Confirmation and the generalized Nagel–Schaffner model of reduction: a Bayesian analysis

Abstract

In their 2010 (Erkenntnis 73:393–412) paper, Dizadji-Bahmani, Frigg, and Hartmann (henceforth ‘DFH’) argue that the generalized version of the Nagel–Schaffner model that they have developed (henceforth ‘the GNS’) is the right one for intertheoretic reduction, i.e. the kind of reduction that involves theories with largely overlapping domains of application. Drawing on the GNS, DFH (Synthese 179:321–338, 2011) presented a Bayesian analysis of the confirmatory relation between the reducing theory and the reduced theory and argued that, post-reduction, evidence confirming the reducing theory also confirms the reduced theory and evidence confirming the reduced theory also confirms the reducing theory, which meets the expectations one has about theories with largely overlapping domains. In this paper, I argue that the Bayesian analysis presented by DFH (Synthese 179:321–338, 2011) faces difficulties. In particular, I argue that the conditional probabilities that DFH introduce to model the bridge law entail consequences that run against the GNS. However, I also argue that, given slight modifications of the analysis that are in agreement with the GNS, one is able to account for these difficulties and, moreover, one is able to more rigorously analyse the confirmatory relation between the reducing and the reduced theory.

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Notes

  1. 1.

    Note that on the GNS one theory reduces to the other in virtue of empirical propositions (i.e. laws that a theory has). However, proponents of the GNS do not commit themselves to the view that “a theory just is [DFH’s italics] a set of laws, i.e. \(\mathcal {\mathbf {T_{_{A}}}}\) is not identified with \(\mathcal{T}_{_{A}}\)” (DFH 2011, p. 323).

  2. 2.

    An example of the bridge law can be found in Sect. 2.2. Another example of the bridge law is \(\mathbf {V}=\mathbf {E}\), where \(\mathbf {V}\) is the light vector from the physical optical theory of light and \(\mathbf {E}\) is the electric force vector from the theory of electromagnetic radiation. This bridge law is used to derive a number of laws of the physical optical theory of light from Maxwell’s equations (see Schaffner 2012, pp. 551–559).

  3. 3.

    In presenting the derivation, I closely follow Feynman et al.  (1964, chapter 39). In parts, I also rely on DFH (2010, pp. 395–396), Dizadji-Bahmani (2011, pp. 31–33, 130–138), Greiner et al.  (1997, pp. 6–11), and Pauli (1973, pp. 94–103).

  4. 4.

    Feynman et al.  (1964, chapter 39) additionally provide a more comprehensive argument for why the mean kinetic energies of the two gases ought to be equal using only the concepts from the kinetic theory of gases and the definition of equilibrium. For the purposes of this paper, however, we need not go into such detail.

  5. 5.

    It is worth pointing out that in both entity identification and property identification we simplify our previously held ontology not by eliminating unnecessary entities or properties of the reduced theory (for instance, eliminating light waves and temperature), but rather by assimilating these entities and properties via identification to the corresponding entities and properties of the reducing theory. So, there are still light waves in the world, but instead of two classes of entities—light waves and electromagnetic waves—there is only one (see Sklar 1967, p. 121, 1993, pp. 361–362).

  6. 6.

    For surveys on Bayesianism see Háyek and Hartmann (2010) and Hartmann and Sprenger (2011). For a critical discussion of Bayesianism see Earman (1992).

  7. 7.

    For an introduction to Bayesian networks see Pearl (1988), Neapolitan (2003), Bovens and Hartmann (2003, pp. 67ff.), DFH (2011, p. 325).

  8. 8.

    Throughout the article, random variables in the network are binary; that is, some random variable A (denoted by italicized letters) can take two values A or \(\lnot \)A (denoted by non-italicized letters).

  9. 9.

    Bayes’ Theorem: \(P(\mathrm {H}\mid \mathrm { E})=\frac{P(\mathrm {E}\mid \mathrm {H})P(\mathrm {H})}{P(\mathrm {E})}=\frac{P(\mathrm {E}\mid \mathrm {H})P(\mathrm {H})}{P(\mathrm {E}\mid \mathrm { H})P(\mathrm {H})+P(\mathrm {E}\mid \lnot \mathrm { H})P(\lnot \mathrm { H})}\).

  10. 10.

