## Saturday, August 10, 2013

### Normal Popper functions and comparative probability

Let Ω be a non-empty set and F be a field of subsets (i.e., set of subsets closed under finite unions and complements). A Popper function on F is a real-valued function P defined for pairs of members of F such that:

1. 0≤P(X|Y)≤P(Y|Y)=1
2. if P(Ω−Y|Y)<1, then P(−|Y) is a finitely additive probability on F
3. P(XY|Z)=P(X|Z)P(Y|XZ)
4. if P(X|Y)=P(Y|X)=1, then P(Z|X)=P(Z|Y).
(This is van Fraassen's axiomatization.) A member B of F is normal provided that P(Ω−B|B)<1. (This condition is equivalent to saying that P(∅|B)<1.) A Popper function is normal provided that every non-empty member of F is normal.

On the other hand, a comparative probability function on F (Paul Bartha has talked about things like this) is a function from pairs of members of F to [0,∞] such that:

1. C(X,X)=1
2. C(X,Y)C(Y,Z)=C(X,Z) provided the left-hand side is defined (0 times ∞ and ∞ times zero are the undefined cases)
3. C(−,Y) is a finitely-additive measure if Y is non-empty.

Alan Hajek, I think, has suggested that one can define a comparative probability function in terms of a Popper function. We can do it as follows. If P is a normal Popper function, then let CP(X,Y)=P(X|XY)/P(Y|XY). And given a comparative probability function, we can define a normal Popper function by PC(X|Y)=CP(XY|Y).

I haven't written out the details, but it looks like we then have:

Proposition. If P is a normal Popper function, then CP is a comparative probability function, and if C is a comparative probability function, then PC is a normal Popper function. Moreover, if P is a normal Popper function and C=CP, then PC=P, while if C is a comparative probability function and P=PC, then CP=C.

Thus, there is a nice one-to-one correspondence between comparative probabilities and normal Popper functions.

Personally, I find comparative probabilities to be easier to prove theorems about than Popper functions, because I find (6) much easier to remember than (3).

#### 1 comment:

Alexander R Pruss said...

Alan Hajek can't remember coming up with the definition of C_P. I wonder where I saw it?