Evidence and the epistemic betterness

Abstract

It seems intuitive that our credal states are improved if we obtain evidence favoring truth over any falsehood. In this regard, Fallis and Lewis have recently provided and discussed some formal versions of such an intuition, which they name ‘the Monotonicity Principle’ and ‘Elimination’. They argue, with those principles in hand, that the Brier rule, one of the most popular rules of accuracy, is not a good measure, and that accuracy-firsters cannot underwrite both probabilism and conditionalization. In this paper, I will argue that their conclusions are somewhat hasty. Specifically, I will demonstrate that there is another version of the Monotonicity Principle that can be satisfied by some additive rules of accuracy, such as the Brier rule. Moreover, it will also be argued that their version of the principle has some undesirable features regarding the epistemic betterness. Therefore, their criticisms can hardly jeopardize accuracy-firsters until any further justification of their versions of the Monotonicity Principle and Elimination is provided.

Appendices

Appendix I: Proofs of the results in Table 1

‘OK’ in Table 1 means that the corresponding accuracy measure satisfies the relevant version of the Monotonicty Principle. On the other hand, ‘NG’ in the table means that the corresponding measure violates the principle in question. Some proofs have already been provided in Fallis and Lewis’s papers of 2016 and 2021. In particular, they have proved what measure of the aforementioned accuracy measures satisfy Monotonicity$^\pi$. However, they have not demonstrated the results related to Monotonicity$^\delta$. In what follows, I will provide such proofs. In particular, I will use some concrete examples to show NGs, and give some mathematical proofs of OKs. All credence functions in what follows are assumed to be coherent and defined over a partition ${\mathbb {H}}=\{H_1, \ldots , H_n \}$.

1.1 Examples showing NGs

The following examples show that neither $\mathfrak {S}^S$ (Example A1) nor $\mathfrak {S}^A$ (Example A2) satisfies Monotonicity$^\delta$.

Example A1

Suppose that $\textbf{s}=(0.7,0.2,0.1)$ and $\textbf{r}=(0.75,0.25,0)$. Then, $\delta _{1}^{{\textbf {s}},{\textbf {r}}}=\delta _{2}^{\textbf{s},\textbf{r}}=0.05$ and $\delta _{3}^{\textbf{s},\textbf{r}}=-0.1$. So, $\delta _{1}^{\textbf{s},\textbf{r}}\ge \delta _{i}^{\textbf{s},\textbf{r}}$ for any i. However, $\mathfrak {S}^S_{1}(\textbf{s})=0.953>0.949=\mathfrak {S}^S_{1}(\textbf{r})$.

Example A2

Suppose that $\textbf{s}=(0.15,0.35,0.35,0.15)$ and $\textbf{r}=(0.2,0.4,0.4,0)$. Then, $\delta _{1}^{\textbf{s},\textbf{r}}=\delta _{2}^{\textbf{s},\textbf{r}}=\delta _{3}^{\textbf{s},\textbf{r}}=0.05$, and $\delta _{4}^{\textbf{s},\textbf{r}}=-0.15$ and so $\delta _{1}^{\textbf{s},\textbf{r}}\ge \delta _{i}^{\textbf{s},\textbf{r}}$ for any i. However, $\mathfrak {S}^A_{1}(\textbf{s})=2.920>2.907=\mathfrak {S}^A_{1}(\textbf{r})$.

1.2 A Proof that $\mathfrak {B}^S$ and $\mathfrak {B}^A$ satisfy monotonicity$^\delta$

Suppose that the antecedent of Monotonicity$^\delta$—that is, $r_{k}-s_{k}=\delta _{k}\ge \delta _{i}=r_{i}-s_{i}$ for any i. (Here I use $\delta _{i}$ rather than $\delta _{i}^{\textbf{s},\textbf{r}}$, for the sake of notational simplicity.) Then, it holds that:

$$\begin{aligned} \mathfrak {B}^S_{k}(\textbf{r})= & {} 2r_{k}-\sum _{i}\left( r_{i}\right) ^{2}\\= & {} 2(s_{k}+\delta _{k})-\sum _{i}(s_{i}+\delta _{i})^{2}\\= & {} \left( 2s_{k}-\sum _{i}s_{i}^{2}\right) +\left( \delta _{k}-\sum _{i}\delta _{i}(\delta _{i}+s_{i})\right) +\left( \delta _{k}-\sum _{i}\delta _{i}s_{i}\right) \\= & {} \mathfrak {B}^S_{k}(\textbf{s})+\left( \delta _{k}-\sum _{i}\delta _{i}r_{i}\right) +\left( \delta _{k}-\sum _{i}\delta _{i}s_{i}\right) . \end{aligned}$$

