.... continued from Pascal's Wager: Objection to Expected Utility
In my research I discovered a seemingly non-statistical argument for
following expected utility. It is found in:
Krantz, Luce, Suppes, and Tversky. "Foundations of Measurement: Volume
I: Additive and Polynomial Representations." Dover Publications, Inc.
Mineola, NY 2007.
The relevant chapter is Chapter 8: Conditional Expected Utility.
It would be hard for me to adequately summarize it, so to follow along
with this post you'll have to have a copy of the book, which I found
at a university library and also bought. It was recently republished
in paperback, so it is available for a reasonable price.
I will copy the axioms of the proof here as best I can in this format,
but again, I have too tenuous an understanding of it even to explain
the meanings of the terms.
^ = intersection
U = union
e = element of
0 = empty set
C = subset of
~, ~ = binary relations, not the numerical kind
1. Closure: (i) If A ^ B = 0, then fA U gB e D;
(ii) If B C A, then the restriction of fA to B is in
D
2. Weak order: >~ is a weak ordering of D
3. Union indifference: If A ^ B = 0 and fA ~ gB, then fA U gB ~ fA
4. Independence: If A ^ B = 0, then fA(1) >~ fA(2) iff
fA(1) U gB >~ fA(2) U gB
5. Compatibility: If A ^ B = 0, fA(i) ~ gB(i), i = 1,2,3,4, fA(1) U
kB(1) ~ fA(2) U kB(2), and hA(1) U gB(1) ~ hA(2) U gB(2), then
fA(3) U kB(1) >~ fA(4) U kB(2) iff hA(1)
U gB(3) >~ hA(2) U gB(4)
6. Archimedean: If A ^ B = 0, N is a sequence of consecutive integers,
not gB(0) ~ gB(1), and fA(i) U gB(1) ~ fA(i+1) U gB(0) for all i, i+1
e N, then either N is finite or {fA(i) | i e N} is unbounded
7. Nullity: (i) if R e N and S C R, then S e N;
(ii) R e N iff for all fAUB e D with A ^ B = 0, fAUB ~
fA, where fA is the restriction of fAUB to A
8. Nontriviality: (i) E - N has at least three pairwise disjoint
elements;
(ii) D/~ has at least two distinct equivalence
classes
9, Restricted solvability: (i) if A and gB are given, then there
exists hA in D for which hA ~ gB.
(ii) if A ^ B = 0 and hA(1) U gB >~
fAUB >~ hA(2) U gB, then there exists hA e D such that hA U gB ~ fAUB.
They go on to state axioms 5 and 6 in a different way:
5'. If {fA(i) | i e N} and {hB(i) | i e N} are any two standard
sequences such that, for some j, j+1 e N, fA(j) ~ hB(j) and fA(j+1) ~
hB(j+1), then for all i e N, fA(i) ~ hB(i).
6'. Any strictly bounded standard sequence is finite.
Axiom 6' is a restatement of axiom 6, and axiom 5' is the statement of
Lemma 5, which is the only place axiom 5 is used in their proof.
Some of these axioms are rather obvious (once you understand what they
mean), and some (pretty much 6 on) are dry mathematics intended mostly
to cover possibilities that can occur in abstract algebra, but
probably are irrelevant to the real world.
Axiom 4 (Independence) seems to have a subtle philosophical argument
to it. Suppose you have to choose between action 1 and action 2, i.e.
fA(1) U gB and fA(2) U gB. Suppose an event will happen in the future.
If event A happens, then depending on your choice you'll get fA(1) or
fA(2). If event B happens, you'll get gB either way. If event A
happened, then afterwards you would prefer to have chosen action 1. If
event B happens, then it doesn't matter, so you may as well wish to
have chosen action 1. Therefore, before the event happens, when you
are deciding between fA(1) U gB and fA(2) U gB, you should prefer
fA(1) U gB. I'm not real confident in this logic but it seems pretty
good.
Axiom 5 just befuddles me. I can barely understand what it means in
terms of the real world. I cannot convince myself one way or the other
whether it is universally true or has counterexamples. Nor does axiom
5' help. It's stated in a less-complicated way, but I still see no
reason why it should necessarily be true. Can anyone familiar with
this proof explain this axiom to me in plain English?
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