Monday, November 21, 2022

Dominance and countably infinite fair lotteries

Suppose we have a finitely-additive probability assignment p (perhaps real, perhaps hyperreal) for a countably infinite lottery with tickets 1, 2, ... in such a way that each ticket has infinitesimal probability (where zero counts as an infinitesimal). Now suppose we want to calculate the expected value or previsio EpU of any bounded wager U on the outcome of the lottery, where we think of the wager as assigning a value to each ticket, and the wager is bounded if there is a finite M such that |U(n)| < M for all n.

Here are plausible conditions on the expected value:

  1. Dominance: If U1 < U2 everywhere, then EpU1 < EpU2.

  2. Binary Wagers: If U is 0 outside A and c on A, then EpU = cP(A).

  3. Disjoint Additivity: If U1 and U2 are wagers supported on disjoint events (i.e., there is no n such
    that U1(n) and U2(n) are both non-zero), then Ep(U1+U2) = EpU1 + EpU2.

But we can’t. For suppose we have it. Let U(n) = 1/(2n). Fix a positive integer m. Let U1(n) be 2 for n ≤ m + 1 and 0 otherwise. Let U2(n) be 1/m for n > m + 1 and 0 for n ≤ m + 1. Then by Binary Wagers and by the fact that each ticket has infinitesimal probability, EpU1 is an infinitesimal α (since the probability of any finite set will be infinitesimal). By Binary Wagers and Dominance, EpU2 ≤ 1/(m+1). Thus by Disjoint Additivity, Ep(U1+U2) ≤ α + 1/(m+1) < 1/m. But U < U1 + U2 everywhere, so by Dominance we have EpU < 1/m. Since 0 < U everywhere, by Dominance and Binary Wagers we have 0 < EpU.

Thus, EpU is a non-zero infinitesimal β. But then β < U(n) for all n, and so by Binary Wagers and Dominance, β < EpU, a contradiction.

I think we should reject Dominance.


IanS said...

You could restrict previsions to real-valued wagers. (This is not entirely arbitrary. What would it mean to win $β?) Then the wager ‘constant β’ would have no prevision. So there would be no contradiction. The best you could do would be ‘constant zero’. This has prevision zero, which is strictly less than β, as expected.

Alexander R Pruss said...

Yeah. I have a more general solution along the same lines. Will post soon.