Note that $\sum _{i}r_{i}=\sum _{i}s_{i}=1$ and $\delta _{k}\ge \delta _{i}$ for any i. Thus, it holds that $\delta _{k}=\sum _{i}\delta _{k}r_{i}\ge \sum _{i}\delta _{i}r_{i}$ and $\delta _{k}=\sum _{i}\delta _{k}s_{i}\ge \sum _{i}\delta _{i}s_{i}$, which implies that $\mathfrak {B}^S_{k}(\textbf{r})\ge \mathfrak {B}^S_{k}(\textbf{s})$. Hence, $\mathfrak {B}^S$ satisfies Monotonicity$^\delta$. As explained, $\mathfrak {B}^S$ is ordinally equivalent to $\mathfrak {B}^A$, and therefore we can also conclude that $\mathfrak {B}^A$ satisfies Monotonicity$^\delta$. $\square$

1.3 A proof that $\mathfrak {L}^S$ and $\mathfrak {L}^A$ satisfy monotonicity$^\delta$

It is very straightforward that $\mathfrak {L}^S$ satisfies Monotonicity$^\delta$. Suppose that $\delta _{i}^{\textbf{s},\textbf{r}}\le \delta _{k}^{\textbf{s},\textbf{r}}$ for any i. Then, it holds that $r_{k}-s_{k}>0$, and so $\mathfrak {L}^S_k(\textbf{r})=\ln {r_k}>\ln {s_k}=\mathfrak {L}^S_k(\textbf{s})$, as Monotonicity$^\delta$ requires. However, it is not easy to show that $\mathfrak {L}^A$ satisfies Monotonicity$^\delta$. In what follows, I will prove this, and provide some relevant examples.

For this purpose, I will show that:

$$\begin{aligned} \delta ^{\textbf{s},\textbf{r}}_{n}\le 0 \le \delta ^{\textbf{s},\textbf{r}}_{n-1} \le \cdots \le \delta ^{\textbf{s},\textbf{r}}_{2}\le \delta ^{\textbf{s},\textbf{r}}_{1} \end{aligned}$$

(*)

entails that $\mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {L}^A_{1}(\textbf{r})$. Note that $\sum _{i} \delta ^{\textbf{s},\textbf{r}}_{i}=0$ since $\textbf{r}$ is coherent. Then, it should hold that, if some $\delta ^{\textbf{s},\textbf{r}}$s are positive, then some other $\delta ^{\textbf{s},\textbf{r}}$s should be negative. In this regard, (*) says that at most one $\delta ^{\textbf{s},\textbf{r}}_i$ has a negative value. However, this assumption does not damage the generality of the proof that follows. This is because a similar proof, mutatis mutandis, can be provided for cases where two or more $\delta ^{\textbf{s},\textbf{r}}_i$s are negative. (Below, I will explain this using Example A7.) Moreover, it is also noteworthy that (*) assumes that $\delta ^{\textbf{s},\textbf{r}}_i$s are weakly decreasing—that is, $\delta ^{\textbf{s},\textbf{r}}_{i+1}\le \delta ^{\textbf{s},\textbf{r}}_{i}$ for any $i=1,\ldots ,n-1$. However, this assumption does not undermine the generality of my proof, either. This is because any two credence function $\textbf{s}$ and $\textbf{r}$, whose $\delta ^{\textbf{s},\textbf{r}}_i$s are not weakly decreasing, can be transformed into two functions $\textbf{s}^\prime$ and $\textbf{r}^\prime$, respectively, so that $\delta ^{\textbf{s}^\prime ,\textbf{r}^\prime }_i$s are weakly decreasing, without any change in their epistemic utilities. Suppose, for instance, that $\textbf{s}=(0.10,0.30,0.20,0.40)$ and $\textbf{r}=(0.15,0.30, 0.25,0.30)$. Note that $\delta ^{\textbf{s},\textbf{r}}_{3}=0.05>0=\delta ^{\textbf{s},\textbf{r}}_{2}$ and so $\delta ^{\textbf{s},\textbf{r}}_i$s are not weakly decreasing. However, when $\textbf{s}$ and $\textbf{r}$, respectively, are transformed into $\textbf{s}^\prime =(0.10,0.20,0.30,0.40)$ and $\textbf{r}^\prime =(0.15,0.3,0.25,0.30)$, $\delta ^{\textbf{s}^\prime ,\textbf{r}^\prime }_i$s become weakly decreasing while $\mathfrak {L}^A_{1}(\textbf{s})=\mathfrak {L}^A_{1}(\textbf{s}^\prime )$ and $\mathfrak {L}^A_{1}(\textbf{r})=\mathfrak {L}^A_{1}(\textbf{r}^\prime )$. It can be said, as a result, that (*) does not undermine the generality of my proof.

