Here’s a fairly simple question about the consequence argument. Still, I think it is an interesting question that focuses on the conditionals in the argument. The rule α is not interestingly in doubt, but I assume both α and β. Let P* be proposition describing a time slice of the universe at some point a billion years ago. Let L all the laws of nature. Let P be a true proposition describing any event after P*. If determinism is true, then P follows from the conjunction of P* and L. ☐ stands for logical necessity. Assume that determinism is true. Here is van Inwagen’s argument.
1. ☐((P* & L) ⊃ P) assumption
2. ☐(P* ⊃ (L ⊃ P)) 1; modal, prop logic
3. N(P* ⊃ (L ⊃ P)) 2; rule β
4. NP* premise
5. N(L ⊃ P) 3, 4; rule α
6. NL premise
7. NP 5, 6; rule α
I’m not so concerned with working out whether rule α or β are valid. But I am concerned about some of the implications of the premises of the argument. Note first that (8) is necessarily true,
8. N(P* ⊃ (L ⊃ P)) ≣ N(P* & L) ⊃ P))
The conditional in (3) allows for strengthening antecedents. So, (9) is also a necessary truth, for any Q whatsoever.
9. N(P* & L) ⊃ P)) ⊃ N(P* & L & Q) ⊃ P))
So, (3), together with (8) and (9) entail (10).
10. N(P* & L & Q) ⊃ P))
Let Q be that God brings time to an end. Or let Q be that a miracle occurs, as certainly happens in some worlds. Or let Q be that an event happens between the occurrences of P* and P that is inconsistent with P, as certainly happens in some worlds. Recall that Np is the proposition that “No one has any choice about the fact that p“. Now no one has a choice about time ending, or miracles occurring or events-inconsistent-with P occurring in the past, in addition to not having a choice about P* and L. And this is true whether or not any of these propositions in Q, L or P* are true.
Consider a world w in which Q is true. It is also true in w that (11), though P* and L are not both true there.
11. N(P* & L & Q)
There is no world in which I have a choice about P* or L, even if they’re false. So, we can derive that (12) is true in w, from (10) and (11).
12. NP & ~P
So, it is consistent with no one ever having a choice about, say, whether I raise my arm that I do not raise my arm. So, NP is consistent with my bringing about ~P. But then NP is consistent with a version of the principle of alternate possibilities that is relevant to my having free will relative to P.
In its classical formulation, Pascal’s Wager contends that we have something like the following payoff matrix:
God exists | No God | |
Believe | +∞ | −a |
Don’t believe | -b | c |
where a,b,c are finite. Alan Hajek, however, observes that it is incorrect to say that if you don’t choose to believe, then the payoff is finite. For even if you don’t now choose to believe, there is a non-zero chance that you will later come to believe, so the expected payoff whether you choose to believe or not is +∞.
Hajek’s criticism has the following unhappy upshot. Suppose that there is a lottery ticket that costs a dollar and has a 9/10 chance of getting you an infinite payoff. That’s a really good deal intuitively: you should rush out and buy the ticket. But the analogue to Hajek’s criticism will say that since there is a non-zero chance that you will obtain the ticket without buying it—maybe a friend will give it to you as a gift—the expected payoff is +∞ whether you buy or don’t buy. So there is no point to buying. So Hajek’s criticism leads to something counterintuitive here, though that won’t surprise Hajek. The point of this post is to develop a rigorous principled response to Hajek’s criticism entailing the intuition that you should go for the higher probability of an infinite outcome over a lower probability of it.
A gamble is a random variable on a probability space. We will consider gambles that take their values in R*=R∪{−∞,+∞}, where R is the real numbers. Say that gambles X and Y are disjoint provided that at no point in the probability space are they both non-zero. We will consider an ordering ≤ on gambles, where X≤Y means that Y is at least as good a deal as X. Write X<Y if X≤Y but not Y≤X. Then we can say Y is a strictly better deal than X. Say that gambles X and Y are probabilistically equivalent provided that for any (Borel measurable) set of values A, P(X∈A)=P(Y∈A). Here are some very reasonable axioms:
- ≤ is a partial preorder, i.e., transitive and reflexive.
