Monday, 30 July 2018

The Dutch Book Argument for Regularity

I've just signed a contract with Cambridge University Press to write a book on the Dutch Book Argument for their Elements in Decision Theory and Philosophy series. So over the next few months, I'm going to be posting some bits and pieces as I get properly immersed in the literature.

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We say that a probabilistic credence function $c : \mathcal{F} \rightarrow [0, 1]$ is regular if $c(A) > 0$ for all propositions $A$ in $\mathcal{F}$ such that there is some world at which $A$ is true.

The Principle of Regularity (standard version) If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, rationality requires that $c$ is regular.

I won't specify which worlds are in the scope of the quantifier over worlds that occurs in the antecedent of this norm. It might be all the logically possible worlds, or the metaphysically possible worlds, or the conceptually possible worlds; it might be the epistemically possible worlds. Different answers will give different norms. But we needn't decide the issue here. We'll just specify that it's the same set of worlds that we quantify over in the Dutch Book argument for Probabilism when we say that, if your credences aren't probabilistic, then there's a series of bets they'll lead you to enter into that will lose you money at all possible worlds.

In this post, I want to consider the almost-Dutch Book Argument for the norm of Regularity. Here's how it goes: Suppose you have a credence $c(A) = 0$ in a proposition $A$, and suppose that $A$ is true at world $w$. Then, recall, the first premise of the standard Dutch Book argument for Probabilism:

Ramsey's Thesis If your credence in a proposition $X$ is $c(X) = p$, then you're permitted to pay £$pS$ for a bet that returns £$S$ if $X$ is true and £$0$ if $X$ is false, for any $S$, positive or negative or zero.

So, since $c(A) = 0$, your credences will permit you to sell the following bet for £0: if $A$, you must pay out £1; if $\overline{A}$, you will pay out £0. But selling this bet for this price is weakly dominated by refusing the bet. Selling the bet at that price loses you money in all $A$-worlds, and gains you nothing in $\overline{A}$-worlds. Whereas refusing the bet neither loses nor gains you anything in any world. Thus, your credences permit you to choose a weakly dominated act. So they are irrational. Or so the argument goes. I call this the almost-Dutch Book argument for Regularity since it doesn't punish you with a sure loss, but rather with a possible loss with no compensating possible gain.

If this argument works, it establishes the standard version of Regularity stated above. But consider the following case. $A$ and $B$ are two logically independent propositions -- It will be rainy tomorrow and It will be hot tomorrow, for instance. You have only credences in $A$ and in the conjunction $AB$. You don't have credences in $\overline{A}$, $A \vee B$, $A\overline{B}$, and so on. What's more, your credences in $A$ and $AB$ are equal, i.e., $c(A) = c(AB)$. That is, you are exactly as confident in $A$ as you are in its conjunction with $B$. Then, in some sense, you violate Regularity, though you don't violate the standard version we stated above. After all, since your credence in $A$ is the same as your credence in $AB$, you must give no credence whatsoever to the worlds in which $A$ is true and $B$ is false. If you did, then you would set $c(AB) < c(A)$. But you don't have a credence in $A\overline{B}$. So there is no proposition true at some worlds to which you assign a credence of 0. Thus, the almost-Dutch Book argument sketched above will not work. We need a different Dutch Book argument for the following version of Regularity:

The Principle of Regularity (full version) If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that there is an extension $c^*$ of $c$ to a full algebra $\mathcal{F}^*$ that contains $\mathcal{F}$ such that $c^* : \mathcal{F}^* \rightarrow [0, 1]$ is regular.

It is this principle that you violate if $c(A) = c(AB)$ when $A$ and $B$ are logically independent. For any probabilistic extension $c^*$ of $c$ that assigns a credence to $A\overline{B}$ must assign it credence 0 even though there is a world at which it is true.

How are we to give an almost-Dutch Book argument for this version of Regularity? There are two possible approaches.

On the first, we strengthen the first premise of the standard Dutch Book argument. Ramsey's Thesis says: if you have credence $c(X) = p$ in $X$, then you are permitted to pay £$pS$ for a bet that pays £$S$ if $X$ and £$0$ if $\overline{X}$. The stronger version says:

Strong Ramsey's Thesis If every extension $c^*$ of $c$ to a full algebra $\mathcal{F}^*$ that contains $\mathcal{F}$ is such that $c^*(X) = p$, then you are permitted to pay £$pS$ for a bet that pays £$S$ if $X$ and £$0$ if $\overline{X}$.

The idea is that, if every extension assigns the same credence $p$ to $X$, then you are in some sense committed to assigning credence $p$ to $X$. And thus, you are permitted to enter into which ever bets you'd be permitted to enter into if you actually had credence $p$.

On the second approach to giving an almost-Dutch Book argument for the full version of the Regularity principle, we actually provide an almost-Dutch Book using just the credences that you do in fact assign. Suppose, for instance, you have credence $c(A) = c(AB) = 0.5$. Then you will sell for £5 a bet that pays out £10 if $A$ and £0 if $\overline{A}$, while you will buy for £5 a bet that pays £10 if $AB$ and £0 if $\overline{AB}$. Then, if $A$ is true and $B$ is true, you will have a net gain of £0, and similarly if $A$ is false. But if $A$ is true and $B$ is false, you will lose £10. Thus, you face the possibility of loss with no possibility of gain. Now, the question is: can we always construct such almost-Dutch Books? And the answer is that we can, as the following theorem shows:

Theorem 1 (Almost-Dutch Book Theorem for Full Regularity) Suppose $\mathcal{F} = \{X_1, \ldots, X_n\}$ is a set of propositions. Suppose $c : \mathcal{F} \rightarrow [0, 1]$ is a credence function that cannot be extended to a regular probability function on a full algebra $\mathcal{F}^*$ that contains $\mathcal{F}$. Then there is a sequence of stakes $S = (S_1, \ldots, S_n)$, such that if, for each $1 \leq i \leq n$, you pay £$(c(X_i) \times S_i)$ for a bet that pays out £$S_i$ if $X_i$ and £0 if $\overline{X_i}$, then the total price you'll pay is at least the pay off of these bets at all worlds, and more than the payoff at some.

That is,
(i) for all worlds $w$,
$$S\cdot (w - c) = S \cdot w - S \cdot c = \sum^n_{i=1} S_iw(X_i) + \sum^n_{i=1} S_ic(X_i) \leq 0$$
(ii) for some worlds $w$,
$$S\cdot (w - c) = S \cdot w - S \cdot c = \sum^n_{i=1} S_iw(X_i) + \sum^n_{i=1} S_ic(X_i) \leq 0$$
where $w(X_i) = 1$ if $X_i$ is true at $w$ and $w(X_i) = 0$ if $X_i$ is false at $w$. We call $w(-)$ the indicator function of $w$.

