An Improved Argument for Superconditionalization

Abstract

Standard arguments for Bayesian conditionalizing rely on assumptions that many epistemologists have criticized as being too strong: (i) that conditionalizers must be logically infallible, which rules out the possibility of rational logical learning, and (ii) that what is learned with certainty must be true (factivity). In this paper, we give a new factivity-free argument for the superconditionalization norm in a personal possibility framework that allows agents to learn empirical and logical falsehoods. We then discuss how the resulting framework should be interpreted. Does it still model norms of rationality, or something else, or nothing useful at all? We discuss five ways of interpreting our results, three that embrace them and two that reject them. We find one of each kind wanting, and leave readers to choose among the remaining three.

Appendix

Lemma 1

Let ${\mathcal {F}}=\{X_1,\dots , X_m\}$ be a set of credal objects over a set of worlds W. Given real numbers $a_1,\dots , a_m \in {\mathbb {R}}$, let $A:W \rightarrow [0,1]$ be an act defined as $A(w)=\sum \limits _{i=1}^m a_iv_w(X_i)$ for all $w\in W$. If $p:W\rightarrow [0,1]$ coherently extends a personally probabilistic credence function $c:{\mathcal {F}}\rightarrow [0,1]$, then:

$$\begin{aligned} \sum \limits _{w\in W}p(w)A(w)=\sum \limits _{i=1}^mc(X_i)a_i \end{aligned}$$

Proof

Consider a $p:W\rightarrow [0,1]$ that coherently extends a personally probabilistic credence function $c:{\mathcal {F}}\rightarrow [0,1]$. For each $1\le i\le m$, define an act $A_i:W\rightarrow [0,1]$ such that $A_i(w)=a_iv_w(X_i)$ for all $w \in W$. So we have that:

$$\begin{aligned} \sum \limits _{w\in W}p(w)A(w)=\sum \limits _{w\in W}p(w)\sum \limits _{i=1}^mA_i(w) \\ =\sum \limits _{i=1}^m\sum \limits _{w\in W}p(w)A_i(w) \end{aligned}$$

Splitting the inner sum according to the truth value of $X_i$, for $1\le i\le m$, it holds that:

$$\begin{aligned} \sum \limits _{w\in W}p(w)A_i(w)= & {} \sum \limits _{w,v_w(X_i)=1}p(w)A_i(w)+\sum \limits _{w,v_w(X_i)=0}p(w)A_i(w)\\= & {} \sum \limits _{w,v_w(X_i)=1}p(w)a_i+\sum \limits _{w,v_w(X_i)=0}p(w)0\\= & {} a_i\sum \limits _{w,v_w(X_i)=1}p(w) \end{aligned}$$

As p coherently extends c, $\sum \limits _{w,v_w(X_i)=1}p(w)=c(X_i)$. Finally, we can conclude that:

$$\begin{aligned} \sum \limits _{w\in W}p(w)A(w)=\sum \limits _{i=1}^mc(X_i)a_i \end{aligned}$$

$\square$

Theorem 3

Let $c:{\mathcal {F}}\rightarrow [0,1]$ be a personally probabilistic credence function and $c'=\langle {c'_1,\dots ,c'_n}\rangle$ be the set of possible future personally probabilistic credence functions defined over ${\mathcal {F}}$.

• (a) If $\langle {c,c'}\rangle$ violates wGRP, then it is vulnerable to a Strong Dutch Strategy.

• (b) If $\langle {c,c'}\rangle$ satisfies wGRP, then it is not vulnerable to a Strong Dutch Strategy.

Proof

• (a) Supposing ${\mathcal {F}}=\{X_1,\dots , X_m\}$, any credence function ${\hat{c}}:{\mathcal {F}}\rightarrow [0,1]$ can be viewed as a vector $\langle {{\hat{c}}(X_1),\dots ,{\hat{c}}(X_m)}\rangle \in {\mathbb {R}}^m$. If $\langle {c,c'}\rangle$ violates wGRP, then c is not in the convex hull of the vectors $c'_1,\dots ,c'_n$. Thus, by the Separating Hyperplane Theorem, there are numbers $a,a' \in {\mathbb {R}}$ and a vector $b\in {\mathbb {R}}^m$ such that, for each $c'_j\in c'$:

  $$\begin{aligned} \sum \limits _{i=1}^m c(X_i)b_i< a<a'<\sum \limits _{i=1}^m c'_j(X_i)b_i \end{aligned}$$