    ’ encodes the information that A and B are conditionally independent given C. By definition, A and B are conditionally independent given C, i.e. , if and only if \(P(\mathrm {A}\mid \text { B, C})=P(\mathrm {A}\mid \mathrm {C})\).

  11. 11.

    d-separation is a property of Bayesian networks by which one can track down all the independences (conditional and unconditional ones) in the Bayesian network: if and only if A and B are d-separated by \(\{C\}\). Two nodes A and B are d-separated by \(\{C\}\) if all the paths in the network between A and B are blocked by \(\{C\}\). For more details on d-separation see Neapolitan (2003, pp. 70ff.).

  12. 12.

    A number of authors support the claim that the reduction relation should be asymmetric: Kuipers (1982), Sarkar (2015, p. 47), Riel (2013), Riel and Gulick (2016, p. 18).

  13. 13.

    Although a, in principle, can take any value in the open interval (0, 1), it seems more plausible that it assumes a rather low value since we do not expect to often find that \(\mathrm {T}^*_\mathrm {P}\) holds and that \(\lnot \mathrm { T}^*_\mathrm {F}\) or \(\lnot \mathrm { B}\) hold.

  14. 14.

    Interestingly, but perhaps unsurprisingly, one can show that in DFH’s original analysis the two theorems hold in exactly the same form (see Theorem 7’ and Theorem 8’ in “Appendix”).

  15. 15.

    Here I use the difference measure d as the measure of degree of confirmation of a hypothesis (\(\mathrm {H}\)) by evidence (\(\mathrm {E}\)): \(d(\mathrm {H,E}):=P(\mathrm {H}\mid \mathrm { E})-P(\mathrm {H})\) (cf. Fitelson 1999, p. 363; Hartmann and Sprenger, 2011, p. 613).

  16. 16.

    The situation does not differ much given somewhat different value assignments for a, b, and \(t_F\).

  17. 17.

    The expression ‘s-g axiom’  stands for semi-graphoid axiom. For more details on semi-graphoid axioms see Pearl (1988, pp. 84ff).

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Acknowledgements

I would like to thank Stephan Hartmann, Benjamin Eva, and anonymous reviewers for helpful and constructive comments that greatly improved the manuscript. I would also like to thank audiences in Lisbon, at the Third Lisbon International Conference on Philosophy of Science and in Dubrovnik, at the Formal Methods and Philosophy II Conference.

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Correspondence to Marko Tešić.

Appendix

Appendix

To show: and entail \(P_1(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {F})=P_1(\mathrm {T}_\mathrm {P})\)Footnote 17

figurea

\(\square \)

Similarly, we get that and entail \(P_1(\mathrm {T}_\mathrm {F}\mid \mathrm {E}_\mathrm {P})=P_1(\mathrm {T}_\mathrm {F})\).

To show: \(P_1(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {F})=P_1(\mathrm {T}_\mathrm {P})\)by d-separation

There is only one possible path between \(E_F\) and \(T_P\), namely \(E_F-T_F-E-T_P\), which is blocked at E by \(\varnothing \). Therefore, . By the definition of independence this translates into \(P_1(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {F})=P_1(\mathrm {T}_\mathrm {P})\). \(\square \)

Similarly, we get that \(P_1(\mathrm {T}_\mathrm {F}\mid \mathrm { E}_\mathrm {P})=P_1(\mathrm {T}_\mathrm {F})\) holds by d-separation before the reduction.

To show: \(P_2(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})=1\)

$$\begin{aligned} P_2(\mathrm {T}^*_\mathrm {P}\mid \lnot \mathrm { T}^*_\mathrm {F})&=0\\ P_2(\mathrm {T}^*_\mathrm {P}\mid \lnot \mathrm { T}^*_\mathrm {F})&=\dfrac{P_2(\mathrm {T}^*_\mathrm {P},\lnot \mathrm { T}^*_\mathrm {F})}{P_2(\lnot \mathrm { T}^*_\mathrm {F})}=0\\ P_2(\mathrm {T}^*_\mathrm {P},\lnot \mathrm { T}^*_\mathrm {F})&=0\\ P_2(\lnot \mathrm { T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})&=\dfrac{P_2(\mathrm {T}^*_\mathrm {P},\lnot \mathrm { T}^*_\mathrm {F})}{P_2(\mathrm {T}^*_\mathrm {P})}=0\\ P_2(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})&=1 \end{aligned}$$

\(\square \)