Proof sketch. First, recall de Finetti's observation that your credence function $c : \mathcal{F} \rightarrow [0, 1]$ is a probability function iff it is in the convex hull of the indicator functions of the possible worlds -- that is, iff $c$ is in $\{w(-) : w \mbox{ is a possible world}\}^+$. Second, note that, if your credence function can't be extended to a regular credence function, it sits on the boundary of this convex hull. In particular, if $W_c = \{w' : c = \sum_w \lambda_w w \Rightarrow \lambda_{w'} > 0\}$, then $c$ lies on the boundary surface created by the convex hull of $W_c$. Third, by the Supporting Hyperplane Theorem, there is a vector $S$ such that $S$ is orthogonal to this boundary surface and thus:
(i) $S \cdot (w-c) = S \cdot w - S \cdot c = 0$ for all $w$ in $W_c$; and
(ii) $S \cdot (w-c) = S \cdot w - S \cdot c < 0$ for all $w$ not in $W_c$.
Fourth, recall that $S \cdot w$ is the total payout of the bets at world $w$ and $S \cdot c$ is the price you'll pay for it. $\Box$

Thursday, 26 July 2018

Dutch Strategy Theorems for Conditionalization and Superconditionalization

I've just signed a contract with Cambridge University Press to write a book on the Dutch Book Argument for their Elements in Decision Theory and Philosophy series. So over the next few months, I'm going to be posting some bits and pieces as I get properly immersed in the literature.

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Many Bayesians formulate the update norm of Bayesian epistemology as follows:

Bayesian Conditionalization  If
(i) your credence function at $t$ is $c : \mathcal{F} \rightarrow [0, 1]$,
(ii) your credence function at a later time $t'$ is $c' : \mathcal{F} \rightarrow [0, 1]$,
(iii) $E$ is the strongest evidence you acquire between $t$ and $t'$,
(iv) $E$ is in $\mathcal{F}$,
then rationality requires that, if $c(E) > 0$, then for all $X$ in $\mathcal{F}$, $$c'(X) = c(X|E) = \frac{c(XE)}{c(E)}$$

I don't. One reason you might fail to conditionalize between $t$ and $t'$ is that you re-evaluate the options between those times. You might disavow the prior that you had at the earlier time, perhaps decide it was too biased in one way or another, or not biased enough; perhaps you come to think that it doesn't give enough consideration to the explanatory power one hypothesis would have were it true, or gives too much consideration to the adhocness of another hypothesis; and so on. Now, it isn't irrational to change your mind. So surely it can't be irrational to fail to conditionalize as a result of changing your mind in this way. On this, I agree with van Fraassen.

Instead, I prefer to formulate the update norm as follows -- I borrow the name from Kenny Easwaran:

Plan Conditionalization If
(i) your credence function at $t$ is $c: \mathcal{F} \rightarrow [0, 1]$,
(ii) between $t$ and $t'$ you will receive evidence from the partition $\{E_1, \ldots, E_n\}$,
(iii) each $E_i$ is in $\mathcal{F}$
(iv) at $t$, your updating plan is $c'$, so that $c'_i : \mathcal{F} \rightarrow [0, 1]$ is the credence function you will adopt if $E_i$,
then rationality requires that, if $c(E_i) > 0$, then for all $X$ in $\mathcal{F}$, $$c'_i(X) = c(X | E_i)$$

I want to do two things in this post. First, I'll offer what I think is a new proof of the Dutch Strategy or Diachronic Dutch Book Theorem that justifies Plan Conditionalization (I haven't come across it elsewhere, though Ray Briggs and I used the trick at the heart of it for our accuracy dominance theorem in this paper). Second, I'll explore how that might help us justify other norms of updating that concern situations in which you don't come to learn any proposition with certainty. We will see that we can use the proof I give to justify the following standard constraint on updating rules: Suppose the evidence I receive between $t$ and $t'$ is not captured by any of the propositions to which I assign a credence -- that is, there is no proposition $e$ to which I assign a credence that is true at all and only the worlds at which I receive the evidence I actually receive between $t$ and $t'$. As a result, there is no proposition $e$ that I learn with certainty as a result of receiving that evidence. Nonetheless, I should update my credence function from $c$ to $c'$ in such a way that it is possible to extend my earlier credence function $c$ to a credence function $c^*$ so that: (i) $c^*$ does assign a credence to $e$, and (ii) my later credence $c'(X)$ in a proposition $X$ is the credence that this extended credence function $c^*$ assigns to $X$ conditional on me receiving evidence $e$ -- that is, $c'(X) = c^*(X | e)$. That is, I should update as if I had assigned a credence to $e$ at the earlier time and then updated by conditionalizing on it.

Here's the Dutch Strategy or Diachronic Dutch Book Theorem for Plan Conditionalization:

Definition (Conditionalizing pair) Suppose $c$ is a credence function and $c'$ is an updating rule defined on $\{E_1, \ldots, E_n\}$. We say that $(c, c')$ is a conditionalizing pair if, whenever $c(E_i) > 0$, then for all $X$, $c'_i(X) = c(X | E_i)$.

Dutch Strategy Theorem Suppose $(c, c')$ is not a conditionalizing pair. Then
(i) there are two acts $A$ and $B$ such that $c$ prefers $A$ to $B$, and
(ii) for each $E_i$, there are two acts $A_i$ and $B_i$ such that $c'_i$ prefers $A_i$ to $B_i$,
and, for each $E_i$, $A + A_i$ has greater utility than $B + B_i$ at all worlds at which $E_i$ is true.

We'll now give the proof of this.

First, we describe a way of representing pairs $(c, c')$. Both $c$ and each $c'_i$ are defined on the same set $\mathcal{F} = \{X_1, \ldots, X_m\}$. So we can represent $c$ by the vector $(c(X_1), \ldots, c(X_m))$ in $[0, 1]^m$, and we can represent each $c'_i$ by the vector $(c'_i(X_1), \ldots, c'_i(X_m))$ in $[0, 1]^m$. And we can represent $(c, c')$ by concatenating all of these representations to give:
$$(c, c') = c \frown c'_1 \frown c'_2 \frown \ldots \frown c'_n$$
which is a vector in $[0, 1]^{m(n+1)}$.

Second, we use this representation to give an alternative characterization of conditionalizing pairs. First, three pieces of notation:
• Let $W$ be the set of all possible worlds.
• For any $w$ in $W$, abuse notation and write $w$ also for the credence function on $\mathcal{F}$ such that $w(X) = 1$ if $X$ is true at $w$, and $w(X) = 0$ if $X$ is false at $w$.
• For any $w$ in $W$, let $$(c, c')_w = w \frown c'_1 \frown \ldots \frown c'_{i-1} \frown w \frown c'_{i+1} \frown \ldots \frown c'_n$$ where $E_i$ is the element of the partition that is true at $w$.
Lemma 1 If $(c, c')$ is not a conditionalizing pair, then $(c, c')$ is not in the convex hull of $\{(c, c')_w : w \in W\}$, which we write $\{(c, c')_w : w \in W\}^+$.