Anyway, I will prove in what follows that (*) entails that $\mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {L}^A_{1}(\textbf{r})$. In particular, I will suggest a way of constructing a sequence of ${\textbf {s}}_1, \ldots , {\textbf {s}}_{n-2}$ such that

$$\begin{aligned} \mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {L}^A_{1}(\textbf{s}_1) \le \cdots \le \mathfrak {L}^A_{1}(\textbf{s}_{n-2}) \le \mathfrak {L}^A_{1}(\textbf{r}), \end{aligned}$$

(**)

under the assumption of (*).

Let me begin with proving the following lemma.

Lemma A3

Suppose that $\textbf{c}=(c_{1},\ldots ,c_{n})$ is a coherent credence function such that $c_{1} >0$ and $c_{i}<1$ for any $i\ne 1$. Then,

$$\begin{aligned} \frac{1}{c_{1}}-\sum _{i=2}^{k}\frac{1}{1-c_{i}}+k>0, \end{aligned}$$

where $k=2,\ldots ,n$.

Proof

Suppose that $\textbf{c}$ is coherent, and that $c_{1} > 0$, and $c_{i}<1$ for any $i\ne 1$. Then, we have that:

$$\begin{aligned} \frac{1}{c_{1}}= & {} 1+\left( 1-c_{1}\right) +\left( 1-c_{1}\right) ^{2} +\cdots =\sum _{j=0}^{\infty }\left( 1-c_{1}\right) ^{j}; \text { and}\\ \frac{1}{1-c_{i}}= & {} 1+\left( c_{i}\right) +\left( c_{i}\right) ^{2} +\cdots =\sum _{j=0}^{\infty }\left( c_{i}\right) ^{j}, \text {for any}\, i\ne 1. \end{aligned}$$

Moreover, it holds that $\left( 1-c_{1}\right) ^{j}=\left( \sum _{i=2}^{n}c_{i}\right) ^{j}\ge \sum _{i=2}^{k}\left( c_{i}\right) ^{j}$ for any natural number j. Then, it follows from the above mathematical facts that

$$\begin{aligned} \sum _{i=2}^{k}\frac{1}{1-c_{i}}= & {} \sum _{i=2}^{k}\sum _{j=0}^{\infty }\left( c_{i}\right) ^{j} =\sum _{j=0}^{\infty }\sum _{i=2}^{k}\left( c_{i}\right) ^{j} = \sum _{i=2}^{k}\left( c_{i}\right) ^{0}+\sum _{j=1}^{\infty }\sum _{i=2}^{k}\left( c_{i}\right) ^{j}\\\le & {} \left( k-1\right) +\sum _{j=1}^{\infty }\left( 1-c_{1}\right) ^{j}=\left( k-1\right) +\frac{1-c_1}{c_1}\\< & {} k+\frac{1}{c_{1}}, \end{aligned}$$

as required. $\square$

Now, I will suggest a particular way of constructing a credence function that is epistemically better relative to $\mathfrak {L}^A$ than ${\textbf {s}}$ at the world where $H_1$ is true. The credence functions so constructed will consist of ${\textbf {s}}_1, \ldots , {\textbf {s}}_{n-2}$ satisfying (**) under the assumption of (*). The following Theorem A4, which can be proved with help of Lemma A3, specifies such a way.

