On the role of explanatory and systematic power in scientific reasoning

Abstract

The paper investigates measures of explanatory power and how to define the inference schema “Inference to the Best Explanation” (IBE). It argues that these measures can also be used to quantify the systematic power of a hypothesis and defines the inference schema “Inference to the Best Systematization” (IBS). It demonstrates that systematic power is a fruitful criterion for theory choice and that IBS is truth-conducive. It also shows that even radical Bayesians must admit that systematic power is an integral component of Bayesian reasoning. Finally, the paper puts the achieved results in perspective with van Fraassen’s famous criticism of IBE.

Appendix

1.1 Proof of Theorem 1

We have to show that $ep^1$–$ep^3$ satisfy requirements 1–3.

Proof for $ep^1$

• 1. $ep^1$ satisfies Requirement 1 trivially. It is defined in terms of probabilities.

• 2. $ep^1$ satisfies Requirement 2 with marker 1:

  $$\begin{aligned} ep_{\Pr }^1(H,E)= \frac{\Pr (E|H)}{\Pr (E)}= {\left\{ \begin{array}{ll} >1, &{} \Pr (E|H )>\Pr (E)\\ =1, &{} \Pr (E|H )=\Pr (E)\\ <1, &{} \Pr (E|H )<\Pr (E) \end{array}\right. } \end{aligned}$$

• 3. $ep^1$ satisfies Requirement 3: if $\Pr (E|H_1)>\Pr (E|H_2)$, then

  $$\begin{aligned} ep_{\Pr }^1(H_1,E)=\frac{\Pr (E|H_1)}{\Pr (E)}>\frac{\Pr (E|H_2)}{\Pr (E)}=ep_{\Pr }^1(H_2,E). \end{aligned}$$

Proof for $ep^2$

• 1. $ep^2$ satisfies Requirement 1 trivially. It is defined in terms of probabilities.

• 2. $ep^2$ satisfies Requirement 2 with marker 0: First note that

  $$\begin{aligned} ep_{\Pr }^2(H,E)= \frac{\Pr (H|E)-\Pr (H|\lnot E)}{\Pr (H|E)+\Pr (H|\lnot E)}={\left\{ \begin{array}{ll} >0, &{} \Pr (H|E )>\Pr (H|\lnot E)\\ =0, &{} \Pr (H|E )=\Pr (H|\lnot E)\\ <0, &{} \Pr (H|E )<\Pr (H|\lnot E) \end{array}\right. } \end{aligned}$$

  Now we only have to see that

  $$\begin{aligned} \Pr (H|E )\begin{array}{rl} >\\ = \\ < \end{array} \Pr (H|\lnot E) \Leftrightarrow \Pr (H|E)\begin{array}{rl} >\\ = \\ < \end{array}\Pr (H)\Leftrightarrow \Pr (E|H)\begin{array}{rl} >\\ = \\ < \end{array}\Pr (E) \end{aligned}$$

• 3. $ep^2$ satisfies Requirement 3: if $\Pr (E|H_1)>\Pr (E|H_2)$, then

  • (a) $\frac{\Pr (E|H_1)}{\Pr (E)}>\frac{\Pr (E|H_2)}{\Pr (E)}$

  • (b) $\Pr (\lnot E|H_1)<\Pr (\lnot E|H_2)$ and, thus, also: $\frac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}<\frac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}$

  (a) and (b) imply that:

  $$\begin{aligned} \left[ \frac{\Pr (E|H_1)}{\Pr (E)}\times \frac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\right]- & {} \left[ \frac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\times \frac{\Pr (E|H_2)}{\Pr (E)}\right] \\> & {} \\ \left[ \frac{\Pr (E|H_2)}{\Pr (E)}\times \frac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\right]- & {} \left[ \frac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\times \frac{\Pr (E|H_1)}{\Pr (E)}\right] \end{aligned}$$

  and that therefore:

  $$\begin{aligned} \bigg [\frac{\Pr (E|H_1)}{\Pr (E)}\times \frac{\Pr (E|H_2)}{\Pr (E)}+\frac{\Pr (E|H_1)}{\Pr (E)}\times \frac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\bigg ]- & {} \bigg [\frac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\times \frac{\Pr (E|H_2)}{\Pr (E)}+\frac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\times \frac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\bigg ]\\> & {} \\ \bigg [\frac{\Pr (E|H_1)}{\Pr (E)}\times \frac{\Pr (E|H_2)}{\Pr (E)}+\frac{\Pr (E|H_2)}{\Pr (E)}\times \frac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\bigg ]- & {} \bigg [\frac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\times \frac{\Pr (E|H_1)}{\Pr (E)}+\frac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\times \frac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\bigg ] \end{aligned}$$

