Handling the infinities in Pascal’s Wager
January 23, 2014 — 12:01

Author: Alexander Pruss  Category: Afterlife Atheism & Agnosticism General  Tags: , , , , ,   Comments: 24

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 XY means that Y is at least as good a deal as X. Write X<Y if XY but not YX. 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(XA)=P(YA). Here are some very reasonable axioms:

1. ≤ is a partial preorder, i.e., transitive and reflexive.
2. If X and Y are real valued and have finite expected values, then XY if and only if E(X)≤E(Y).
3. If X and Y are defined on the same probability space and X(ω)≤Y(ω) for every point ω, then XY.
4. If X and Y are disjoint, and so are W and Z, and if XW and YZ, then X+YW+Z. If further X<W, then X+Y<W+Z.
5. If X and Y are probabilistically equivalent, then XY and YX.

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:

1. P(X=+∞)<P(Y=+∞)
2. P(X=−∞)≥P(Y=−∞)
3. 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≤xP(E), there is an event FE with P(F)=x. By Axiom 5, without loss of generality {XA} and {YA} 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≤xP(XA), just choose a∈[0,1] such that aP(XA)=x and let F={XA&Ua}. 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 X1 and Y1 on [0,1] with uniform distribution by X1(x)=−∞ if xP(A) and X1(x)=0 otherwise, and Y1(x)=−∞ if xP(B) and Y1(x)=0 otherwise. Since P(A)≥P(B) by (7), it follows that X1(x)≤Y1(x) everywhere and so X1Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXBY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+∞)<P(Y2=+∞), X2* and Y2* have finite expected values and X2 and Y2 never have the value −∞. We must show that X2Y2. Let C={X2}=+∞. By subdivisibility, let D be a subset of {Y2}=+∞ with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+∞)=P(Y2=+∞)−P(X2=+∞)>0.

Choose a finite N sufficiently large that NP(Y3=+∞)>E(X3)−E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+∞)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3Y4. By Axiom 1 it follows that Y3>X3. but DY2CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Pascal’s Wager in a social context
December 18, 2013 — 11:11

Author: Alexander Pruss  Category: Atheism & Agnosticism  Tags: , , , , , , ,   Comments: 4

One of our graduate students, Matt Wilson, suggested an analogy between Pascal’s Wager and the question about whether to promote or fight theistic beliefs in a social context (and he let me cite this here).

This made me think. (I don’t know what of the following would be endorsed by Wilson.) The main objections to Pascal’s Wager are:

1. Difficulties in dealing with infinite utilities. That’s merely technical (I say).
2. Many gods.
3. Practical difficulties in convincing oneself to sincerely believe what one has no evidence for.
4. The lack of epistemic integrity in believing without evidence.
5. Would God reward someone who believes on such mercenary grounds?
6. The argument just seems too mercenary!

Do these hold in the social context, where I am trying to decide whether to promote theism among others? If theistic belief non-infinitesimally increases the chance of other people getting infinite benefits, without any corresponding increase in the probability of infinite harms, then that should yield very good moral reason to promote theistic belief. Indeed, given utilitarianism, it seems to yield a duty to promote theism.

But suppose that instead of asking what I should do to get myself to believe the question is what I should try to get others to believe. Then there are straightforward answers to the analogue of (3): I can offer arguments for and refute arguments against theism, and help promote a culture in which theistic belief is normative. How far I can do this is, of course, dependent on my particular skills and social position, but most of us can do at least a little, either to help others to come to believe or at least to maintain their belief.

Moreover, objection (4) works differently. For the Wager now isn’t an argument for believing theism, but an argument for increasing the number of people who believe. Still, there is force to an analogue to (4). It seems that there is a lack of integrity in promoting a belief that one does not hold. One is withholding evidence from others and presenting what one takes to be a slanted position (for if one thought that the balance of the evidence favored theism, then one wouldn’t need any such Wager). So (4) has significant force, maybe even more force than in the individual case. Though of course if utilitarianism is true, that force disappears.

Objections (5) and (6) disappear completely, though. For there need be nothing mercenary about the believers any more, and the promoter of theistic beliefs is being unselfish rather than mercenary. The social Pascal’s Wager is very much a morally-based argument.

Objections (1) and (2) may not be changed very much. Though note that in the social context there is a hedging-of-the-bets strategy available for (2). Instead of promoting a particular brand of theism, one might instead fight atheism, leaving it to others to figure out which kind of theist they want to be. Hopefully at least some theists get right the brand of theism—while surely no atheist does.

I think the integrity objection is the most serious one. But that one largely disappears when instead of considering the argument for promoting theism, one considers the argument against promoting atheism. For while it could well be a lack of moral integrity to promote one-sided arguments, there is no lack of integrity in refraining from promoting one’s beliefs when one thinks the promotion of these beliefs is too risky. For instance, suppose I am 99.99% sure that my new nuclear reactor design is safe. But 99.9999% is just not good enough for a nuclear reactor design! I therefore might choose not promote my belief about the safety of the design, even with the 99.9999% qualifier, because politicians and reporters who aren’t good in reasoning about expected utilities might erroneously conclude not just that it’s probably safe (which it probably is), but that it should be implemented. And the harms of that would be too great. Prudence might well require me to be silent about evidence in cases where the risks are asymmetrical, as in the nuclear reactor case where the harm of people coming to believe that it’s safe when it’s unsafe so greatly outweighs the harm of people coming to believe that it’s unsafe when it’s safe. But the case of theism exhibits a similar asymmetry.

Thus, consistent utilitarian atheists will promote theism. (Yes, I think that’s a reductio of utilitarianism!) But even apart from utilitarianism, no atheist should promote atheism.