Proof of Lemma 1. If $(c, c')$ is in $\{(c, c')_w : w \in W\}^+$, then there are $\lambda_w \geq 0$ such that

(1) $\sum_{w \in W} \lambda_w = 1$,
(2) $c(X) = \sum_{w \in W} \lambda_w w(X)$
(3) $c'_i(X) = \sum_{w \in E_i} \lambda_w w(X) + \sum_{w \not \in E_i} \lambda_w c'_i(X)$.

By (2), we have $\lambda_w = c(w)$. So by (3), we have $$c'_i(X) = c(XE_i) + (1-c(E_i))c'_i(X)$$ So, if $c(E_i) > 0$, then $c'_i(X) = c(X | E_i)$.

Third, we use this alternative characterization of conditionalizing pairs to specify the acts in question. Suppose $(c, c')$ is not a conditionalizing pair. Then $(c, c')$ is outside $\{(c, c')_w : w \in W\}^+$. Now, let $(p, p')$ be the orthogonal projection of $(c, c')$ into $\{(c, c')_w : w \in W\}^+$. Then let $(S, S') = (c, c') - (p, p')$. That is, $S = c - p$ and $S'_i = c'_i - p'_i$. Now pick $w$ in $W$. Then the angle between $(S, S')$ and $(c, c')_w - (c, c')$ is obtuse and thus
$$(S, S') \cdot ((c, c')_w - (c, c')) = -\varepsilon_w < 0$$

Thus, define the acts $A$, $B$, $A'_i$ and $B'_i$ as follows:
• The utility of $A$ at $w$ is $S \cdot (w - c) + \frac{1}{3}\varepsilon_w$:
• The utility of $B$ at $w$ is 0;
• The utility of $A'_i$ at $w$ is $S'_i \cdot (w - c'_i) + \frac{1}{3}\varepsilon_w$;
• The utility of $B'_i$ at $w$ is 0.
Then the expected utility of $A$ by the lights of $c$ is $\sum^w c(w)\frac{1}{3}\varepsilon_w > 0$, while the expected utility of $B$ is 0, so $c$ prefers $A$ to $B$. And the expected utility of $A'_i$ by the lights of $c'_i$ is $\sum_w c'_i(w)\frac{1}{3}\varepsilon_w > 0$, while the expected utility of $B'_i$ is 0, so $c'_i$ prefers $A'_i$ to $B'_i$. But the utility of $A + A'_i$ at $w$ is
$$S \cdot (w - c) + S'_i \cdot (w - c'_i) + \frac{2}{3}\varepsilon_w = (S, S') \cdot ((c, c')_w - (c, c')) + \frac{2}{3}\varepsilon_w = - \frac{1}{3}\varepsilon_w < 0$$
where $E_i$ is true at $w$. While the utility of $B + B'_i$ at $w$ is 0.

This completes our proof. $\Box$

You might be forgiven for wondering why we are bothering to give an alternative proof for a theorem that is already well-known. David Lewis proved the Dutch Strategy Theorem in a handout for a seminar at Princeton in 1972, Paul Teller then reproduced it (with full permission and acknowledgment) in a paper in 1973, and Lewis finally published his handout in 1997 in his collected works. Why offer a new proof?

It turns out that this style of proof is actually a little more powerful. To see why, it's worth comparing it to an alternative proof of the Dutch Book Theorem for Probabilism, which I described in this post (it's not original to me, though I'm afraid I can't remember where I first saw it!). In the standard Dutch Book Theorem for Probabilism, we work through each of the axioms of the probability calculus, and say how you would Dutch Book an agent who violates it. The axioms are: Normalization, which says that $c(\top) = 1$ and $c(\bot) = 0$; and Additivity, which says that $c(A \vee B) = c(A) + c(B) - c(AB)$. But consider an agent with credences only in the propositions $\top$, $A$, and $A\ \&\ B$.  Her credences are: $c(\top) = 1$, $c(A) = 0.4$, $c(A\ \&\ B) = 0.7$. Then there is no axiom of the probability calculus that she violates. And thus the standard proof of the Dutch Book Theorem is no help in identifying any Dutch Book against her. Yet she is Dutch Bookable. And she violates a more expansive formulation of Probabilism that says, not only are you irrational if your credence function is not a probability function, but also if your credence function cannot be extended to a probability function. So the standard proof of the Dutch Book Theorem can't establish this more expansive version. But the alternative proof I mentioned above can.

Now, something similar is true of the alternative proof of the Dutch Strategy Theorem that I offered above (I happened upon this while discussing Superconditionalizing with Jason Konek, who uses similar techniques in his argument for J-Kon, the alternative to Jeffrey's Probability Kinematics that he proposes in his paper, 'The Art of Learning', which was runner-up for last year's Sander's Prize in Epistemology). In Lewis' proof of that theorem: First, if you violate Plan Conditionalization, there must be $E_i$ and $X$ such that $c(E_i) > 0$ and $c'_i(X) \neq c(X|E_i)$. Then you place bets on $XE_i$, $\overline{E_i}$ at the earlier time $t$, and a bet on $X$ at $t'$. These bets then together lose you money in any world at which $E_i$ is true. Now, it might seem that you must have the required credences to make those bets just in virtue of violating Plan Conditionalization. But imagine the following is true of you: between $t$ and $t'$, you'll obtain evidence from the partition $\{E_1, \ldots, E_n\}$. And, at $t'$, you'll update on this evidence using the rule $c'$. That is, if $E_i$, then you'll adopt the new credence function $c'_i$ at time $t'$. Now, you don't assign credences to the propositions in $\{E_1, \ldots, E_n\}$. Perhaps this is because you don't have the conceptual resources to formulate these propositions. So while you will update using the rule $c'$, this is not a rule you consciously or explicitly adopt, since to state it would require you to use the propositions in $\{E_1, \ldots, E_n\}$. So it's more like you have a disposition to update in this way. Now, how might we state Plan Conditionalization for such an agent? We can't demand that $c'_i(X) = c(X|E_i)$, since $c(X | E_i)$ is not defined. Rather, we demand that there is some extension $c^*$ of $c$ to a set of propositions that does include each $E_i$ such that $c'_i(X) = c^*(X | E_i)$. Thus, we have:

Plan Superconditionalization If
(i) your credence function at $t$ is $c : \mathcal{F} \rightarrow [0, 1]$,
(ii) between $t$ and $t'$ you will receive evidence from the partition $\{E_1, \ldots, E_n\}$,
(iii) at $t$, your updating plan is $c'$, so that $c'_i : \mathcal{F} \rightarrow [0, 1]$ is the credence function you plan to adopt if $E_i$,
then rationality requires that there is some extension $c^*$ of $c$ for which, if $c^*(E_i) > 0$, then for all $X$, $$c'_i(X) = c^*(X | E_i)$$

And it turns out that we can adapt the proof above for this purpose. Say that $(c, c')$ is a superconditionalizing pair if there is an extension $c^*$ of $c$ such that, if $c^*(E_i) > 0$, then for all $X$, $c'_i(X) = c^*(X | E_i)$. Then we can prove that if $(c, c')$ is not a superconditionalizing pair, then $(c, c')$ is not in $\{(c, c')_w : w \in W\}^+$. Here's the proof from above adapted to our case: If $(c, c')$ is in $\{(c, c')_w : w \in W\}^+$, then there are $\lambda_w \geq 0$ such that

(1) $\sum_{w \in W} \lambda_w = 1$,
(2) $c(X) = \sum_{w \in W} \lambda_w w(X)$
(3) $c'_i(X) = \sum_{w \in E_i} \lambda_w w(X) + \sum_{w \not \in E_i} \lambda_w c'_i(X)$.

