Probability from decision theory

There exists a problem in the world, that is even more pressing than the position of cheese in the cheeseburger emoji: namely that nobody™ understands the Deutsch-Wallace theorem. I’ve talked to a lot of people about it, and the usual reaction I get is that they have heard of it, are vaguely interested in how can one prove the Born rule, but have no idea how Deutsch and Wallace actually did it.

It’s hard to blame them. The original paper by Deutsch is notoriously idiosyncratic: he even neglected to mention that one of his assumptions was the Many-Worlds interpretation1! Several people wrote papers trying to understand it: Barnum et al. mistakenly concluded that Deutsch was simply wrong, Gill made a valiant effort but gave up without a conclusion, and Wallace finally succeeded, clarifying Deutsch’s proof and putting it in context.

Wallace was not successful, however, in popularising the theorem. I think this is because his paper is a 27-page mess. It did not help, either, that Wallace quickly moved on to improving and formalising Deutsch’s theorem, providing an even more complicated proof from weaker assumptions, leaving the community with no easy entry point into this confusing literature.

To fill this hole, then, I’m writing two “public service” blog posts. The first (this one) is to explain how to derive probabilities from decision theory, and the second is to show how this decision-theoretical argument, together with the Many-Worlds interpretation, yields the Born rule.

Unlike Deutsch, I’m going to use a standard decision theory, taken from the excellent “The Foundations of Causal Decision Theory” by James Joyce. We’re going to consider a simple betting scenario, where an agent – called Amir – decides how much to pay to take part in a game where he receives $a$ euros if event $E$ happens, and $b$ euros if event $\lnot E$ happens2. The game is then defined by the vector $(a,b)$, and the maximal price Amir accepts to pay for it is its value $V(a,b)$3.

The first rationality axiom we demand is that if the game is certain to pay him back $c$ euros, he must assign value $c$ to the game. This means that the Amir is indifferent to betting per se, he doesn’t demand some extra compensation to go through the effort of betting, nor does he accept to lose money just to experience the thrill of betting (unlike real gambling addicts, I must say). The axiom is then

  • Constancy: $V(c,c) = c$.

The second axiom we demand is that if for a pair of games $(a,b)$ and $(c,d)$ it happens that $a \ge c$ and $b \ge d$, that is, if in both cases where $E$ happens or $\lnot E$ happens the first game pays a reward that is larger or equal than the second game, then Amir must value the first game no less than the second game. The axiom is then

  • Dominance: if $(a,b) \ge (c,d)$ then $V(a,b) \ge V(c,d)$.

The third and last axiom we need sounds very innocent: if Amir is willing to pay $V(a,b)$ to play the game with rewards $(a,b)$, and thinks that playing for rewards $(c,d)$ is worth $V(c,d)$, then the price he should pay for getting the rewards $(a+c,b+d)$ must be $V(a,b) + V(c,d)$. In other words: it shouldn’t matter if tickets for the game with rewards $(a+c,b+d)$ are sold at once, or broken down into first a ticket for rewards $(a,b)$ followed by a ticket for rewards $(c,d)$. The axiom is then

  • Additivity: $V(a+c,b+d) = V(a,b) + V(c,d)$.

One problem with Additivity is that real agents don’t behave like this. People usually assign values such that $V(a+c,b+d) < V(a,b) + V(c,d)$, because if you have nothing then 10€ might be the difference between life and death, whereas if you already have 10,000€ then 10€ is just a nice gift. Besides not matching reality, this linear utility function implied by Additivity causes pathological decisions such as the St. Petersburg paradox or Pascal’s Wager. But these problems do not appear if the amounts at stake are small compared to Amir’s wealth, which we can assume to be the case, and Additivity makes for a rather simple and elegant decision theory, so we’ll use it anyway4. After all, I’m not writing for the people whose objection to the Deutsch-Wallace theorem is that Deutsch’s decision theory implies linear utilities, but rather for those whose objection is “What the hell is going on?”.

Now, to work. First we shall show how Additivity allows us to write the value of any game as a function of the value of the elementary games $(1,0)$ and $(0,1)$. Additivity immediately implies that
\[V(a,b) = V(a,0) + V(0,b),\]and that for any positive integer $n$
\[V(na,0) = nV(a,0).\]Taking now $a=1/n$, the previous equation gives us that \[V(1,0) = nV(1/n,0),\] or that \[V(1/n,0) = \frac1n V(1,0).\] Considering $m$ such games, we have now that \[V(m/n,0) = \frac{m}{n} V(1,0)\] for any positive rational $m/n$. We can extend this to all rationals if we remember that by Constancy $V(0,0) = 0$ and that by Additivity
\[V(0,0) = V(m/n,0) + V(-m/n,0).\]Now one could extend this argument to all reals by taking some continuity assumption, but I don’t think it is interesting to do so. I’d rather assume that one can only have rational amounts of euros5. Anyway, now we have shown that for all rational $a$ and $b$ we have that
\[V(a,b) = aV(1,0) + bV(0,1).\]What is left to see is that the values of the elementary games $(1,0)$ and $(0,1)$ behave like probabilities. If we consider Constancy with $c=1$ we have that \[V(1,0) + V(0,1) = 1,\] so these “probabilities” are normalised. If we now consider Dominance, we get that
\[ V(1,0) \ge V(0,0) = 0,\]so the “probabilities” are positive. Is there anything left to show? Well, if you are a Bayesian, no. The probability of an event $E$ is defined as the price a rational agent would pay for a lottery ticket that gives then 1€ if $E$ happens and nothing otherwise. Bayesians have the obligation to show that these probabilities to obey the usual Kolmogorov axioms, but on the interpretational side there is nothing left to explain.

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