Theorem A4

Suppose that $\textbf{s}=(s_{1},\ldots ,s_{n})$ is a coherent credence function. Let’s define a credence function $\textbf{s}_{x,k}$ as follows:

$$\begin{aligned} \textbf{s}_{x,k}=(s_{1}+x,\ldots ,s_{k}+x,s_{k+1},\ldots ,s_{n-1},s_{n}-kx), \end{aligned}$$

where x is a real number, and k is a natural number such that $1\le k \le n-1$ and $0\le x\le s_{n}/k$. Then, $\mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {L}^A_{1}(\textbf{s}_{x,k})$ for any $x\in [0,s_{n}/k]$.

Proof

It is not hard to find that this theorem is true when $\textbf{s}$ is maximally opinionated-that is, $s_{i}=1$ for some i. On the one hand, when $s_{i}=1$ for some $i<n$, x should be zero and so the theorem is trivially true. On the other hand, when $s_{n}=1$, it holds that $\mathfrak {L}^A_{1}(\textbf{s})=\ln {0}+\sum _{i=2}^{n-1}\ln (1-0)+\ln {(1-1)}=-\infty$, and so the theorem is true. Moreover, the theorem is also trivially true when $s_{n}=0$.

Now, consider the cases in which $\textbf{s}$ is not maximally opinionated and $s_{n}\ne 0$. Note that $\mathfrak {L}^A_{1}(\textbf{s}_{x,k})$ can be regarded as a function of x. Let f be such a function. That is,

$$\begin{aligned} f(x)=\ln \left( s_{1}+x\right) +\sum _{i=2}^{k}\ln \left( 1-s_{i}-x\right) +\sum _{i=k+1}^{n-1}\ln \left( 1-s_{i}\right) +\ln \left( 1-s_{n}+kx\right) . \end{aligned}$$

This function is continuous and continuously differentiable on $[0,s_{n}/k]$. Then, we obtain the following equations:

$$\begin{aligned} f^{\prime }(x)= & {} \frac{1}{s_{1}+x}-\sum _{i=2}^{k}\frac{1}{1-s_{i}-x}+\frac{k}{1-s_{n}+kx}.\\ f^{\prime \prime }(x)= & {} -\frac{1}{\left( s_{1}+x\right) ^{2}}-\sum _{i=2}^{k} \frac{1}{\left( 1-s_{i}-x\right) ^{2}}-\frac{k^{2}}{\left( 1-s_{i}+kx\right) ^{2}}. \end{aligned}$$

Note that $f^{\prime \prime }(x)<0$ for any $x\in [0,s_{n}/k]$, and so $f^{\prime }(x)$ is a decreasing function of $x\in [0,s_{n}/k]$. So, if

then it is guaranteed that f(x) is an increasing function of $x\in [0,s_{n}/k]$, and hence that, for any $x\in [0,s_{n}/k]$, $f(0)=\mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {L}^A_{1}(\textbf{s}_{x,k})=f(x)$, which is the conclusion that we want to reach.

To prove ($\dagger$), let me first define another credence function $\textbf{c}=(c_{1},\ldots ,c_{n})$, as follows:

$$\begin{aligned} c_{i}={\left\{ \begin{array}{ll} s_{i}+s_{n}/k &{} \text { if }1\le i\le k;\\ s_{i} &{} \text { if }k<i\le n-1;\\ 0 &{} \text { if }i=n. \end{array}\right. } \end{aligned}$$