Define the following extension $c^*$ of $c$: $c^*(w) = \lambda_w$. Then, by (3), we have $$c'_i(X) = c^*(XE_i) + (1-c^*(E_i))c'_i(X)$$ So, if $c^*(E_i) > 0$, then $c'_i(X) = c^*(X | E_i)$, as required. $\Box$

Now, this is a reasonably powerful version of conditionalization. For instance, as Skyrms showed here, if we make one or two further assumptions on the extension of $c$ to $c^*$, we can derive Richard Jeffrey's Probability Kinematics from Plan Superconditionalization. That is, if the evidence $E_i$ will lead you to set your new credences across the partition $\{B_1, \ldots, B_k\}$ to $q_1, \ldots, q_k$, respectively, so that $c'_i(B_j) = q_j$, then your new credence $c'_i(X)$ must be $\sum^k_{j=1} c(X | B_j)q_j$, as Probability Kinematics demands. Thus, Plan Superconditionalization places a powerful constraint on updating rules for situations in which the proposition stating your evidence is not one to which you assign a credence. Other cases of this sort include the Judy Benjamin problem and the many cases in which MaxEnt is applied.

Wednesday, 25 July 2018

Deadline for PhD position in formal epistemology & law extended

This position is still available. Deadline extended to September 7, 2018.

On the Expected Utility Objection to the Dutch Book Argument for Probabilism

I've just signed a contract with Cambridge University Press to write a book on the Dutch Book Argument for their Elements in Decision Theory and Philosophy series. So over the next few months, I'm going to be posting some bits and pieces as I get properly immersed in the literature. The following came up while thinking about Brian Hedden's paper 'Incoherence without Exploitability'.

What is Probabilism?

Probabilism says that your credences should obey the axioms of the probability calculus. Suppose $\mathcal{F}$ is the algebra of propositions to which you assign a credence. Then we let $0$ represent the lowest possible credence you can assign, and we let $1$ represent the highest possible credence you can assign. We then represent your credences by your credence function $c : \mathcal{F} \rightarrow [0, 1]$, where, for each $A$ in $\mathcal{F}$, $c(A)$ is your credence in $A$.

Probabilism
If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that:
(P1a) $c(\bot) = 0$, where $\bot$ is a necessarily false proposition;
(P1b) $c(\top) = 1$, where $\top$ is a necessarily true proposition;
(P2) $c(A \vee B) = c(A) + c(B)$, for any mutually exclusive propositions $A$ and $B$ in $\mathcal{F}$.

This is equivalent to:

Partition Probabilism
If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that, for any two partitions $\mathcal{X} = \{X_1, \ldots, X_m\}$ and $\mathcal{Y} = \{Y_1, \ldots, Y_n\}$,$$\sum^m_{i=1} c(X_i) = 1= \sum^n_{j=1} c(Y_j)$$

The Dutch Book Argument for Probabilism

The Dutch Book Argument for Probabilism has three premises. The first, which I will call Ramsey's Thesis and abbreviate RT, posits a connection between your credence in a proposition and the prices you are rationally permitted or rationally required to pay for a bet on that proposition. The second, known as the Dutch Book Theorem, establishes that, if you violate Probabilism, there is a set of bets you might face, each with a price attached, such that (i) by Ramsey's Thesis, for each bet, you are rationally required to pay the attached price for it, but (ii) the sum of the prices of the bets exceeds the highest possible payout of the bets, so that, having paid each of those prices, you are guaranteed to lose money. The third premise, which we might call the Domination Thesis, says that credences are irrational if they mandate you to make a series of decisions (i.e, paying certain prices for the bets) that is guaranteed to leave you worse off than another series of decisions (i.e., refusing to pay those prices for the bets)---in the language of decision theory, paying the attached price for each of the bets is dominated by refusing each of the bets, and credences that mandate you to choose dominated options are irrational. The conclusion of the Dutch Book Argument is then Probabilism. Thus, the argument runs:

The Dutch Book Argument for Probabilism
(DBA1) Ramsey's Thesis
(DBA2) Dutch Book Theorem
(DBA3) Domination Thesis
Therefore,
(DBAC) Probabilism

The argument is valid. The second premise is a mathematical theorem. Thus, if the argument fails, it must be because the first or third premise is false, or both. In this paper, we focus on the first premise, and the expected utility objection to it. So, let's set out that premise in a little more detail.

In what follows, we assume that (i) you are risk-neutral, and (ii) that there is some quantity such that your utility is linear in that quantity---indeed, we will speak as if your utility is linear in money, but that is just for ease of notation and familiarity; any quantity would do. Neither (i) nor (ii) is realistic, and indeed these idealisations are the source of other objections to Ramsey's Thesis. But they are not our concern here, so we will grant them.

Ramsey's Thesis (RT) Suppose your credence in $X$ in $c(X)$. Consider a bet that pays you £$S$ if $X$ is true and £0 if $X$ is false, where $S$ is a real number, either positive, negative, or zero---$S$ is called the stake of the bet. You are offered this bet for the price £$x$, where again $x$ is a real number, either positive, negative, or zero. Then:
(i) If $x < c(X) \times S$, you are rationally required to pay £$x$ to enter into this bet;
(ii) If $x = c(X) \times S$, you are rationally permitted to pay £$x$ and rationally permitted to refuse;
(iii) If $x > c(X) \times S$, you are rationally required to refuse.

Roughly speaking, Ramsey's Thesis says that, the more confident you are in a proposition, the more you should be prepared to pay for a bet on it. More precisely, it says: (a) if you have minimal confidence in that proposition (i.e. 0), then you should be prepared to pay nothing for it; (b) if you have maximal confidence in it (i.e. 1), then you should be prepared to pay the full stake for it; (c) for levels of confidence in between, the amount you should be prepared to pay increases linearly with your credence.