  Now consider the constant acts $A:W\rightarrow [0,1]$ and $A':W\rightarrow [0,1]$ such that $A(w)=a$ and $A'(w)=-a'$ for all $w \in W$. Furthermore, consider the act $B: W \rightarrow [0,1]$ defined in the following way: $B(w)= \sum \limits _{i=1}^m b_iv_w(X_i)$. That is, for each world $w\in W$, determine those $X_i$ that are true and sum the corresponding $b_i$ to obtain B(w). If all $X_i \in {\mathcal {F}}$ are false in w, then $B(w)=0$. Now, define the act $B':W \rightarrow [0,1]$ via $B'(w)=-B(w)$ for all $w\in W$. By Lemma 1, for any $p:W\rightarrow [0,1]$ that coherently extends c we have that $\sum _w p(w)B(w)=\sum \limits _{i=1}^mc(X_i)b_i$. As $\sum _wp(w)A(w)=a$, c prefers A to B. Analogously, by Lemma 1, each $c'_j$ prefers $A'$ to $B'$. Finally, note that $B(w)+B'(w)=0>a-a'=A(w)+A'(w)$ for any $w\in W$, therefore $\langle {c,c'}\rangle$ is vulnerable to a Strong Dutch Strategy.

• (b) Assume there are weights $\lambda _j\in [0,1]$, summing up to one, such that $c(X)=\sum _j\lambda _jc'_j(X)$ for all $X \in {\mathcal {F}}$. To prove by contradiction, suppose there are acts $A,A',B,B'$ such that c prefers A to B, each $c_j$ prefers $A'$ to $B'$, but $A(w)+A'(w)<B(w)+B'(w)$ at any $w \in W$. For each $c'_j$, consider a credence function $p'_j:W\rightarrow [0,1]$ that coherently extends it, thus preferring $A'$ to $B'$. Defining $p:W\rightarrow [0,1]$ via $p(w)=\sum _j\lambda _jp'_j(w)$ for all $w\in W$, we have that $\langle {p,p'}\rangle$ satisfies wGRP. Note that p coherently extends c, hence preferring A to B. Consequently, $\langle {p,p'}\rangle$ is vulnerable to a Strong Dutch Strategy, which contradicts Theorem 1(b).

$\square$

Theorem 4

Let $c:{\mathcal {F}}\rightarrow [0,1]$ be a personally probabilistic credence function and $c'=\langle {c'_1,\dots ,c'_n}\rangle$ be the set of possible future personally probabilistic credence functions defined over ${\mathcal {F}}$.

• (a) If $\langle {c,c'}\rangle$ violates wGRP, then it is accuracy dominated.

• (b) If $\langle {c,c'}\rangle$ satisfies wGRP, then it is not accuracy dominated.

Proof

(a) See the ($\rightarrow$)-part of the proof of Theorem 2 (Pettigrew, 2023), just interpreting W as a set of personally possible worlds.

(b) Suppose $\langle {c,c'}\rangle$ satisfies wGRP, so that there are $\lambda _1,\dots ,\lambda _n\in [0,1]$ such that $\sum \limits _j\lambda _j=1$ and $c(X)=\sum \limits _{j=1}^n\lambda _jc_j(X)$ for any $X\in {\mathcal {F}}$. To prove by contradiction, assume $\langle {c,c'}\rangle$ is accuracy dominated: there are credence functions $c^*,c^*_1,\dots ,c^*_n$, defined on ${\mathcal {F}}$, such that ${\mathfrak {I}}(c,w)+{\mathfrak {I}}(c'_j,w)>{\mathfrak {I}}(c^*,w)+{\mathfrak {I}}(c^*_j,w)$ for all $w\in W$ and $1\le j\le n$. Since accuracy is measured with an additive strictly proper ${\mathfrak {I}}$, there is a strictly proper scoring rule s such that, for any credence function ${\hat{c}}$, ${\mathfrak {I}}({\hat{c}},w)=\sum \limits _{X\in {\mathcal {F}}}s(v_w(X),{\hat{c}}(X))$. For s is strictly proper, we have that, for any $X\in {\mathcal {F}}$ and any $1\le j\le n$:

$$\begin{aligned}{} & {} c(X)s(1,c(X))+(1-c(X))s(0,c(X))\nonumber \\{} & {} \quad \le c(X)s(1,c^*(X))+(1-c(X))s(0,c^*(X)) \end{aligned}$$

(1)

$$\begin{aligned}{} & {} c'_j(X)s(1,c'_j(X))+(1-c'_j(X))s(0,c'_j(X))\nonumber \\{} & {} \quad \le c'_j(X)s(1,c^*_j(X))+(1-c'_j(X))s(0,c^*_j(X)) \end{aligned}$$

(2)