  and

  $$\begin{aligned} \bigg [\frac{\Pr (E|H_1)}{\Pr (E)}-\frac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\bigg ]\times & {} \bigg [\frac{\Pr (E|H_2)}{\Pr (E)}+\frac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\bigg ]\\> & {} \\ \bigg [\frac{\Pr (E|H_2)}{\Pr (E)}-\frac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\bigg ]\times & {} \bigg [\frac{\Pr (E|H_1)}{\Pr (E)}+\frac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\bigg ] \end{aligned}$$

  which implies:

  $$\begin{aligned} \frac{\bigg [\dfrac{\Pr (E|H_1)}{\Pr (E)}-\dfrac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\bigg ]}{\bigg [\dfrac{\Pr (E|H_1)}{\Pr (E)}+\dfrac{\Pr (\lnot E|H_1)}{\Pr (\lnot E)}\bigg ]} >\frac{\bigg [\dfrac{\Pr (E|H_2)}{\Pr (E)}-\dfrac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\bigg ]}{\bigg [\dfrac{\Pr (E|H_2)}{\Pr (E)}+\dfrac{\Pr (\lnot E|H_2)}{\Pr (\lnot E)}\bigg ]} \end{aligned}$$

  We can reformulate this as follows:

  $$\begin{aligned} \frac{\bigg [\dfrac{\Pr (H_1|E)}{\Pr (H_1)}-\dfrac{\Pr (H_1|\lnot E)}{\Pr (H_1)}\bigg ]}{\bigg [\dfrac{\Pr (H_1|E)}{\Pr (H_1)}+\dfrac{\Pr (H_1|\lnot E)}{\Pr (H_1)}\bigg ]} >\frac{\bigg [\dfrac{\Pr (H_2|E)}{\Pr (H_2)}-\dfrac{\Pr (H_2|\lnot E)}{\Pr (H_2)}\bigg ]}{\bigg [\dfrac{\Pr (H_2|E)}{\Pr (H_2)}+\dfrac{\Pr (H_2|\lnot E)}{\Pr (H_2)}\bigg ]} \end{aligned}$$

  Finally, by cancelling $\Pr (H_1)$, respectively $\Pr (H_2)$ out of these formulae we get the desired result:

  $$\begin{aligned} \frac{\big [\Pr (H_1|E)-\Pr (H_1|\lnot E)\big ]}{\big [\Pr (H_1|E)+\Pr (H_1|\lnot E)\big ]} >\frac{\big [\Pr (H_2|E)-\Pr (H_2|\lnot E)\big ]}{\big [\Pr (H_2|E)+\Pr (H_2|\lnot E)\big ]} \end{aligned}$$

Proof for $ep^3$

• 1. $ep^3$ satisfies Requirement 1 trivially. It is defined in terms of probabilities.

• 2. $ep^3$ satisfies Requirement 2 with marker 0: First note that

  $$\begin{aligned} ep_{\Pr }^3(H, E)= {\left\{ \begin{array}{ll}\frac{\Pr (E|H)-\Pr (E)}{1- \Pr (E)} &{} \hbox { if } \Pr (E|H )\ge \Pr (E)>0\\ \frac{\Pr (E|H)-\Pr (E)}{\Pr (E)} &{} \hbox { if } \Pr (E|H )< \Pr (E)\\ \end{array}\right. } \end{aligned}$$

  Thus,

  

• 3. $ep^2$ satisfies Requirement 3: if $\Pr (E|H_1)>\Pr (E|H_2)$, then

  • (a) $ep_{\Pr }^3(H_1,E)=\frac{\Pr (E|H_1)-\Pr (E)}{1-\Pr (E)}>ep_{\Pr }^3(H_2,E)=\frac{\Pr (E|H_2)-\Pr (E)}{1-\Pr (E)}$, if $\Pr (E|H_1)\ge \Pr (E)$ and $\Pr (E|H_2)\ge \Pr (E)$.