The Expected Utility Objection

We turn now to the objection to Ramsey's Thesis (RT) we wish to treat here. Hedden (2013) begins by pointing out that we have a general theory of how credences and utilities should guide action:

Given a set of options available to you, expected utility theory says that your credences license you to choose the option with the highest expected utility, defined as:
$$\mathrm{EU}(A) = \sum_i P(O_i|A) \times U(O_i)$$
On this view, we should evaluate which bets your credences license you to accept by looking at the expected utilities of those bets. (Hedden, 2013, 485)

He considers the objection that this only applies when credences satisfy Probabilism, but rejects it:

In general, we should judge actions by taking the sum of the values of each possible outcome of that action, weighted by one's credence that the action will result in that outcome. This is a very intuitive proposal for how to evaluate actions that applies even in the context of incoherent credences. (Hedden, 2013, 486)

Thus, Hedden contends that we should always choose by maximising expected utility relative to our credences, whether or not those credences are coherent. Let's call this principle Maximise Subjective Expected Utility and abbreviate it MSEU. He then observes that MSEU conflicts with RT. Consider, for instance, Cináed, who is 60% confident it will rain and 20% confident it won't. According to RT, he is rationally required to sell for £65 a bet in which he pays out £100 if it rains and £0 if is doesn't. But the expected utility of this bet for him is$$0.6 \times (-100 + 65) + 0.2 \times (-0 + 65) = -8$$That is, it has lower expected utility than refusing to sell the bet, since his expected utility for doing that is$$0.6 \times 0 + 0.2 \times 0 = 0$$So, while RT says you must sell that bet for that price, MSEU says you must not. So RT and MSEU are incompatible, and Hedden claims that we should favour MSEU. There are two ways to respond to this. On the first, we try to retain RT in some form in spite of Hedden's objection---I call this the permissive response below. On the second, we try to give a pragmatic argument for Probabilism using MSEU instead of RT---I call this the bookless response below. In the following sections, I will consider these in turn.

The Permissive Response

While Hedden is right to say that maximising expected utility in line with Maximise Subjective Expected Utility (MSEU) is intuitively rational even when your credences are incoherent, so is Ramsey's Thesis (RT). It is certainly intuitively correct that, to quote Hedden, ''we should judge actions by taking the sum of the values of each possible outcome of that action, weighted by one's credence that the action will result in that outcome.'' But it is also intuitively correct that, to quote from our gloss of Ramsey's Thesis above, ''(a) if you have minimal confidence in that proposition (i.e. 0), then you should be prepared to pay nothing for it; (b) if you have maximal confidence in it (i.e. 1), then you should be prepared to pay the full stake for it; (c) for levels of confidence in between, the amount you should be prepared to pay increases linearly with your credence.'' What are we to do in the face of this conflict between our intuitions?

One natural response is to say that choosing in line with RT is rationally permissible and choosing in line with MSEU is also rationally permissible. When your credences are coherent, the dictates of MSEU and RT are the same. But when you are incoherent, they are sometimes different, and in that situation you are allowed to follow either. In particular, faced with a bet and proposed price, you are permitted to pay that price if it is permitted by RT and you are permitted to pay it if it is permitted by MSEU.

If this is right, then we can resurrect the Dutch Book Argument with a permissive version of RT as the first premise:

Permissive Ramsey's Thesis Suppose your credence in $X$ in $c(X)$. Consider a bet that pays you £$S$ if $X$ is true and £0 if $X$ is false. You are offered this bet for the price £$x$. Then:
(i) If $x \leq c(X) \times S$, you are rationally permitted to pay £$x$ to enter into this bet.

And we could then amend the third premise---the Domination Thesis (DBA3)---to ensure we could still derive our conclusion. Instead of saying that credences are irrational if they mandate you to make a series of decisions that is guaranteed to leave you worse off than another series of decisions, we might say that credences are irrational if they permit you to make a series of decisions that is guaranteed to leave you worse off than another series of decisions. In the language of decision theory, instead of saying only that credences that mandate you to choose dominated options are irrational, we say also that credences that permit you to choose dominated options are irrational. We might call this the Permissive Domination Thesis.

Now, by weakening the first premise in this way, we respond to Hedden's objection and make the premise more plausible. But we strengthen the third premise to compensate and perhaps thereby make it less plausible. However, I imagine that anyone who accepts one of the versions of the third premise---either the Domination Thesis or the Permissive Domination Thesis---will also accept the other. Having credences that mandate dominated choices may be worse than having credences that permit such choices, but both seem sufficient for irrationality. Perhaps the former makes you more irrational than the latter, but it seems clear that the ideally rational agent will have credences that do neither. And if that's the case, then we can replace the standard Dutch Book Argument with a slight modification:

The Permissive Dutch Book Argument for Probabilism
(PDBA1) Permissive Ramsey's Thesis
(PDBA2) Dutch Book Theorem
(PDBA3) Permissive Domination Thesis
Therefore,
(PDBAC) Probabilism

The Bookless Response

Suppose you refuse even the permissive version of RT, and insist that coherent and incoherent agents alike should choose in line with MSEU. Then what becomes of the Dutch Book Argument? As we noted above, Hedden shows that it fails---MSEU is not sufficient to establish the conclusion. In particular, Hedden gives an example of an incoherent credence function that is not Dutch Bookable via MSEU. That is, there are no sets of bets with accompanying prices such that (a) MSEU will demand that you pay each of those prices, and (b) the sum of those prices is guaranteed to exceed the sum of the payouts of that set of bets. However, as we will see, accepting individual members of such a set of bets is just one way to make bad decisions based on your credences.

Consider Hedden's example. In it, you assign credences to propositions in the algebra built up from three possible worlds, $w_1$, $w_2$, and $w_3$. Here are some of your credences:
• $c(w_1 \vee w_2) = 0.8$ and $c(w_3) = 0$
• $c(w_1) = 0.7$ and $c(w_2 \vee w_3) = 0$
Now, consider the following two options, $A$ and $B$, whose utilities in each state of the world are set out in the following table:

Then notice first that $A$ dominates $B$---that is, the utility of $A$ is higher than $B$ in every possible state of the world. But, using your incoherent credences, you assign a higher expected utility to $B$ than to $A$. Your expected utility for $A$---which must be calculated relative to your credences in $w_1$ and $w_2 \vee w_3$, since  the utility of $A$ given $w_1 \vee w_2$ is undefined---is $0.7 \times 78 + 0 \times 77 = 54.6$. And your expected utility for $B$---which must be calculated relative to your credences in $w_1 \vee w_2$ and $w_3$, since the utility of $B$ given $w_2 \vee w_3$ is undefined---is $0.8 \times 74 + 0 \times 75 = 59.2$. So, while Hedden might be right that MSEU won't leave you vulnerable to a Dutch Book, it will leave you vulnerable to choosing a dominated option. And since what is bad about entering a Dutch Book is that it is a dominated option---it is dominated by the option of refusing the bets---the invulnerability to Dutch Books should be no comfort to you.