If we replace those c(X) out of the scope of s(.) by $\sum _j\lambda _jc_j(X)$ in Expression (1) (note also that $\sum _j\lambda _j=1$) and, in Expression (2), multiply both sides by $\lambda _j$ before summing for all j, we obtain, respectively:

$$\begin{aligned}{} & {} \sum \limits _j\lambda _jc'_j(X)s(1,c(X))+\sum \limits _j\lambda _j(1-c'_j(X))s(0,c(X))\nonumber \\{} & {} \quad \le \sum \limits _j\lambda _jc'_j(X)s(1,c^*(X))+\sum \limits _j\lambda _j(1-c'_j(X))s(0,c^*(X)) \end{aligned}$$

(3)

$$\begin{aligned}{} & {} \sum \limits _j\lambda _j\big [c'_j(X)s(1,c'_j(X))+(1-c'_j(X))s(0,c'_j(X))\big ]\nonumber \\{} & {} \quad \le \sum \limits _j\lambda _j\big [ c'_j(X)s(1,c^*_j(X))+(1-c'_j(X))s(0,c^*_j(X))\big ] \end{aligned}$$

(4)

Grouping the summations in j in each side of Expression (3), it can be added to Expression (4) to obtain:

$$\begin{aligned}{} & {} \sum \limits _j\lambda _j\big [c'_j(X)(s(1,c(X))+s(1,c'_j(X)))+(1-c'_j(X))(s(0,c(X))+s(0,c'_j(X)))\big ]\nonumber \\{} & {} \quad \le \sum \limits _j\lambda _j\big [c'_j(X)(s(1,c^*(X))+s(1,c^*_j(X)))\nonumber \\{} & {} \qquad +(1-c'_j(X))(s(0,c^*(X))+s(0,c^*_j(X)))\big ] \end{aligned}$$

(5)

Since $\langle {c,c'}\rangle$ is accuracy dominated by $c^*$ and $\langle {c^*_1,\dots ,c^*_n}\rangle$, ${\mathfrak {I}}(c,w)+{\mathfrak {I}}(c'_j,w)>{\mathfrak {I}}(c^*,w)+{\mathfrak {I}}(c^*_j,w)$ for all $w\in W$ and $1\le j\le n$. Multiplying each side of this inequality by $\lambda _jp_j(w)$, for a $p_j$ that coherently extends $c'_j$, and summing for all $w \in W$ and all $1\le j \le n$, we obtain:

$$\begin{aligned} \sum \limits _j\lambda _j\sum \limits _w p_j(w)\big [{\mathfrak {I}}(c,w)+{\mathfrak {I}}(c'_j,w)\big ]> \sum \limits _j\lambda _j\sum \limits _w p_j(w)\big [{\mathfrak {I}}(c^*,w)+{\mathfrak {I}}(c^*_j,w)\big ] \end{aligned}$$

(6)

Recall that, for any ${\hat{c}}:{\mathcal {F}}\rightarrow [0,1]$, ${\mathfrak {I}}({\hat{c}},w)=\sum \limits _{X\in {\mathcal {F}}}s(v_w(X),{\hat{c}}(X))$. Thus, for any ${\hat{c}}$, ${\mathfrak {I}}({\hat{c}},w)$ can be rewritten as:

$$\begin{aligned}{} & {} {\mathfrak {I}}({\hat{c}},w) = \sum \limits _{X\in {\mathcal {F}}}v_w(X)s(1,{\hat{c}}(X))+\sum \limits _{X\in {\mathcal {F}}}(1-v_w(X))s(0,{\hat{c}}(X))\nonumber \\{} & {} \quad = \sum \limits _{X\in {\mathcal {F}}}v_w(X)s(1,{\hat{c}}(X)) -\sum \limits _{X\in {\mathcal {F}}}v_w(X)s(0,{\hat{c}}(X))+\sum \limits _{X\in {\mathcal {F}}}s(0,{\hat{c}}(X)) \end{aligned}$$

(7)

If ${\hat{c}}$ is fixed, the first two summations in Expression (7) can be viewed as acts in the format $\sum _ia_iv_w(X_i)$. Hence, applying Lemma 1 to $\sum _w p_j(w){\mathfrak {I}}(c,w)$ yields:

$$\begin{aligned} \sum \limits _{X\in {\mathcal {F}}}c'_j(X)s(1,c(X))-\sum \limits _{X\in {\mathcal {F}}}c'_j(X)s(0,c(X))+\sum \limits _w p_j(w)\sum \limits _{X\in {\mathcal {F}}}s(0,c(X)) \end{aligned}$$