  • (b) $ep_{\Pr }^3(H_1,E)=\frac{\Pr (E|H_1)-\Pr (E)}{\Pr (E)}>ep_{\Pr }^3(H_2,E)=\frac{\Pr (E|H_2)-\Pr (E)}{\Pr (E)}$, if $\Pr (E|H_1)<\Pr (E)$ and $\Pr (E|H_2)<\Pr (E)$.

  • (c) $ep_{\Pr }^3(H_1,E)=\frac{\Pr (E|H_1)-\Pr (E)}{1-\Pr (E)}>ep_{\Pr }^3(H_2,E)=\frac{\Pr (E|H_2)-\Pr (E)}{\Pr (E)}$, if $\Pr (E|H_1)\ge \Pr (E)$ and $\Pr (E|H_2)<\Pr (E)$.

1.2 Proof of Theorem 2

Let W be a set of possible worlds and let $\mathcal {A}$ be some algebra over W. The elements of $\mathcal {A}$ are interpreted as propositions. Let $e_0,\ldots , e_n,\ldots$ be a sequence of propositions of $\mathcal {A}$ which separates W, and let $e^w_i =e_i$ if $w\vDash e_i$ and $\lnot e_i$ otherwise. Let $\Pr$ be a strict (or regular) probability function on $\mathcal {A}$. Let $\Pr ^*$ be the unique probability function on the smallest $\sigma$-field $\mathcal {A}^*$ containing the field $\mathcal {A}$ satisfying $\Pr ^*(A)=\Pr (A)$ for all $A\in \mathcal {A}$. Then there is a $W^\prime \subseteq W$ with $\Pr ^*(W^\prime )=1$ so that the following holds for every $w\in W^\prime$ and all hypotheses $H \in \mathcal {A}$ and for all $\mathfrak {sp}_{\Pr }$ satisfying Requirements 1–3.

Then, according to the Gaifman–Snir Theorem (1982), there is a $W^\prime \subseteq W$ with $\Pr ^*(W^\prime )=1$ so that the following holds for every $w\in W^\prime$ and all theories H of $\mathcal {A}$:

$$\begin{aligned} \lim _{n\implies \infty }\Pr (H|E^w_n)=\mathcal {I}(H,w) \end{aligned}$$

where $\mathcal {I}(H,w)=1$, if $w\vDash H$ and 0 otherwise.

• 1. Suppose $w\vDash H_1$ and $w\vDash \lnot H_2$. Then $\lim _{n\implies \infty }\Pr (H_1|E^w_n)=1$ and $\lim _{n\implies \infty }\Pr (H_2|E^w_n)=0$ which implies that $\exists n \forall m\ge n:\Pr (H_1|E^w_n)>\Pr (H_1) \& \Pr (H_2|E^w_n)<\Pr (H_2)$. The latter entails by symmetry of probabilistic relevance that $\exists n \forall m\ge n:\Pr (E^w_n|H_1)>\Pr (E^w_n) \& \Pr (E^w_n|H_2)<\Pr (E^w_n)$ and that therefore $\exists n \forall m\ge n:\Pr (E^w_n|H_1)>\Pr (E^w_n|H_2)$. Thus, with Requirement 3 on measures of explanatory and systematic power we can conclude that: $\exists n \forall m\ge n: [\mathfrak {sp}_{\Pr }(H_1, E^w_m )>\mathfrak {sp}_{\Pr }(H_2 , E^w_m )]$.

• 2. Suppose $w\vDash H_1\cap H_2$ and $H_1\vDash H_2$, but $H_2\nvDash H_1$. We already know that

  $$\begin{aligned} \lim _{n\rightarrow \infty }\left[ \frac{\Pr (H|E_n^w)}{\Pr (H)}\right] = \dfrac{1}{\Pr (H)},\hbox { if }\lim _{n\implies \infty }\Pr (H|E^w_n)=1. \end{aligned}$$

  and thus that

  $$\begin{aligned} \lim _{n\rightarrow \infty }\left[ \frac{\Pr (E_n^w|H)}{\Pr (E_n^w)}\right] = \dfrac{1}{\Pr (H)},\hbox { if }\lim _{n\implies \infty }\Pr (H|E^w_n)=1. \end{aligned}$$