Note that $\textbf{c}$ is coherent, and that $c_{1} >0$ and $c_{i}<1$ for any $i\ne 1$. (Recall that we assume that $\textbf{s}$ is not maximally opinionated. So, $s_i <1$ for any i.) Then, we obtain, with the help of Lemma A3, that:

$$\begin{aligned} f^{\prime }(s_{n}/k)= & {} \frac{1}{s_{1}+s_{n}/k}-\sum _{i=2}^{k}\frac{1}{1-s_{i}-s_{n}/k}+k.\\= & {} \frac{1}{c_{1}}-\sum _{i=2}^{k}\frac{1}{1-c_{i}}+k>0, \end{aligned}$$

where $k=2,\ldots ,n-1$. Hence, it can be said that, when $k=2,\ldots ,n-1$, the decreasing function $f^{\prime }$ on $[0,s_{n}/k]$ has a positive minimum value, and so f(x)—i.e., $\mathfrak {L}^A_{1}(\textbf{s}_{x,k})$—is an increasing function of $x\in [0,s_{n}/k]$. Therefore, we have that $\mathfrak {L}^A_{1}(\textbf{s}_{x,k})\ge \mathfrak {L}^A_{1}(\textbf{s})$ for any $x\in [0,s_{n}/k]$ when $k=2,\ldots ,n-1$.

What about the case in which $k=1$? It is easy to find that $\mathfrak {L}^A_{1}(\textbf{s}_{x,1})\ge \mathfrak {L}^A_{1}(\textbf{s})$ for any $x\in [0,s_{n}]$. What is called ‘Truth-directedness’ entails that $\textbf{s}_{x,1}=(s_{1}+x,s_{2},\ldots ,s_{n-1},s_{n}-x)$ is epistemically better than $\textbf{s}=(s_{1},s_{2},\ldots ,s_{n-1},s_{n})$ at the world where $H_1$ is true. Note that, while $\textbf{s}_{x,1}$ is closer to the truth hypothesis $H_1$ than ${\textbf {s}}$, $\textbf{s}_{x,1}$ is not closer to any false hypotheses than ${\textbf {s}}$. As a result, we have that $\mathfrak {L}^A_{1}(\textbf{s}_{x,k})\ge \mathfrak {L}^A_{1}(\textbf{s})$ for any $x\in [0,s_{n}/k]$ when $k=1,\ldots ,n-1$, as required. $\square$

Using this theorem, we can prove our main result—that is, $\mathfrak {L}^A$ satisfies Monotonicity$^\delta$. For the proof, it may be helpful to consider a concrete example.

Example A5

Suppose that $\textbf{s}=(0.10,0.20,0.30,0.40)$ and $\textbf{r}=(0.25,0.30,0.33,0.12)$. Note that $0\le \delta ^{\textbf{s},\textbf{r}}_{i}\le \delta ^{\textbf{s},\textbf{r}}_{1}$ for any $i<4$, and $\delta ^{\textbf{s},\textbf{r}}_{4}<0$. Regarding these functions, we can provide two other credence functions $\textbf{s}^{1}$ and $\textbf{s}^{2}$, as follows:

$$\begin{aligned} \textbf{s}= & {} (0.10,0.20,0.30,0.40)\\ \textbf{s}^{1}=\textbf{s}_{0.03,3}= & {} (0.13,0.23,\underline{0.33},0.31)\\ \textbf{s}^{2}=\textbf{s}_{0.07,2}^{1}= & {} (0.20,\underline{0.30},\underline{0.33},0.17)\\ \textbf{s}^{3}=\textbf{s}_{0.05,1}^{2}= & {} (\underline{0.25},\underline{0.30},\underline{0.33},0.12)=\textbf{r} \end{aligned}$$

Here, a credence function $\textbf{s}^{i}_{x,k}$ is generated from the corresponding credence function $\textbf{s}^i$ in accordance with the definition in Theorem A2. Then, it can be said, with the help of the theorem, that $\mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {L}^A_{1}(\textbf{s}^{1})\le \mathfrak {L}^A_{1} (\textbf{s}^{2})\le \mathfrak {L}^A_{1}(\textbf{r})$, as required by Monotonicity$^\delta$.

In a similar way to this example, we can prove the main result, which is formulated as follow.

Theorem A6

Suppose that $\textbf{s}=(s_{1},\ldots ,s_{n})$ and $\textbf{r}=(r_{1},\ldots ,r_{n})$ are coherent credence functions. Suppose also that $\delta ^{\textbf{s},\textbf{r}}_{n}\le 0 \le \delta ^{\textbf{s},\textbf{r}}_{n-1}\le \cdots \le \delta ^{\textbf{s},\textbf{r}}_{2}\le \delta ^{\textbf{s},\textbf{r}}_{1}$. Then, $\mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {\mathfrak {L}^A}_{1}(\textbf{r})$.