- If X and Y are real valued and have finite expected values, then X≤Y if and only if E(X)≤E(Y).
- If X and Y are defined on the same probability space and X(ω)≤Y(ω) for every point ω, then X≤Y.
- If X and Y are disjoint, and so are W and Z, and if X≤W and Y≤Z, then X+Y≤W+Z. If further X<W, then X+Y<W+Z.
- If X and Y are probabilistically equivalent, then X≤Y and Y≤X.
For any random variable X, let X* be the random variable that has the same value as X where X is finite and has value zero where X is infinite (positively or negatively).
The point of the above axioms is to avoid having to take expected values where there are infinite payoffs in view.
Theorem. Assume Axioms 1-5. Suppose that X and Y are gambles with the following properties:
- P(X=+∞)<P(Y=+∞)
- P(X=−∞)≥P(Y=−∞)
- X* and Y* have finite expected values
Then: X<Y.
It follows that in the lottery case, as long as the probability of getting a winning ticket without buying is smaller than the probability of getting a winning ticket when buying, you should buy. Likewise, if choosing to believe has a greater probability of the infinite payoff than not choosing to believe, and has no greater probability of a negative infinite payoff, and all the finite outcomes are bounded, you should choose to believe.
Proof of Theorem: Say that an event E is continuous provided that for any 0≤x≤P(E), there is an event F⊆E with P(F)=x. By Axiom 5, without loss of generality {X∈A} and {Y∈A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0≤x≤P(X∈A), just choose a∈[0,1] such that aP(X∈A)=x and let F={X∈A&U≤a}. Ditto for Y.)
Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=−∞} and B={Y=−∞}. Define the random variables X_{1} and Y_{1} on [0,1] with uniform distribution by X_{1}(x)=−∞ if x≤P(A) and X_{1}(x)=0 otherwise, and Y_{1}(x)=−∞ if x≤P(B) and Y_{1}(x)=0 otherwise. Since P(A)≥P(B) by (7), it follows that X_{1}(x)≤Y_{1}(x) everywhere and so X_{1}≤Y_{1} by Axiom 3. But AX and BY are probabilistically equivalent to X_{1} and Y_{1} respectively, so by Axiom 5 we have AX≤BY. If we can show that A^{c}X<B^{c}Y then the conclusion of our Theorem will follow from the second part of Axiom 4.
Let X_{2}=A^{c}X and Y_{2}=B^{c}Y. Then P(X_{2}=+∞)<P(Y_{2}=+∞), X_{2}* and Y_{2}* have finite expected values and X_{2} and Y_{2} never have the value −∞. We must show that X_{2}≤Y_{2}. Let C={X_{2}=+∞}. By subdivisibility, let D be a subset of {Y_{2}=+∞} with P(D)=P(C). Then CX_{2} and DY_{2} are probabilistically equivalent, so CX_{2}≤DY_{2} by Axiom 5. Let X_{3}=C^{c}X_{2} and Y_{3}=D^{c}Y_{3}. Observe that X_{3} is everywhere finite. Furthermore P(Y_{3}=+∞)=P(Y_{2}=+∞)−P(X_{2}=+∞)>0.
Choose a finite N sufficiently large that NP(Y_{3}=+∞)>E(X_{3})−E(Y_{3}*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y_{4} be a random variable that agrees with Y_{3} everywhere where Y_{3} is finite, but equals N where Y_{3} is infinite. Then E(Y_{4})=NP(Y_{3}=+∞)+E(Y_{3}*)>E(X_{3}). Thus, Y_{4}>X_{3} by Axiom 2. But Y_{3} is greater than or equal to Y_{4} everywhere, so Y_{3}≥Y_{4}. By Axiom 1 it follows that Y_{3}>X_{3}. but DY_{2}≥CX_{2} and X_{2}=CX_{2}+X_{3} and Y_{2}=DY_{2}+Y_{3}, so by Axiom 4 we have Y_{2}>X_{2}, which was what we wanted to prove.