)∈U)=α by (4). Thus, 2α = α, and so α = 0.
In other words, if we require (1)-(5) to hold, then the probability of every single point outcome of either experiment must be exactly zero. In particular, it is not possible for the probability of a single point outcome to be a positive infinitesimal.
Cognoscenti of these kinds of arguments will recognize (3) and (4) as special cases of conglomerability, and are likely to say that we cannot expect conglomerability when dealing with infinitesimal probabilities. Maybe so: but (3) and (4) are only a special case of conglomerability, and they feel particularly intuitive to me, in that we are partitioning the sample space I2 on the basis of the values of one of the two independent random variables that generate the sample space. The setup—say, two independent spinners—seems perfectly natural and unparadoxical, the partitions seem perfectly natural, and the set U to which we apply (3) and (4) is also a perfectly natural set, a union of two line segments. Yet even in this very natural setup, the friend of infinitesimal probabilities has to embrace a counterintuitive violation of (3) and (4).
Well, I can’t argue with your intuition.:-)
ReplyDeleteBut in a generic independent cross product space, sloping lines are not as ‘natural’ as they could be. The natural (to me!) sets would be 'rectangles’ (i.e. a subset of A set crossed with a subset of B). The natural algebra would consist of sets that could be expressed as a finite union of such sets.
To define general sloping straight lines (as in (2) and in the definition of U), you need special properties of A and B (viz. that you can define offset-linear functions between them.)
So I’m not seeing (3) and (4) as more natural or intuitive than general conglomerability. If they were restricted to ‘natural’ sets as described above, maybe they would be more intuitive. But that would exclude your example U.
All that said, conglomerability is highly intuitive in any case.
"Let U consist of two line segments, one from (0, 0) to (1, 1/2) and the other from (1/2, 0) to (1, 1). "
ReplyDeleteI think you mean (0,1/2) rather than (1/2,0).
Andrew: Good catch!
ReplyDeleteIan: It's not so much the naturalness of the set U that makes (3) and (4) more intuitive to me as the naturalness of the partitions.
Your argument is fine but I don't see why it's necessary: doesn't the usual proof that the measure of a point is zero suffice?
ReplyDeleteIf x is any point, and if a>0 is any number, even an infinitesimal, then x is contained in an interval of length a. If the measure/probability is uniform, then this means that
P(x) <= a.
Letting a->0 shows that P(x) = 0.
But I'm not a probabilist, I'm probably missing some subtlety...