Now, this raises the question: For which incoherence credences is it guaranteed that MSEU won't lead you to choose a dominated option? Is it all incoherent credences, in which case we would have a new Dutch Book Argument for Probabilism from MSEU rather than RT? Or is it some subset? Below, we prove a theorem that answers that. First, a weakened version of Probabilism:

Bounded Probabilism If $c : \mathcal{F}\rightarrow [0, 1]$ is your credence function, then rationality requires that:
(BP1a) $c(\bot) = 0$, where $\bot$ is a necessarily false proposition;
(BP1b) There is $0 < M \leq 1$ such that $c(\top) = M$, where $\top$ is a necessarily true proposition;
(BP2) $c(A \vee B) = c(A) + c(B)$, if $A$ and $B$ are mutually exclusive.

Bounded Probabilism says that you should have lowest possible credence in necessary falsehoods, some positive credence---not necessarily 1---in necessary truths, and your credence in a disjunction of two incompatible propositions should be the sum of your credences in the disjuncts.

Theorem 1 The following are equivalent:
(i) $c$ satisfies Bounded Probabilism
(ii) For all options $A$, $B$, if $A$ dominates $B$, then $\mathrm{EU}_c(A) > \mathrm{EU}_c(B)$.

The proof is in the Appendix below. Thus, even without Ramsey's Thesis or the permissive version described above, you can still give a pragmatic argument for a norm that lies very close to Probabilism, namely, Bounded Probabilism. On its own, this argument cannot say what is wrong with someone who gives less than the highest possible credence to necessary truths, but it does establish the other requirements that Probabilism imposes. To see just how close to Probabilism lies Bounded Probabilism, consider the following two norms, which are equivalent to it:

Scaled Probabilism  If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that there is $0 < M \leq 1$ and a probability function $p : \mathcal{F} \rightarrow [0, 1]$ such that $c(-) = M \times p(-)$.

Bounded Partition Probabilism  If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that, for any two partitions $\mathcal{X} = \{X_1, \ldots, X_m\}$ and $\mathcal{Y} = \{Y_1, \ldots, Y_n\}$,$$\sum^m_{i=1} c(X_i) = \sum^n_{j=1} c(Y_j)$$Then

Lemma 2 The following are equivalent:
(i) Bounded Probabilism
(ii) Scaled Probabilism
(iii) Bounded Partition Probabilism

As before, the proof is in the Appendix.

So, on its own, MSEU can deliver us very close to Probabilism. But it cannot establish (P1b), namely, $c(\top) = 1$. However, I think we can also appeal to a highly restricted version of the Permissive Ramsey's Thesis to secure (P1b) and push us all the way to Probabilism.

Consider Dima and Esther. They both have minimal confidence---i.e. 0---that it won't rain tomorrow. But Dima has credence 0.01 that it will rain, while Esther has credence 0.99 that it will. If we permit only actions that maximise expected utility, then Dima and Esther are required to pay exactly the same prices for bets on rain---that is, Dima will be required to pay a price exactly when Esther is. After all, if £$S$ is the payoff when it rains, £0 is the payoff when it doesn't, and $x$ is a proposed price, then $0.01\times (S- x) + 0 \times (0-x) \geq 0$ iff $0.99 \times (S-x) + 0 \times (0-x) \geq 0$ iff $S \geq x$. So, according to MSEU, Dima and Esther are rationally required to pay anything up to the stake of the bet for such a bet. But this is surely wrong. It is surely at least permissible for Dima to refuse to pay a price that Esther accepts. It is surely permissible for Esther to pay £99 for a bet on rain that pays £100 if it rains and £0 if it doesn't, while Dima refuses to pay anything more than £1 for such a bet, in line with Ramsey's Thesis. Suppose Dima were offered such a bet for the price of £99, and suppose she then defended her refusal to pay that price saying, 'Well, I only think it's 1% likely to rain, so I don't want to risk such a great loss with so little possible gain when I think the gain is so unlikely'. Then surely we would accept that as a rational defence.

In response to this, defenders of MSEU might concede that RT is sometimes the correct norm of action when you are incoherent, but only in very specific cases, namely, those in which you have a positive credence in a proposition, minimal credence (i.e. 0) in its negation, and you are considering the price you might pay for a bet on that proposition. In all other cases---that is, in any case in which your credences in the proposition and its negation are both positive, or in which you are considering an action other than a bet on a proposition---you should use MSEU. I have some sympathy with this. But, fortunately, this restricted version is all we need. After all, it is precisely by applying Ramsey's Thesis to such a case that we can produce a Dutch Book against someone with $c(\bot) = 0$ and $c(\top) < 1$---we simply offer to pay them £$c(\top) \times 100$ for a bet in which they will pay out £100 if $\top$ is true and £0 if it is false; this is then guaranteed to lose them £$100 \times (1-c(X))$, which is positive. Thus, we end up with a disjunctive pragmatic argument for Probabilism: if $c(\bot) = 0$ and $c(\top) < 1$, then RT applies and we can produce a Dutch Book against you; if you violate Probabilism in any other way, then you violate Bounded Probabilism and we can then produce two options $A$ and $B$ such that $A$ dominates $B$, but your credences, via MSEU, dictate that you should choose $B$ over $A$. This, then, is our bookless pragmatic argument for Probabilism:

Bookless Pragmatic Argument for Probabilism
(BPA1) If $c$ violates Probabilism, then either (i) $c(\bot) = 0$ and $c(\top) < 1$, or (ii) $c$ violates Bounded Probabilism.
(BPA2) If $c(\bot) = 0$ and $c(\top) < 1$, then RT applies, and there is a bet on $\top$ such that you are required by RT to pay a higher price for that bet than its guaranteed payoff. Thus, there are options $A$ and $B$ (namely, refuse the bets and pay the price), such that $A$ dominates $B$, but RT demands that you choose $B$ over $A$.
(BPA3) If $c$ violates Bounded Probabilism, then by Theorem 1, there are options $A$ and $B$ such that $A$ dominates $B$, but RT demands that you choose $B$ over $A$. Therefore, by (BPA1), (BPA2), and (BPA3),
(BPA4) If $c$ violates Probabilism, then there are options $A$ and $B$ such that $A$ dominates $B$, but rationality requires you to choose $B$ over $A$.
(BPA5) Dominance Thesis
Therefore,
(BPAC) Probabilism

Conclusion

The Dutch Book Argument for Probabilism assumes Ramsey's Thesis, which determines the prices an agent is rationally required to pay for a bet. Hedden argues that Ramsey's Thesis is wrong. He claims that Maximise Subjective Expected Utility  determines those prices, and it often disagrees with RT. In our Permissive Dutch Book Argument, I suggested that, in the face of that disagreement, we might be permissive: agents are permitted to pay any price that is required or permitted by RT and they are permitted to pay any price that is required or permitted by MSEU. In our Bookless Pragmatic Argument, I then explored what we might do if we reject this permissive response and insist that only prices permitted or required by MSEU are permissible. I showed that, in that case, we can give a pragmatic argument for Bounded Probabilism, which comes close to Probabilism, but doesn't quite reach; and I showed that, if we allow RT in the very particular cases in which it agrees better with intuition than MSEU does, we can give a pragmatic argument for Probabilism.