Proof

I will use ‘$\delta _i$’ instead of $\delta ^{\textbf{s},\textbf{r}}_{i}$’ for notational simplicity. Suppose that two credence functions $\textbf{s}$ and $\textbf{r}$ satisfy all assumptions of this theorem. Let’s define recursively a credence function $\textbf{s}^{i}$, as follows:

• $\textbf{s}^{1}=\textbf{s}_{\delta _{n-1},n-1}$;

• $\textbf{s}^{i+1}=\textbf{s}_{\delta _{n-i-1}-\delta _{n-i},n-i-1}^{i}$ for $i=1,\ldots ,n-2$.

Here, $\textbf{s}^i_{x,k}$ is generated from the corresponding credence function $\textbf{s}^i$ in accordance with the definition in Theorem A4. Then, it follows from Theorem A4 that:

Here, it can also be shown that $\textbf{s}^{n-1}=\textbf{r}$. In particular, it holds that: for any i,

$$\begin{aligned} {\textbf {s}}^{n-1}(H_i )=s_{i}^{n-1}&= \cdots =s_{i}^{n-i}=\left( \delta _{i}-\delta _{i+1}\right) +s_{i}^{n-i-1}\\&= \left( \delta _{i}-\delta _{i+1}\right) +\left( \delta _{i+1}-\delta _{i+2}\right) +s_{i}^{n-i-2}\\&= \cdots \\&= \left( \delta _{i}-\delta _{i+1}\right) +\left( \delta _{i+1}-\delta _{i+2}\right) +\cdots +\left( \delta _{n-2}-\delta _{n-1}\right) +s_{i}^{1}\\&= \left( \delta _{i}-\delta _{i+1}\right) +\left( \delta _{i+1}-\delta _{i+2}\right) +\cdots +\left( \delta _{n-2}-\delta _{n-1}\right) +\delta _{n-1}+s_{i}\\&= \delta _{i}+s_{i} = r_{i}={\textbf {r}}(H_i ). \end{aligned}$$

To understand, it may be useful to pay attention to the underlined numbers in Example A5, where $n=4$ and it holds, for instance, that:

$$\begin{aligned} {\textbf {s}}^3 (H_2 )&=s_{2}^{3}= s_{2}^{2}=(\delta _2 - \delta _3 )+s_2^1\\&=(\delta _2 - \delta _3 )+\delta _3+s_{3}=(0.10-0.03)+0.03+0.20\\&=0.10+0.20=\delta _{2}+s_{2} = r_{2}={\textbf {r}}(H_2 ) \end{aligned}$$

Therefore, we obtain that $\textbf{s}^{n-1}=\textbf{r}$, and so it follows from ($\ddagger$) that $\mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {L}^A_{1}(\textbf{r})$, as required. $\square$

As mentioned, Theorem A6 assumes that there is at most one $\delta ^{\textbf{s},\textbf{r}}_i$ that has a negative value. As shown in the following example, however, this assumption does not undermine the generality of Theorem A6 and its proof.

Example A7

Suppose that $\textbf{s}=(0.10,0.20,0.30,0.40)$ and $\textbf{r}=(0.40,0.30,0.20,0.10)$. Note that there are two negative $\delta ^{\textbf{s},\textbf{r}}_i$s—that is, $\delta ^{\textbf{s},\textbf{r}}_3 = -0.1$ and $\delta ^{\textbf{s},\textbf{r}}_4= -0.3$. Be that as it may, we can provide two credence functions $\textbf{s}^{1}$ and $\textbf{s}^{2}$ such that $\mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {L}^A_{1} (\textbf{s}^{1})\le \mathfrak {L}^A_{1}(\textbf{s}^{2})\le \mathfrak {L}^A_{1}(\textbf{r})$, as follows:

$$\begin{aligned} \textbf{s}^{1}=\textbf{s}_{0.10,2}= & {} (0.20,0.30,0.30,0.20)\\ \textbf{s}^{2}=\textbf{s}_{0.10,1}^{1}= & {} (0.30,0.30,0.30,0.10). \end{aligned}$$