  The latter implies that

  $$\begin{aligned} \lim _{n\rightarrow \infty }\left[ \frac{\Pr (E_n^w|H_1)}{\Pr (E_n^w)}\right] = \dfrac{1}{\Pr (H_1)}>\lim _{n\rightarrow \infty }\left[ \frac{\Pr (E_n^w|H_2)}{\Pr (E_n^w)}\right] = \dfrac{1}{\Pr (H_2)} \end{aligned}$$

  since $H_1\vDash H_2$ implies that $H_2\nvDash H_1$, $\Pr (H_1)<\Pr (H_2)$. This means that

  $$\begin{aligned} \lim _{n\rightarrow \infty }\left[ \Pr (E_n^w|H_1)\right] >\lim _{n\rightarrow \infty }\left[ \Pr (E_n^w|H_2)\right] \end{aligned}$$

  and that therefore $\exists n \forall m\ge n:\Pr (E^w_n|H_1)>\Pr (E^w_n|H_2)$. Thus, with Requirement 3 on measures of explanatory and systematic power we can conclude that:

  $$\begin{aligned} \exists n \forall m\ge n: [\mathfrak {sp}_{\Pr }(H_1, E^w_m )>\mathfrak {sp}_{\Pr }(H_2 , E^w_m )]. \end{aligned}$$

where $E^w_m=\bigcap _{0\le i\le m}e^w_i$.

1.3 Proof of Theorem 3

Let W be a set of possible worlds and let $\mathcal {A}$ be some algebra over W. The elements of $\mathcal {A}$ are interpreted as propositions. Let $e_0,\ldots , e_n,\ldots$ be a sequence of propositions of $\mathcal {A}$ which separates W, and let $e^w_i =e_i$ if $w\vDash e_i$ and $\lnot e_i$ otherwise. Let $\Pr$ be a strict (or regular) probability function on $\mathcal {A}$. Let $\Pr ^*$ be the unique probability function on the smallest $\sigma$-field $\mathcal {A}^*$ containing the field $\mathcal {A}$ satisfying $\Pr ^*(A)=\Pr (A)$ for all $A\in \mathcal {A}$. Then there is a $W^\prime \subseteq W$ with $\Pr ^*(W^\prime )=1$ so that the following holds for every $w\in W^\prime$ and all hypotheses $H \in \mathcal {A}$ and for all $\mathfrak {sp}_{\Pr }$ satisfying Requirements 1–3.

• 1. Suppose there is a $H_j\in \{H_1, \ldots , H_n\}$ such that $w\vDash H_j$. Then according to Theorem 2, for every false hypothesis $H_i\in \{H_1, \ldots , H_n\}$ and all for all $\mathfrak {sp}_{\Pr }$ satisfying Requirements 1–3: $\exists n \forall m\ge n: [\mathfrak {sp}_{\Pr }(H_j, E^w_m )>\mathfrak {sp}_{\Pr }(H_i , E^w_m )]$ (note Requirement 3 is the crucial requirement here). Since there are only finitely many false hypotheses in $\{H_1, \ldots , H_n\}$ we can conclude that:

  $$\begin{aligned}&\exists n \forall m\ge n \hbox { such that }\exists H_j\in \{H_1, \ldots , H_n\}\hbox { with }w\vDash H_j\hbox { and } \forall H_i\in \{H_1, \ldots , H_n\}\hbox { with }w\vDash \lnot H_i:\\&\quad [\mathfrak {sp}_{\Pr }(H_j, E^w_m )>\mathfrak {sp}_{\Pr }(H_i , E^w_m )] \end{aligned}$$

  Thus, with Definition 9 we can conclude that: $\exists n \forall m\ge n$ such that if $H_i$ is the best systematization for $E^w_m$ with respect to the set of hypotheses $\{H_1, \ldots , H_n\}$, then $w\vDash H_i$. For Definitions 10 and 11 we can show the same since according to Theorem 2 the true hypothesis will be more probable than the false ones after finitely many steps of observation and for every observation thereafter.

• 2. The proof for the second part proceeds along the same lines.

where $E^w_m=\bigcap _{0\le i\le m}e^w_i$.

