If we assume that a false proposition is true, we can prove anything (ex falso quodlibet). Bertrand Russell, so the story goes, once mentioned this in class. A student raised his hand and challenged: in that case prove that 1=0 implies that you’re the Pope. Russell promptly obliged, see below. Bernoulli 1738, as discussed in an earlier post, contains two contradictory definitions of expected utility theory. This contradiction amounts to a false proposition, and that means any statement can be proved using expected utility theory. As an illustration, let’s prove that Bertrand Russell is the Pope.

Let’s start with Russell’s proof that 1=0 implies he’s the Pope. Russell said the following.

False Proposition:

(Eq.1) 1=0

Theorem 1: I am the Pope.

Proof: Add 1 to both sides of (Eq.1): then we have 2 = 1. The set containing just me and the Pope has 2 members. But 2 = 1, so it has only 1 member; therefore, I am the Pope.

QED.

This makes our work a lot easier. We only have to prove that 1=0, using expected utility theory, and then, by Theorem 1, we know that Bertrand Russell is the Pope.

Here’s the strategy. We identify the contradiction in Bernoulli 1738 and show that it implies 1=0. The contradiction will be of the following form: “ v=1 and v=0.” Since v=v, this implies that 1=0. So we’ll go for that. We find a place where Bernoulli says v=1, and then we’ll find another place where he says v=0. That’s all we need to do, the rest was done by Russell.

On p. 24 Bernoulli writes the following:

At least since Laplace 1814, this has been interpreted as follows: the value of an uncertain prospect is the expected change in utility induced by it (Bernoulli converts this utility change into an equivalent certain monetary change, but we won’t do that, to keep things simple). In symbols, if u is the utility function, v is the value of the proposition, and \langle \cdot \rangle the expectation operator, we have

(Eq. 2) v=\langle \Delta u \rangle

To really keep things simple, let’s work with a trivial gamble: our initial wealth is x, we have to pay a fee F, and we are guaranteed (probability 1) to receive a payout G. According to (Eq.2) the value of this trivial gamble is then simply

(Eq.3) v = u(x+G-F) -u(x)

If this number is positive we should take the gamble, if it’s negative we should stay away from it. To be specific, let’s use the logarithmic utility function proposed by Bernoulli, u(x)=\ln x, and the following parameters:

latex F=\$1 G=\$3.43 (\text{or, more precisely, } \$\frac{e+1+\sqrt{e^2+2e-3}}{2}) x= \$1.41 (\text{or, more precisely, } \frac{G-F}{e-1})Evaluate (Eq.3) with these parameters, and you’ll find that v=1.

Later in the paper, on p.27, Bernoulli contradicts himself (referring to an equation on p.26, see this longer blog post for details).

Bernoulli accompanied this statement with a figure (original Latin version (1738), German version (1896), English version (1954) [update 2020-07-30: an earlier version of this post wrongly stated that the figure was not included in the original 1738 version. This has now been corrected.]). Written as an equation this gives us a different expression for the value , namely

(Eq.4) v= u(x+G)-u(x) – [u(x)-u(x-F)].

Evaluating this expression with our chosen parameters yields v=0. Since in both cases we have evaluated the same quantity — the value of the same prospect to the same person of the same wealth x, according to Bernoulli’s expected utility theory — we have shown that according to expected utility theory v=1 and v=0. Since that implies 1=0, using Russell’s Theorem 1, we have also shown the following:

Expected utility theory proves that Bertrand Russell is the Pope.

QED.

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