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.