Appendix: Proof of Theorem 1

Theorem 1 The following are equivalent:
(i) $c$ satisfies Bounded Probabilism
(ii) For all options $A$, $B$, if $A$ dominates $B$, then $\mathrm{EU}_c(A) > \mathrm{EU}_c(B)$.

($\Rightarrow$) Suppose $c$ satisfies Bounded Probabilism. Then, by Lemma 2, there is $0 < M \leq 1$ and a probability function $p$ such that $c(-) = M \times p(-)$. Now suppose $A$ and $B$ are actions. Then
• $\mathrm{EU}_c(A) = \mathrm{EU}_{M \times p}(A) = M \times \mathrm{EU}_p(A)$
• $\mathrm{EU}_c(B) = \mathrm{EU}_{M \times p}(B) = M \times \mathrm{EU}_p(B)$
Thus, $\mathrm{EU}_c(A) > \mathrm{EU}_c(B)$ iff $\mathrm{EU}_p(A) > \mathrm{EU}_p(B)$. And we know that, if $A$ dominates $B$ and $p$ is a probability function, then $\mathrm{EU}_p(A) > \mathrm{EU}_p(B)$.

($\Leftarrow$) Suppose $c$ violates Bounded Probabilism. Then there are partitions $\mathcal{X} = \{X_1, \ldots, X_m\}$ and $\mathcal{Y} = \{Y_1, \ldots, Y_n\}$ such that $$\sum^m_{i=1} c(X_i) = x < y = \sum^n_{j=1} c(Y_j)$$We will now define two acts $A$ and $B$ such that $A$ dominates $B$, but $\mathrm{EU}_c(A) < \mathrm{EU}_c(B)$.
• For any $X_i$ in $\mathcal{X}$, $$U(A, X_i) = y - i\frac{y-x}{2(m + 1)}$$
• For any $Y_j$ in $\mathcal{Y}$,$$U(B, Y_j) = x + j\frac{y-x}{2(n + 1)}$$
Then the crucial facts are:
• For any two $X_i \neq X_j$ in $\mathcal{X}$,$$U(A, X_i) \neq U(A, X_j)$$
• For any two $Y_i \neq Y_j$ in $\mathcal{Y}$,$$U(B, Y_i) \neq U(B, Y_j)$$
• For any $X_i$ in $\mathcal{X}$ and $Y_j$ in $\mathcal{Y}$, $$x < U(B, Y_j) < \frac{x+y}{2} < U(A, X_i) < y$$
So $A$ dominates $B$, but$$\mathrm{EU}_c(A) = \sum^m_{i=1} c(X_i) U(A, X_i) < \sum^m_{i=1} c(X_i) \times y = xy$$
while$$\mathrm{EU}_c(B) = \sum^n_{j=1} c(Y_i) U(B, Y_j) > \sum^n_{j=1} c(Y_j) \times x = yx$$So $\mathrm{EU}_c(B) > \mathrm{EU}_c(A)$, as required.

Thursday, 19 July 2018

What is Probabilism?

I've just signed a contract with Cambridge University Press to write a book on the Dutch Book Argument for their Elements in Decision Theory and Philosophy series. So over the next few months, I'm going to be posting some bits and pieces as I get properly immersed in the literature.

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Probabilism is the claim that your credences should satisfy the axioms of the probability calculus. Here is an attempt to state the norm more precisely, where $\mathcal{F}$ is the algebra of propositions to which you assign credences and $c$ is your credence function, which is defined on $\mathcal{F}$, so that $c(A)$ is your credence in $A$, for each $A$ in $\mathcal{F}$.

Probabilism (initial formulation)
• (Non-Negativity) Your credences should not be negative. In symbols: $c(A) \geq 0$, for all $A$ in $\mathcal{F}$.
• (Normalization I) Your credence in a necessarily false proposition should be 0. In symbols: $c(\bot) = 0$.
• (Normalization II) Your credence in a necessarily true proposition should be 1. In symbols: $c(\top) = 1$.
• (Finite Additivity) Your credence in the disjunction of two mutually exclusive propositions should be the sum of your credences in the disjuncts. In symbols: $c(A \vee B) = c(A) + c(B)$.
This sort of formulation is fairly typical. But I think it's misleading in various ways.

As is often pointed out, 0 and 1 are merely conventional choices. Like utilities, we can measure credences on different scales. But what are they conventional choices for? It seems to me that they must represent the lowest possible credence you can have and the highest possible credence you can have, respectively. After all, what we want Normalization I and II to say is that we should have lowest possible credence in necessary falsehoods and highest possible credence in necessary truths. It follows that Non-Negativity is not a normative constraint on your credences, which is how it is often presented. Rather, it follows immediately from the particular representation of our credences that we have chosen to. Suppose we chose a different representation, where -1 represents the lowest possible credence and 1 represents the highest. Then Normalization I and II would say that $c(\bot) = -1$ and $c(\top) = 1$, so Non-Negativity would be false.

One upshot of this is that Non-Negativity is superfluous once we have specified the representation of credences that we are using. But another is that Probabilism incorporates not only normative claims, such as Normalization I and II and Finite Additivity, but also a metaphysical claim, namely, that there is a lowest possible credence that you can have and a highest possible credence that you can have. Without that, we couldn't specify the representation of credences in such a way that we would want to sign up to Normalization I and II. Suppose that, for any credence you can have, there is a higher one than you could have. Then there is no credence that I would want to demand you have in a necessary truth--for any I demanded, it would be better for you to have one higher. So I either have to say that all credences in necessary falsehoods are rationally forbidden, or all are rationally permitted, or I pick some threshold above which any credence is rationally permitted. And the same goes, mutatis mutandis, for credences in necessary falsehoods. I'm not sure what the norm of credences would be if our credences were unbounded in one or other or both directions. But it certainly wouldn't be Probabilism.

So Non-Negativity is not a normative claim, but rather a trivial consequence of a metaphysical claim together with a conventional choice of representation. The metaphysical claim is that there is a minimal and a maximal credence; the representation choice is that 0 will represent the minimal credence and 1 will represent the maximal credence.

Next, suppose we make a different conventional choice. Suppose we pick real numbers $a$ and $b$, and we say that $a$ represents minimal credence and $b$ represents maximal credence. Then clearly Normalization I becomes $c(\bot) = a$ and Normalization II becomes $c(\top) = b$. But what of Finite Additivity? This looks problematic. After all, if $a = 10$ and $b = 30$, and $c(A) = 20 = c(\overline{A})$, then Finite Addivitity demands that $c(\top) = c(A \vee \overline{A}) = c(A) + c(\overline{A}) = 40$, which is greater than the maximal credence. So Finite Additivity makes an impossible demand on an agent who seems to have perfectly rational credences in $A$ and $\overline{A}$, given the representation.