1.4 Proof of Theorem 6

$$\begin{aligned}&\Pr _{t_0}(H|E)=\frac{\dfrac{e^{\tanh ^{-1}\left[ {sp^2}_{t_0}(H,E)\right] } \Pr _{t_0}(E)}{e^{\tanh ^{-1}\left[ {sp^2}_{t_0}(H,E)\right] } \Pr _{t_0}(E)+e^{\tanh ^{-1}{sp^2}_{t_0}(H,\lnot E)} \Pr _{t_0}(\lnot E)}}{\Pr _{t_0}(E)}\times \Pr _{t_0}(H)\\&\quad =\frac{\dfrac{e^{\frac{1}{2}\left[ \log \left[ {sp^2}_{t_0}(H,E)+1\right] -\log \left[ 1-{sp^2}_{t_0}(H,E)\right] \right] } \Pr _{t_0}(E)}{e^{\frac{1}{2}\left[ \log \left[ {sp^2}_{t_0}(H,E)+1\right] -\log \left[ 1-{sp^2}_{t_0}(H,E)\right] \right] } \Pr _{t_0}(E)+e^{\frac{1}{2}\left[ \log \left[ {sp^2}_{t_0}(H,\lnot E)+1\right] -\log \left[ 1-{sp^2}_{t_0}(H,\lnot E)\right] \right] } \Pr _{t_0}(\lnot E)}}{\Pr _{t_0}(E)}\\&\qquad \times \Pr _{t_0}(H)\\&\quad =\frac{\dfrac{e^{\frac{1}{2}\left[ \log \left[ \frac{{sp^2}_{t_0}(H,E)+1}{1-{sp^2}_{t_0}(H,E)}\right] \right] } \Pr _{t_0}(E)}{e^{\frac{1}{2}\left[ \log \left[ \frac{{sp^2}_{t_0}(H,E)+1}{1-{sp^2}_{t_0}(H,E)}\right] \right] } \Pr _{t_0}(E)+e^{\frac{1}{2}\left[ \log \left[ \frac{{sp^2}_{t_0}(H,\lnot E)+1}{1-{sp^2}_{t_0}(H,\lnot E)}\right] \right] } \Pr _{t_0}(\lnot E)}}{\Pr _{t_0}(E)}\times \Pr _{t_0}(H) \end{aligned}$$

Now we know that

$$\begin{aligned} \frac{\frac{\Pr _{t_0}(\lnot E|H)}{\Pr _{t_0}(\lnot E)}}{\frac{\Pr _{t_0}(E|H)}{\Pr _{t_0}(E)}}&=\frac{e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|\lnot E)}{\Pr _{t_0}(H| E)}\right] }}{e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] }}\quad (\hbox {by the definition of }e\hbox { and }\log )\\ \frac{\Pr _{t_0}(\lnot E|H)}{\Pr _{t_0}(E|H)}&=\frac{\Pr _{t_0}(\lnot E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|\lnot E)}{\Pr _{t_0}(H| E)}\right] }}{\Pr _{t_0}(E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] }}\\ \frac{\Pr _{t_0}(\lnot E|H)}{\Pr _{t_0}(E|H)}+\frac{\Pr _{t_0}(E|H)}{\Pr _{t_0}(E|H)}&=\frac{\Pr _{t_0}(\lnot E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|\lnot E)}{\Pr _{t_0}(H| E)}\right] }}{\Pr _{t_0}(E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] }}+\frac{\Pr _{t_0}(E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] }}{\Pr _{t_0}(E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] }}\\ \frac{1}{\Pr _{t_0}(E|H)}&=\frac{\Pr _{t_0}(\lnot E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|\lnot E)}{\Pr _{t_0}(H| E)}\right] }+\Pr _{t_0}(E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] }}{\Pr _{t_0}(E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] }}\\ \Pr _{t_0}(E|H)&=\frac{\Pr _{t_0}(E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] }}{\Pr _{t_0}(\lnot E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|\lnot E)}{\Pr _{t_0}(H| E)}\right] }+\Pr _{t_0}(E)\times e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] }}\\ \Pr _{t_0}(E|H)&=\frac{e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] } \Pr _{t_0}(E)}{e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|E)}{\Pr _{t_0}(H|\lnot E)}\right] } \Pr _{t_0}(E)+e^{\frac{1}{2}\log \left[ \frac{\Pr _{t_0}(H|\lnot E)}{\Pr _{t_0}(H| E)}\right] } \Pr _{t_0}(\lnot E)} \end{aligned}$$

and since $\frac{{sp^2}_{t_0}(H,E)+1}{1-{sp^2}_{t_0}(H,E)}=\frac{\Pr (H|E)}{\Pr (H|\lnot E)}$ we can conclude that

$$\begin{aligned} \Pr _{t_0}(H|E)&=\frac{\Pr _{t_0}(E|H)}{\Pr _{t_0}(E)}\times \Pr _{t_0}(H) \end{aligned}$$