As $\sum \limits _w p_j(w)=1$, a bit of algebraic manipulation results in:

$$\begin{aligned} \sum \limits _{X\in {\mathcal {F}}}\big [c'_j(X)s(1,c(X))+(1-c'_j(X))s(0,c(X))\big ] \end{aligned}$$

Analogously, we can apply Lemma 1 to each $\sum _w p_j(w){\mathfrak {I}}(\cdot ,w)$ resulting from Expression (6), obtaining:

$$\begin{aligned}{} & {} \sum \limits _j\sum \limits _{X\in {\mathcal {F}}}\lambda _j\big [c'_j(X)(s(1,c(X))+s(1,c'_j(X)))\nonumber \\{} & {} \qquad +(1-c'_j(X))(s(0,c(X))+s(0,c'_j(X)))\big ]\nonumber \\{} & {} \quad >\sum \limits _j\sum \limits _{X\in {\mathcal {F}}}\lambda _j\big [c'_j(X)(s(1,c^*(X))+s(1,c^*_j(X)))\nonumber \\{} & {} \qquad +(1-c'_j(X))(s(0,c^*(X))+s(0,c^*_j(X)))\big ] \end{aligned}$$

(8)

But note that this is just the negation of Expression (5) summed for all $X\in {\mathcal {F}}$, which is a contradiction, completing the proof. $\square$

Theorem 5

Let c be a credence function. Let $c'=\langle {c'_1,\dots ,c'_n}\rangle$ be an updating rule for a partition $\{E_1,\dots ,E_n\}$ of W, respecting it. If the pair $\langle {c,c'}\rangle$ satisfies the Weak General Reflection Principle, then $\langle {c,c'}\rangle$ is superconditioning.

Proof

As $c'$ respects the partition $\{E_1,\dots , E_n\}$, for each $1\le i\le n$ there is a function $p_i:W\rightarrow [0,1]$, with $\sum \limits _{w\in W}p_i(w)=1$, such that $c'_i(X)=\sum \limits _{w \in W}p_i(w)v_w(X)$ for all $X \in {\mathcal {F}}$ and $\sum \limits _{w \in E_i}p_i(w)=1$. wGRP implies that, for some $\lambda _1,\dots ,\lambda _n\in [0,1]$ with $\sum \limits _{i=1}^n\lambda _i=1$, we have that $c(X)=\sum \limits _{i=1}^n \lambda _ic'_i(X)$ for all $X \in {\mathcal {F}}$. Thus, $c(X)=\sum \limits _{i=1}^n \lambda _i\sum \limits _{w \in W}p_i(w)v_w(X)=\sum \limits _{w\in W}\sum \limits _{i=1}^n\lambda _ip_i(w)v_w(X)$ for all $X \in {\mathcal {F}}$. Let the function $p:W\rightarrow [0,1]$ be such that $p(w)=\sum \limits _{i=1}^n\lambda _ip_i(w)$ for all $w \in W$. Note that $\sum \limits _{w \in W}p(w)=1$.

Now consider an element $E_j$ of the partition. We have, for all $X \in {\mathcal {F}}$, that $\sum \limits _{w\in E_j}p(w)v_w(X)=\sum \limits _{w\in E_j}v_w(X)\sum \limits _{i=1}^n\lambda _ip_i(w)$. Given that, for any $w\in E_j$, $p_i(w)=0$ whenever $i\ne j$, as the partition is respected, it follows that $\sum \limits _{w\in E_j}v_w(X)\sum \limits _{i=1}^n\lambda _ip_i(w)=\sum \limits _{w\in E_j}v_w(X)\lambda _jp_j(w)=\lambda _j\sum \limits _{w\in W}v_w(X)p_j(w)$. For all $E_j$, we also have that $\sum \limits _{w\in E_j}p(w)=\sum \limits _{w\in E_j}\sum \limits _{i=1}^n\lambda _ip_i(w)=\sum \limits _{w\in E_j}\lambda _jp_j(w)$, thus $\sum \limits _{w \in E_j}p(w)=\lambda _j\sum \limits _{w \in E_j}p_j(w)=\lambda _j$. Finally, for each $1\le j \le n$ with $\sum \limits _{w \in E_j}p(w)=\lambda _j>0$, we obtain, for all $X \in {\mathcal {F}}$:

$$\begin{aligned} c_j'(X)=\sum _{w \in W}p_j(w)v_w(X)= \frac{\lambda _j\sum \limits _{w \in W}p_j(w)v_w(X)}{\lambda _j} =\frac{\sum \limits _{w\in E_j}p(w)v_w(X)}{\sum \limits _{w\in E_j}p(w)} \end{aligned}$$

Consequently, $\langle {c,c'}\rangle$ is superconditioning. $\square$