The reason is that Finite Additivity, formulated as we formulated it above, is peculiar to very specific representations of credences, such as the standard one on which 0 stands for minimal credence and 1 stands for maximal credence. The correct formulation of Finite Additivity in general says: $c(A \vee B) = c(A) + c(B) - c(A\ \&\ B)$, for any propositions $A$, $B$ in $\mathcal{F}$. Thus, in the case we just gave above, if $c(A\ \&\ \overline{A}) = 10$, in keeping with the relevant version of Normalization I, we have $c(A \vee \overline{A}) = 20 + 20 - 10 = 30$, as required. So we see that it's wrong to say that Probabilism says that your credence in the disjunction of two mutually exclusive propositions should be the sum of your credences in the disjuncts--that's actually only true on some representation of your credences (namely, those for which 0 represents minimal credence).

Bringing all of this together, I propose the following formulation of Probabilism:

Probabilism (revised formulation)
• (Bounded credences) There is a lowest possible credence you can have; and there is a highest possible credence you can have.
• (Representation) We represent the lowest possible credence you have using $a$, and we represent the highest possible credence you can have using $b$.
• (Normalization I) Your credence in a necessarily false proposition should be the lowest possible credence you can have. In symbols: $c(\bot) = a$.
• (Normalization II) Your credence in a necessarily true proposition should be the highest possible credence you can have. In symbols: $c(\top) = b$.
• (Finite Additivity) $c(A \vee B) = c(A) + c(B) - c(A\ \&\ B)$, for any propositions $A$, $B$ in $\mathcal{F}$.
We call such a credence function a probability$_{a, b}$ function. How can we be sure this is right? Here are some considerations in its favour:

Switching representations
(i) Suppose $c(-)$ is a probability$_{a, b}$ function. Then $\frac{1}{b-a}c(-) - \frac{a}{b-a}$ is a probability function (or probability$_{0, 1}$ function).
(ii) Suppose $c(-)$ is a probability function and $a, b$ are real numbers. Then $c(-)(b-a) + a$ is a probability$_{a, b}$ function.

Dutch Book Argument
The standard Dutch Book Argument for Probabilism assumes that, if you have credence $p$ in proposition $X$, then you will pay £$pS$ for a bet that pays £$S$ if $X$ and £$0$ if $\overline{X}$. But this assumes that you have credences between 0 and 1, inclusive. What is the corresponding assumption if you represent credences in a different scale? Shorn of its conventional choice of representation, the assumption is: (a) you will pay £$0$ for a bet on $X$ if you have minimal credence in $X$; (b) you will pay £$S$ for a bet on $X$ if you have maximal credence in $X$; (c) the price you will pay for a bet on $X$ increases linearly with your credence in $X$. Translated into a framework in which we measure credence on a scale from $a$ to $b$, the assumption is then: you will pay £$\frac{p-a}{b-a}S$ for a bet that pays £$S$ if $X$ and £$0$ if $\overline{X}$. And, with this assumption, we can find Dutch Books against any credence function that isn't a probability$_{a, b}$ function.

Accuracy Dominance Argument
The standard Accuracy Dominance Argument for Probabilism assumes that, for each world, the ideal or vindicated credence function at that world assigns 0 to all falsehoods and 1 to all truths. Of course, if we represent minimal credence by $a$ and maximal credence by $b$, then we'll want to change that assumption. We'll want to say instead that the ideal or vindicated credence function at a world assigns $a$ to falsehoods and $b$ to truths. Once we say that, for any credence function that isn't a probability$_{a, b}$ function, there is another credence function that is closer to the ideal credence function at all worlds.

So, the usual arguments for having a credence function that is a probability function when you represent your credences on a scale from 0 to 1 can be repurposed to argue that you should have a credence function that is a probability$_{a, b}$ function when you represent your credences on a scale from $a$ to $b$. And that gives us good reason to think that the second formulation of Probabilism above is correct.

Monday, 16 July 2018

Yet another assistant professorship in formal philosophy @ University of Gdansk

Some time ago the Chair of Logic, Philosophy of Science and Epistemology had an opening in formal philosophy that since then has been filled. Now, another position (leading to a permanent position upon second renewal) is available (so, there'll be three tenure-track faculty members working on formal philosophy). Details.

Lecturer Position in Logic and Philosophy of Language (MCMP)

The Ludwig-Maximilians-University Munich is one of the largest and most prestigious universities in Germany.

Ludwig-Maximilians-University Munich is seeking applications for one

Lecturer Position (equivalent to Assistant Professorship)
in Logic and Philosophy of Language
(for three years, with the possibility of extension)

at the Chair of Logic and Philosophy of Language (Professor Hannes Leitgeb) and the Munich Center for Mathematical Philosophy (MCMP) at the Faculty of Philosophy, Philosophy of Science and Study of Religion. The position, which is to start on December 1, 2018, is for three years with the possibility of extension for another three years.

The appointee will be expected (i) to do philosophical research, especially in logic and philosophy of language, (ii) to teach five hours a week in areas relevant to the chair, and (iii) to participate in the administrative work of the MCMP.

The successful candidate will have a PhD in philosophy or logic, will have teaching experience in philosophy and logic, and will have carried out research in logic and related areas (such as philosophy of logic, philosophy of language, philosophy of mathematics, formal epistemology).

Your workplace is centrally located in Munich and is very easy to reach by public transport. We offer you an interesting and responsible job with good training and development opportunities.
The employment takes place within the TV-L scheme.
The position is initially limited to November 30, 2021.
Furthermore, given equal qualification, severely physically challenged applicants will be preferred.
There is the possibility of part-time employment.
The application of women is strongly welcome.

Applications (including CV, certificates, list of publications, list of courses taught, a writing sample and a description of planned research projects (1000-1500 words)) should be sent either by email (ideally all requested documents in just one PDF document) or by mail to

Ludwig-Maximilians-Universität München
Faculty of Philosophy, Philosophy of Science and Study of Religion
Chair of Logic and Philosophy of Language / MCMP
Geschwister-Scholl-Platz 1
80539 München

by September 1, 2018. If possible, we very much prefer applications by email.

In addition, we ask for two letters if reference, which must be sent by the reviewers directly to the above address (e-mail preferred).

For further questions you can contact by e-mail office.leitgeb@lrz.uni-muenchen.de.

The German description of the position is to be found at https://www.uni-muenchen.de/aktuelles/stellenangebote/wissenschaft/20180704161330.html

Tuesday, 3 July 2018

Entia et Nomina 2018 (August 28-29, Gdansk)

This is the seventh conference in the Entia et NominA series (previous editions took place in Poland, Belgium and India), which  features workshops for researchers in formally and analytically oriented philosophy, in particular in epistemology, logic, and philosophy of science.  The distinctive format of the workshop requires participants to distribute extended abstracts or full papers a couple of weeks before the workshop and to prepare extended comments on another participant's paper.

Invited speakers
- Zalan Gyenis (Jagiellonian University, Poland)
- Masashi Kasaki (Nagoya University, Japan)
- Martin Smith (University of Edinburgh, Scotland)

Dates: