Correcting credences with chances

Abstract

Lewis’s Principal Principle is widely recognized as a rationality constraint that our credences should satisfy throughout our epistemic life. In practice, however, our credences often fail to satisfy this principle because of our various epistemic limitations. Facing such violations, we should correct our credences in accordance with this principle. In this paper, I will formulate a way of correcting our credences, which will be called the Adams Correcting Rules and then show that such a rule yields non-commutativity between conditionalizing and correcting. With the help of the notion of ‘accuracy’, then, I attempt to provide a vindication of the Adams Correcting Rule and show how we can respond to the non-commutativity in question.

Appendix

The following proofs and derivations have various conditional credences and chances. For presentational simplicity, I will assume in what follows that such conditional credences and chances are all well defined. I think this assumption yields no confusion.

1.1 A derivation of ACR0 from PP0 and GAC

Suppose that $C_{CH}$ satisfies PP0, and that $C_{CH}$ is updated from C by means of GAC. Then we have that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{CH}(A)&=\sum _{U_{i}\in \mathcal {U},w_{j}\in \mathcal {W}}C(U_{i})C(A|w_{j}U_{i})C_{CH}(w_{j}|U_{i})\\&=\sum _{U_{i}\in \mathcal {U},w_{j}\in \mathcal {W}}C(U_{i})C(A|w_{j}U_{i})ch_{i}(w_{j})\\&=\sum _{U_{i}\in \mathcal {U}}C(U_{i})\sum _{w_{j}\in \mathcal {W}}C(A|w_{j}U_{i})ch_{i}(w_{j})\\&=\sum _{U_{i}\in \mathcal {U}}C(U_{i})\sum _{w_{j}\in A}C(A|w_{j}U_{i})ch_{i}(w_{j})\\&\quad +\sum _{U_{i}\in \mathcal {U}}C(U_{i})\sum _{w_{j}\in \lnot A}C(A|w_{j}U_{i})ch_{i}(w_{j})\\&=\sum _{U_{i}\in \mathcal {U}}C(U_{i})\sum _{w_{j}\in A}ch_{i}(w_{j})=\sum _{U_{i}\in \mathcal {U}}C(U_{i})ch_{i}(A), \end{aligned}$$

as required.

1.2 A derivation of ACR+ from PP+ and GAC

Suppose that $C_{E,CH}$ satisfies PP+, and that $C_{E,CH}$ is updated from $C_{E}$ by means of GAC. Then we have that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{E,CH}(A)&=\sum _{U_{i}\in \mathcal {U},w_{j}\in \mathcal {W}}C_{E}(U_{i})C_{E}(A|w_{j}U_{i})C_{E,CH}(w_{j}|U_{i})\\&=\sum _{U_{i}\in \mathcal {U},w_{j}\in \mathcal {W}}C_{E}(U_{i})C_{E}(A|w_{j}U_{i})ch_{i}(w_{j}|E)\\&=\sum _{U_{i}\in \mathcal {U}}C_{E}(U_{i})\sum _{w_{j}\in \mathcal {W}}C_{E}(A|w_{j}U_{i})ch_{i}(w_{j}|E)\\&=\sum _{U_{i}\in \mathcal {U}}C_{E}(U_{i})\sum _{w_{j}\in A}C_{E}(A|w_{j}U_{i})ch_{i}(w_{j}|E)\\&\quad +\sum _{U_{i}\in \mathcal {U}}C_{E}(U_{i})\sum _{w_{j}\in \lnot A}C_{E}(A|w_{j}U_{i})ch_{i}(w_{j}|E)\\&=\sum _{U_{i}\in \mathcal {U}}C_{E}(U_{i})\sum _{w_{j}\in A}ch_{i}(w_{j}|E)=\sum _{U_{i}\in \mathcal {U}}C_{E}(U_{i})ch_{i}(A|E), \end{aligned}$$

as required.

1.3 A proof of Proposition 1

Suppose that our credences are updated by means of Conditionalization, ACR0, and ACR+. It is straightforward that $C_{E,CH}(w|U_{i})=ch_{i}(w|E)$ for any $w\in \mathcal {W}$, and so $C_{E,CH}(A|U_{i})=ch_{i}(A|E)$ for any $A\subseteq \mathcal {W}$. Similarly, we have that $C_{CH}(w|U_{i})=ch_{i}(w),$ and so that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{CH}(A|U_{i})=\sum _{w\in A}C_{CH}(w|U_{i})=\sum _{w\in A}ch_{i}(w)=ch_{i}(A). \end{aligned}$$

Then, Conditionalization and the above equation imply that, for any $A\subseteq \mathcal {W}$ and $U_{i}\in \mathcal {U}$,

$$\begin{aligned} C_{CH,E}(A|U_{i})=C_{CH}(A|U_{i}E)=\frac{C_{CH}(AE|U_{i})}{C_{CH}(A|U_{i})}=ch_{i}(A|E). \end{aligned}$$

Then, we have Proposition 1. Done.

1.4 A proof of Proposition 2

Note that $U_{i}\subseteq \mathcal {W}$ for any $U_{i}\in \mathcal {U}$. Thus, it is straightforward that (a) implies (b). Now, let me prove that (b) implies (a). For this purpose, let’s assume (b)—that is, $C_{E,CH}(U_{i})=C_{CH,E}(U_{i})$ for any $U_{i}\in \mathcal {U}$. From this assumption and Proposition 1, then, it follows that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{E,CH}(A)&=\sum _{U_{i}\in \mathcal {U}}C_{E,CH}(U_{i})C_{E,CH}(A|U_{i})=\sum _{U_{i}\in \mathcal {U}}C_{E,CH}(U_{i})ch_{i}(A|E)\\&=\sum _{U_{i}\in \mathcal {U}}C_{CH,E}(U_{i})ch_{i}(A|E)=\sum _{U_{i}\in \mathcal {U}}C_{CH,E}(U_{i})C_{CH,E}(A|U_{i})\\&=C_{CH,E}(A), \end{aligned}$$

as required.

1.5 A proof of Proposition 3

According to ACR+ and Conditionalization, we have that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{E,CH}(A)=\sum _{U_{i}\in \mathcal {U}}C(U_{i}|E)ch_{i}(A|E). \end{aligned}$$

(1)

Similarly, it follows from Conditionalization and ACR0 that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{CH,E}(A)&=C_{CH}(A|E)=\frac{C_{CH}(AE)}{C_{CH}(E)}\nonumber \\&=\frac{\sum _{U_{i}\in \mathcal {U}}C(U_{i})ch_{i}(AE)}{\sum _{U_{i}\in \mathcal {U}}C(U_{i})ch_{i}(E)}. \end{aligned}$$

(2)

First, let me prove that if E is a disjunction of some $U_{i}$s, then it holds that $C_{E,CH}(A)=C_{CH,E}(A)$ for any $A\subseteq \mathcal {W}$. Suppose that E is a disjunction of some $U_{i}$s. Let $\mathcal {U}_{E}$ be the set of the disjuncts in question. Then, it holds that $U_{i}$ implies E when $U_{i}\in \mathcal {U}_{E}$, and $U_{i}$ implies $\lnot E$ otherwise. Moreover, $ch_{i}(E)=1$ when $U_{i}\in \mathcal {U}_{E}$, and $ch_{i}(E)=0$ otherwise (Note that Self-esteem was assumed). From this, it follows that $ch_{i}(AE)=ch_{i}(A|E)=ch_{i}(A)$ when $U_{i}\in \mathcal {U}_{E}$, and $ch_{i}(AE)=0$ otherwise. Then, we have that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{CH,E}(A)&=\frac{\sum _{U_{i}\in \mathcal {U}}C(U_{i})ch_{i}(AE)}{\sum _{U_{i}\in \mathcal {U}}C(U_{i})ch_{i}(E)}\\&=\frac{\sum _{U_{i}\in \mathcal {U}_{E}}C(U_{i})ch_{i}(AE)+\sum _{U_{i}\notin \mathcal {U}_{E}}C(U_{i})ch_{i}(AE)}{\sum _{U_{i}\in \mathcal {U}_{E}}C(U_{i})ch_{i}(E)+\sum _{U_{i}\notin \mathcal {U}_{E}}C(U_{i})ch_{i}(E)}\\&=\frac{\sum _{U_{i}\in \mathcal {U}_{E}}C(U_{i})ch_{i}(AE)}{\sum _{U_{i}\in \mathcal {U}_{E}}C(U_{i})}=\frac{\sum _{U_{i}\in \mathcal {U}_{E}}C(U_{i})ch_{i}(A|E)}{\sum _{U_{i}\in \mathcal {U}_{E}}C(U_{i})}\\&=\frac{\sum _{U_{i}\in \mathcal {U}}C(U_{i}E)ch_{i}(A|E)}{C(E)}=\sum _{U_{i}\in \mathcal {U}}C(U_{i}|E)ch_{i}(A|E)=C_{E,CH}(A), \end{aligned}$$

as required.

Now, let us prove that if there is a $U_{i}$ such that $E\subseteq U_{i}$, then it holds that $C_{E,CH}(A)=C_{CH,E}(A)$ for any $A\subseteq \mathcal {W}$. Suppose that there is an ur-chance proposition that is a subset of E. Let $U_{k}$ be such a proposition. Then, it holds that $ch_{i}(AE)=ch_{i}(E)=0$ when $U_{i}\ne U_{k}$, and that $C(U_{k}|E)=1$. Then, we have that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{CH,E}(A)&=\frac{\sum _{U_{i}}C(U_{i})ch_{i}(AE)}{\sum _{U_{i}}C(U_{i})ch_{i}(E)}\\&=\frac{\sum _{U_{i}=U_{k}}C(U_{i})ch_{i}(AE)+\sum _{U_{i}\ne U_{k}}C(U_{i})ch_{i}(AE)}{\sum _{U_{i}=U_{k}}C(U_{i})ch_{i}(E)+\sum _{U_{i}\ne U_{k}}C(U_{i})ch_{i}(E)}\\&=\frac{C(U_{k})ch_{k}(AE)}{C(U_{k})ch_{k}(E)}=ch_{k}(A|E);\\ C_{E,CH}(A)&=\sum _{U_{i}}C(U_{i}|E)ch_{i}(A|E)\\&=\sum _{U_{i}=U_{k}}C(U_{i}|E)ch_{i}(A|E)+\sum _{U_{i}\ne U_{k}}C(U_{i}|E)ch_{i}(AE)\\&=C(U_{k}|E)ch_{k}(A|E)=ch_{k}(A|E). \end{aligned}$$

Thus, it holds that $C_{E,CH}(A)=C_{CH,E}(A)$ for any $A\subseteq \mathcal {W}$. Done.

1.6 Several calculations related to Proposition 4 and Fig. 1

As assumed in the example related to Fig. 1, $\mathcal {U}=\{U_{1},U_{2}\}$. Then, it follows from (1) and (2) in the proof of Proposition 3 that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{CH}(A)&=C(U_{1})ch_{1}(A)+C(U_{2})ch_{2}(A), \end{aligned}$$

(4a)

$$\begin{aligned} C_{CH,E}(A)&=\frac{C(U_{1})ch_{1}(AE)+C(U_{2})ch_{2}(AE)}{C(U_{1})ch_{1}(E)+C(U_{2})ch_{2}(E)}, \end{aligned}$$

(4b)

$$\begin{aligned} C_{E}(A)&=\frac{C(AE)}{C(E)},\,\text {and} \end{aligned}$$

(4c)

$$\begin{aligned} C_{E,CH}(A)&=C(U_{1}|E)ch_{1}(A|E)+C(U_{2}|E)ch_{2}(A|E). \end{aligned}$$

(4d)

Note that $E\equiv w_{1}\vee w_{2}\vee w_{4}\vee w_{5}$, $U_{1}\equiv w_{1}\vee w_{2}\vee w_{3}$, and $U_{2}\equiv w_{4}\vee w_{5}\vee w_{6}$. With the help of (4a)–(4d), the chance assignments of $ch'$ and $ch^{*}$, and the initial credence assignment of C, we have that, for any $A\subseteq \mathcal {W}$,

$$\begin{aligned} C_{CH}(A)&=\frac{1}{2}ch_{1}(A)+\frac{1}{2}ch_{2}(A),\\ C_{CH,E}(A)&=\frac{12}{17}\left( ch_{1}(AE)+ch_{2}(AE)\right) ,\\ C_{E}(A)&=\frac{10}{7}C(AE),\,\text {and}\\ C_{E,CH}(A)&=\frac{9}{14}ch_{1}(AE)+\frac{16}{21}ch_{2}(AE). \end{aligned}$$

Now, we can derive the probability assignments of $C_{CH}$, $C_{CH,E}$, $C_{E}$, and $C_{E,CH}$. For example,

$$\begin{aligned} C_{CH}(w_{1})&=\frac{1}{2}ch_{1}(w_{1})+\frac{1}{2}ch_{2}(w_{1})=\frac{1}{2}ch_{1}(w_{1})=\frac{1}{6};\\ C_{CH,E}(w_{1})&=\frac{12}{17}\left( ch_{1}(w_{1}E)+ch_{2}(w_{1}E)\right) =\frac{12}{17}ch_{1}(w_{1})=\frac{4}{17};\\ C_{E}(w_{1})&=\frac{10}{7}C(w_{1}E)=\frac{10}{7}C(w_{1})=\frac{1}{7};\,\text {and}\\ C_{E,CH}(w_{1})&=\frac{9}{14}ch_{1}(w_{1}E)+\frac{16}{21}ch_{2}(w_{1}E)=\frac{9}{14}ch_{1}(w_{1})=\frac{3}{14}. \end{aligned}$$

These results conform with the probability assignments in Fig. 1.

1.7 Proofs of Propositions 5 and 6

Let me start with noting that it follows from GAC and the probability calculus that: for any $F_{k}\in \mathcal {F}$,

$$\begin{aligned} C_{\mathcal {E}|\mathcal {F}}(F_{k})&=\sum _{E_{i}\in \mathcal {E},F_{j}\in \mathcal {F}}C(F_{j})C{}_{\mathcal {E}|\mathcal {F}}(E_{i}|F_{j})C(F_{k}|E_{i}F_{j})\\&=\sum _{E_{i}\in \mathcal {E},F_{j}=F_{k}}C(F_{j})C{}_{\mathcal {E}|\mathcal {F}}(E_{i}|F_{j})C(F_{k}|E_{i}F_{j})\\&\quad +\sum _{E_{i}\in \mathcal {E},F_{j}\ne F_{k}}C(F_{j})C{}_{\mathcal {E}|\mathcal {F}}(E_{i}|F_{j})C(F_{k}|E_{i}F_{j})\\&=\sum _{E_{i}\in \mathcal {E}}C(F_{k})C{}_{\mathcal {E}|\mathcal {F}}(E_{i}|F_{k})=C(F_{k}). \end{aligned}$$

That is, when some experience directly changes the relevant conditional credences, and so C is updated to $C_{\mathcal {E}|\mathcal {F}}$ by means of GAC, the credences in the conditioning propositions, namely $F_{i}$s, remain the same. As explained above, the correction of $C_{E}$ to $C_{E,CH}$ by means of ACR+ can be regarded as a belief updating by means of GAC, in which the conditioning propositions are the ur-chance propositions $U_{i}$s. So, the credences in $U_{i}$s remain the same through the correction in question. More formally, it holds that: for any $E\subseteq \mathcal {W}$ and $U_{i}\in \mathcal {U}$,

$$\begin{aligned} C_{E,CH}(U_{i})=C_{E}(U_{i})=C(U_{i}|E). \end{aligned}$$

(A)

Now, with this mathematical feature of the Adams Correcting Rule in hand, we can prove Propositions 5 and 6.

A proof of Proposition 5

Let $\mathbb {R}$ be a set of updating rules for a chance-free credence function $C_{E}$. All members of $\mathbb {R}$ correct $C_{E}$ to a chance-fed function. Let $\mathbf {R}$ be the Adams Correcting Rule, i.e., ACR+. Then, it holds that $C_{E}^{\mathbf {R}}=C_{E,CH}$. Moreover, let $\mathbf {R}^{*}$ be a correcting rule in $\mathbb {R}$. Then, Immodesty implies that,

$$\begin{aligned} EI_{C_{E}^{\mathbf {R}}}[C_{E}^{\mathbf {R}},\mathcal {U}]\le EI_{C_{E}^{\mathbf {R}}}[C_{E}^{\mathbf {R}^{*}},\mathcal {U}], \end{aligned}$$

(5a)

where equality holds if and only if $C_{E}^{\mathbf {R}}(U_{i})=C_{E}^{\mathbf {R}^{*}}(U_{i})$ for any $U_{i}\in \mathcal {U}$. Note that $C_{E}^{\mathbf {R}}=C_{E,CH}$. Thus, it follows from (A) that

$$\begin{aligned} EI_{C_{E}^{\mathbf {R}}}[C_{E}^{\mathbf {R}},\mathcal {U}]&=\sum _{U_{j}\in \mathcal {U}}C_{E,CH}(U_{j})\mathfrak {I}_{\mathcal {U}}(C_{E}^{\mathbf {R}},V_{j})\nonumber \\&=\sum _{U_{j}\in \mathcal {U}}C_{E}(U_{j})\mathfrak {I}_{\mathcal {U}}(C_{E}^{\mathbf {R}},V_{j})=EI_{C_{E}}[C_{E}^{\mathbf {R}},\mathcal {U}]. \end{aligned}$$

(5b)

Similarly, we also have that:

$$\begin{aligned} EI_{C_{E}^{\mathbf {R}}}[C_{E}^{\mathbf {R}^{*}},\mathcal {U}]&=EI_{C_{E}}[C_{E}^{\mathbf {R}^{*}},\mathcal {U}]. \end{aligned}$$

(5c)

Now, (5a), (5b), and (5c) jointly imply that

$$\begin{aligned} EI_{C_{E}}[C_{E}^{\mathbf {R}},\mathcal {U}]\le EI_{C_{E}}[C_{E}^{\mathbf {R}^{*}},\mathcal {U}], \end{aligned}$$

where equality holds if and only if $C_{E}^{\mathbf {R}}(U_{i})=C_{E}^{\mathbf {R}^{*}}(U_{i})$ for any $U_{i}\in \mathcal {U}$. In a similar way to the proof of Proposition 2, on the other hand, we can prove that the following two propositions are equivalent to each other: (i) $C_{E}^{\mathbf {R}}(U_{i})=C_{E}^{\mathbf {R}^{*}}(U_{i})$ for any $U_{i}\in \mathcal {U}$; (ii) $C_{E}^{\mathbf {R}}=C_{E}^{\mathbf {R}^{*}}$. Finally, we have that

$$\begin{aligned} EI_{C_{E}}[C_{E}^{\mathbf {R}},\mathcal {U}]\le EI_{C_{E}}[C_{E}^{\mathbf {R}^{*}},\mathcal {U}], \end{aligned}$$

where equality holds if and only if $C_{E}^{\mathbf {R}}=C_{E}^{\mathbf {R}^{*}}$. Done.

A proof of Proposition 6

We can prove Proposition 6 in the very similar way to the proof of Proposition 5. Let $\mathbb {R}_{\mathcal {E}}$ be a set of updating rules on $\mathcal {E}$ for a chance-free credence function C. All members of $\mathbb {R}_{\mathcal {E}}$ correct C to a chance-fed function such that $C_{E_{i}}^{\mathbf {R}}$ satisfies the Principal Principle and assigns $E_{i}$ to 1, for any $\mathbf {R}_{\mathcal {E}}\in \mathbb {R}_{\mathcal {E}}$ and $E_{i}\in \mathcal {E}$. Let $\mathbf {R}_{\mathcal {E}}$ be the Conditionalizing-first Rule on $\mathcal {E}$ for C such that $C_{E_{i}}^{\mathbf {R}}=C_{E_{i},CH}$ for any $E_{i}\in \mathcal {E}$. Lastly, let $\mathbf {R}_{\mathcal {E}}^{*}$ be an updating rule in $\mathbb {R}_{\mathcal {E}}$. Then, Immodesty implies that: for any $E_{i}\in \mathcal {E}$,

$$\begin{aligned} EI_{C_{E_{i}}^{\mathbf {R}}}[C_{E_{i}}^{\mathbf {R}},\mathcal {U}]\le EI_{C_{E_{i}}^{\mathbf {R}}}[C_{E_{i}}^{\mathbf {R}^{*}},\mathcal {U}], \end{aligned}$$

(6a)

where equality holds if and only if $C_{E_{_{i}}}^{\mathbf {R}}(U_{j})=C_{E_{i}}^{\mathbf {R}^{*}}(U_{j})$ for any $U_{j}\in \mathcal {U}$, which is equivalent to $C_{E_{i}}^{\mathbf {R}}=C_{E_{i}}^{\mathbf {R}^{*}}$. Note that $C_{E_{i}}^{\mathbf {R}}=C_{E_{i},CH}$ for any $E_{i}\in \mathcal {E}$. Similar to (5b) and (5c), it follows from (A) that, for any $E_{i}\in \mathcal {E}$,

$$\begin{aligned} EI_{C_{E_{i}}^{\mathbf {R}}}[C_{E_{i}}^{\mathbf {R}},\mathcal {U}]&=\sum _{U_{j}\in \mathcal {U}}C_{E_{i},CH}(U_{j})\mathfrak {I}_{\mathcal {U}}(C_{E_{i}}^{\mathbf {R}},V_{j})\nonumber \\&=\sum _{U_{j}\in \mathcal {U}}C_{E_{i}}(U_{j})\mathfrak {I}_{\mathcal {U}}(C_{E}^{\mathbf {R}},V_{j})\nonumber \\&=\sum _{U_{j}\in \mathcal {U}}C(U_{j}|E_{i})\mathfrak {I}_{\mathcal {U}}(C_{E}^{\mathbf {R}},V_{j}). \end{aligned}$$

(6b)

By the same token, it also holds that: for any $E_{i}\in \mathcal {E}$,

$$\begin{aligned} EI_{C_{E_{i}}^{\mathbf {R}}}[C_{E_{i}}^{\mathbf {R}^{*}},\mathcal {U}]&=\sum _{U_{j}\in \mathcal {U}}C(U_{j}|E_{i})\mathfrak {I}_{\mathcal {U}}(C_{E}^{\mathbf {R}^{*}},V_{j}). \end{aligned}$$

(6c)

Then, (6b) and the relevant definitions imply that:

$$\begin{aligned} EI_{C}[\mathbf {R}_{\mathcal {E}},\mathcal {U}]&=\sum _{E_{i}\in \mathcal {E},U_{j}\in \mathcal {U}}C(E_{i}U_{j})\mathfrak {I}_{\mathcal {U}}(C_{E_{i}}^{\mathbf {R}},V_{j})\nonumber \\&=\sum _{E_{i}\in \mathcal {E}}\left( \sum _{U_{j}\in \mathcal {U}}C(E_{i}U_{j})\mathfrak {I}_{\mathcal {U}}(C_{E_{i}}^{\mathbf {R}},V_{j})\right) \nonumber \\&=\sum _{E_{i}\in \mathcal {E}}C(E_{i})\left( \sum _{U_{j}\in \mathcal {U}}C(U_{j}|E_{i})\mathfrak {I}_{\mathcal {U}}(C_{E_{i}}^{\mathbf {R}},V_{j})\right) \nonumber \\&=\sum _{E_{i}\in \mathcal {E}}C(E_{i})EI_{C_{E_{i}}^{\mathbf {R}}}[C_{E_{i}}^{\mathbf {R}},\mathcal {U}]. \end{aligned}$$

(6d)

Similarly, it follows from (6c) and the relevant definition that:

$$\begin{aligned} EI_{C}[\mathbf {R}_{\mathcal {E}}^{*},\mathcal {U}]=\sum _{E_{i}\in \mathcal {E}}C(E_{i})EI_{C_{E_{i}}^{\mathbf {R}}}[C_{E_{i}}^{\mathbf {R}^{*}},\mathcal {U}]. \end{aligned}$$

(6e)

Now, (6a), (6d), and (6e) imply that:

$$\begin{aligned} EI_{C}[\mathbf {R}_{\mathcal {E}},\mathcal {U}]\le EI_{C}[\mathbf {R}_{\mathcal {E}}^{*},\mathcal {U}], \end{aligned}$$

where equality holds if and only if $\mathbf {R}_{\mathcal {E}}=\mathbf {R}_{\mathcal {E}}^{*}$. Done.

1.8 Correcting chance-free credences in accordance with the new principle

As mentioned in Sect. 2, the original version of the Principal Principle suffers from the Big Bad Bug—however, the New Principle does not. That said, my discussions go through even if we accept the New Principle rather than the original version. In particular, we can derive the new versions of ACR0 and ACR+, which require us to correct in accordance with the New Principle. Let C and $C_{E}$ be chance-free credence functions, and $C_{CH}$ and $C_{E,CH}$ be chance-fed credence functions that are updated from C and $C_{E}$, respectively, in accordance with the New Principle. Then, we can formulate such versions, as follows:

• New ACR0: $C_{CH}(A)=\sum _{U_{i}\in \mathcal {U}}C(U_{i})ch_{i}(A|U_{i})$, for any $A\subseteq \mathcal {W}$.

• New ACR+: $C_{E,CH}(A)=\sum _{U_{i}\in \mathcal {U}}C_{E}(U_{i})ch_{i}(A|EU_{i})$, for any $A\subseteq \mathcal {W}$.

Note that these versions follow, in a similar way to ACR0 and ACR+, from GAC and the New Principle. In particular, we can derive these versions by replacing the unconditional chance function $ch_{i}(\cdot )$ in ACR0 and ACR+ with the conditional chance function $ch_{i}(\cdot |U_{i})$. In a similar way, moreover, the Propositions corresponding to Propositions 1–6 also follow from New ACR0 and New ACR+. Therefore, we can say that my discussions do not depend on which version of the Principal Principle we accept.