Note first that Theorem A4 ensures that $\mathfrak {L}^A_{1}(\textbf{s})\le \mathfrak {L}^A_{1}(\textbf{s}^{1})\le \mathfrak {L}^A_{1}(\textbf{s}^{2})$. Now, consider the following two credence functions:

$$\begin{aligned} \textbf{s}^{2*}= & {} (0.30,0.30,0.10,0.30)\\ \textbf{s}^{3*}=\textbf{s}_{0.10,1}^{2*}= & {} (0.40,0.30,0.10,0.20). \end{aligned}$$

Here, $\textbf{s}^{2*}$ is the credence function generated from $\textbf{s}^2$ by switching $s^2_3(=0.30)$ with $s^2_4(=0.10)$. According to Theorem A4, it holds that $\mathfrak {L}^A_{1}(\textbf{s}^{2*})\le \mathfrak {L}^A_{1}(\textbf{s}^{3*})$. On the other hand, the definition of the additive Logarithmic rule entails that $\mathfrak {L}^A_{1}(\textbf{s}^2)=\mathfrak {L}^A_{1}(\textbf{s}^{2*})$ and $\mathfrak {L}^A_{1}(\textbf{s}^{3*})=\mathfrak {L}^A_{1}(\textbf{r})$. Therefore, we have that $\mathfrak {L}^A_{1}(\textbf{s}^{2})\le \mathfrak {L}^A_{1}(\textbf{r})$.

This example clearly shows that there is a way of proving, without the assumption in question, that $\mathfrak {S}^A$ satisfies Monotonicity$^\delta$.

Appendix II: Proofs regarding weak monotonicity and weak elimination

In this appendix, I will prove some propositions that are relevant to the discussions in Sect. 4. Especially, it is demonstrated here that Monotonicity$^\delta$ entails Weak Monotonicity, and that all of the strictly proper accuracy measures satisfy Weak Elimination.

1.1 A Proof that monotonicity$^\delta$ entails weak monotonicity

Suppose that $\textbf{s}$ and $\textbf{r}$ are coherent credence functions over a partition ${\mathbb {H}}=\left\{ H_{1},\ldots ,H_{n}\right\}$. For our purpose, it is sufficient to prove that: for any i,

$$\begin{aligned} \pi _{i}^{\textbf{s},\textbf{r}} = r_i /s_i&\le r_k/s_k=\pi _{k}^{\textbf{s},\textbf{r}},\text { and } \end{aligned}$$

(a)

$$\begin{aligned} \bar{\pi }_{k}^{\textbf{s},\textbf{r}}=(1-r_k )/(1-s_k )&\le (1-r_i )/(1-s_i )=\bar{\pi }_{i}^{\textbf{s},\textbf{r}} \end{aligned}$$

(b)

entails $\delta _{k}^{\textbf{s},\textbf{r}} \ge \delta _{i}^{\textbf{s},\textbf{r}}.$ Note that (b) entails that:

$$\begin{aligned} s_k - r_k \ge s_i -r_i +(r_k s_i -r_i s_k ). \end{aligned}$$

Thus, we have that $s_k - r_k = \delta _{k}^{\textbf{s},\textbf{r}} \ge \delta _{i}^{\textbf{s},\textbf{r}} = s_i -r_i$ since (a) says that $r_k s_i -r_i s_k \ge 0$. This reasoning holds for any i. Therefore, it can be concluded that Monotonicity$^\delta$ entails Weak Monotonicity. On the other hand, if at least one of (a) and (b) does not follow from the condition that $\delta _{1}^{\textbf{s},\textbf{r}} \ge \delta _{i}^{\textbf{s},\textbf{r}}$ for any i, then it can be said that the converse does not hold. Suppose that ${\textbf {s}}=(0.7,0.2,0.1)$ and ${\textbf {r}}=(0.75,0.25,0)$. Then, we obtain that $\delta _{1}^{\textbf{s},\textbf{r}} \ge \delta _{i}^{\textbf{s},\textbf{r}},$ for any i. However, we have that $\pi _{1}^{\textbf{s},\textbf{r}}=15/14 < 5/4=\pi _{i}^{\textbf{s},\textbf{r}}$—that is, (a) does not hold. Hence, we can conclude that the converse in question does not hold. $\square$

1.2 A Proof that the strictly proper accuracy measures satisfy weak elimination

Suppose that $\textbf{s}$ and $\textbf{r}$ are coherent credence functions over a partition ${\mathbb {H}}=\left\{ H_{1},\ldots ,H_{n}\right\}$. Suppose also that $s_{n}>0=r_{n}$, and

$$\begin{aligned} s_{i}/s_{j}&=r_{i}/r_{j};\text { and } \end{aligned}$$

(1)

$$\begin{aligned} (1-s_{i})/(1-s_{j})&=(1-r_{i})/(1-r_{j}) \end{aligned}$$

(2)

for any i, j ($\ne n$). It is the case that $n>2$. If not, ${\textbf {r}}$ cannot be coherent. Similarly, it cannot be the case that $r_i =0$ for any $i(\ne n)$, since ${\textbf {r}}$ is coherent. Note that (1) entails that there is a real number $\pi$ such that $r_{i}=\pi s_{i}$ for any $i(\ne n)$. Similarly, it follows from (2) that there is a real number $\bar{\pi }$ such that $(1-r_{i})=\bar{\pi } (1-s_{i})$ for any $i(\ne 1)$. And, these consequences entail that $s_{i}=(1-\bar{\pi })/(1-\pi )$ for any $i(\ne n)$. As a result, we have that $s_{i}=s_{j}$ for any i, j ($\ne n$), which says that the old credences are evenly distributed over the hypotheses, except for the hypothesis $H_{n}$.

Suppose now that $\mathfrak {A}$ is a strictly proper accuracy measure. From the above result, it follows that there is a real number r such that $r_{i}=r$ for any $i(\ne n)$. This is because $r_{i}=\pi s_{i}$ for any $i(\ne n)$, and $s_{i}=s_{j}$ for any i, j ($\ne n$). As a result, it holds that $\mathfrak {A}_{i}(\textbf{s})=\mathfrak {A}_{j}(\textbf{s})$ and $\mathfrak {A}_{i}(\textbf{r})=\mathfrak {A}_{j}(\textbf{r})$ for any i and j ($\ne n$). Therefore, we have that: for any $k(\ne n)$,

$$\begin{aligned} \sum _{i}r_{i}\mathfrak {A}_{i}(\textbf{r})&=\sum _{i\ne n}r_{i}\mathfrak {A}_{i}(\textbf{r}) +r_{n}\mathfrak {A}_{n}(\textbf{r})=(n-1) r \mathfrak {A}_{k}(\textbf{r}) +0\cdot \mathfrak {A}_{n}(\textbf{r})=(n-1) r \mathfrak {A}_{k}(\textbf{r});\\ \sum _{i}r_{i}\mathfrak {A}_{i}(\textbf{s})&=\sum _{i\ne n}r_{i}\mathfrak {A}_{i}(\textbf{s}) +r_{n}\mathfrak {A}_{n}(\textbf{s})=(n-1) r \mathfrak {A}_{k}(\textbf{s}) +0\cdot \mathfrak {A}_{n}(\textbf{s})=(n-1) r \mathfrak {A}_{k}(\textbf{s}). \end{aligned}$$

As assumed, $\mathfrak {A}$ is strictly proper. Hence, we have that: for any $k(\ne n)$,

$$\begin{aligned} (n-1) r \mathfrak {A}_{k}(\textbf{r})=\sum _{i}r_{i}\mathfrak {A}_{i}(\textbf{r})>\sum _{i}r_{i}\mathfrak {A}_{i}(\textbf{s})=(n-1) r \mathfrak {A}_{k}(\textbf{s}). \end{aligned}$$

As mentioned, $n\ne 1$ and $r\ne 0$. Therefore, we obtain that $\mathfrak {A}_k (\textbf{s})<\mathfrak {A}_k (\textbf{r})$ for any $k(\ne n)$. $\square$