To show: \(P_2(\mathrm {T}^*_\mathrm {F}\mid \lnot \mathrm { T}^*_\mathrm {P})=0\)

$$\begin{aligned} P_2(\mathrm {T}^*_\mathrm {P}\mid \mathrm { T}^*_\mathrm {F})&=1\\ P_2(\lnot \mathrm { T}^*_\mathrm {P}\mid \mathrm { T}^*_\mathrm {F})&=\dfrac{P_2(\lnot \mathrm {T}^*_\mathrm {P},\mathrm {T}^*_\mathrm {F})}{P_2(\mathrm {T}^*_\mathrm {F})}=0\\ P_2(\lnot \mathrm { T}^*_\mathrm {P},\mathrm {T}^*_\mathrm {F})&=0\\ P_2(\mathrm {T}^*_\mathrm {F}\mid \lnot \mathrm { T}^*_\mathrm {P})&=\dfrac{P_2(\lnot \mathrm {T}^*_\mathrm {P},\mathrm {T}^*_\mathrm {F})}{P_2(\lnot \mathrm { T}^*_\mathrm {P})}=0 \end{aligned}$$

\(\square \)

To show: \(P_2(\mathrm {T}^*_\mathrm {F})=P_2(\mathrm {T}^*_\mathrm {P})\)

$$\begin{aligned} P_2(\mathrm {T}^*_\mathrm {P}\mid \mathrm { T}^*_\mathrm {F})&=\dfrac{P_2(\mathrm {T}^*_\mathrm {P},\mathrm {T}^*_\mathrm {F})}{P_2(\mathrm {T}^*_\mathrm {F})}=1\\ P_2(\mathrm {T}^*_\mathrm {P},\mathrm {T}^*_\mathrm {F})&=P_2(\mathrm {T}^*_\mathrm {F})\\ P_2(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})&=\dfrac{P_2(\mathrm {T}^*_\mathrm {P},\mathrm {T}^*_\mathrm {F})}{P_2(\mathrm {T}^*_\mathrm {P})}=\dfrac{P_2(\mathrm {T}^*_\mathrm {F})}{P_2(\mathrm {T}^*_\mathrm {P})}=1\\ P_2(\mathrm {T}^*_\mathrm {F})&=P_2(\mathrm {T}^*_\mathrm {P}) \end{aligned}$$

\(\square \)

I adopt the following convention: \(\overline{z}:=1-z\).

Theorem 3\(\ E_\mathrm {F}\) confirms \(T_\mathrm {P}\) iff \((p_F - q_F) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)>0\).

Proof

$$\begin{aligned} P_3(\mathrm {T}_\mathrm {P}\vert \mathrm { E}_\mathrm {F})&= \dfrac{P_3(\mathrm {T}_\mathrm {P}, \mathrm { E}_\mathrm {F})}{P_3(\mathrm {E}_\mathrm {F})}\\ P_3(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {F})&= \sum _{T^*_P,T^*_F,T_F} P_3(\mathrm {T}_\mathrm {P}\mid T^*_P)P_3(T^*_P\mid T^*_F)P_3(T^*_F\mid T_F)P_3(T_F)\\&\quad \cdot P_3(\mathrm {E}_\mathrm {F}\mid T_F)\\&= p_F \, t_F \, (p^*_P \, p^*_F + a \, p^*_P \, \overline{p^*_F} + \overline{a} \, q^*_P \, \overline{p^*_F})\\&\quad + q_F \, \overline{t_F} \, (p^*_P \, q^*_F + a \, p^*_P \, \overline{q^*_F} + \overline{a} \, q^*_P \, \overline{q^*_F})\\ P_3(\mathrm {E}_\mathrm {F})&= \sum _{T_F} P_3(\mathrm {E}_\mathrm {F}\mid T_F)P_3(T_F)\\&= p_F \, t_F + q_F \, \overline{t_F}\\ P_3(\mathrm {T}_\mathrm {P})&= \sum _{T^*_P,T^*_F,T_F} P_3(\mathrm {T}_\mathrm {P}\mid T^*_P)P_3(T^*_P\mid T^*_F)P_3(T^*_F\mid T_F)P_3(T_F)\\&= t_F \, (p^*_P \, p^*_F + a \, p^*_P \, \overline{p^*_F} + \overline{a} \, q^*_P \, \overline{p^*_F})\\&\quad + \overline{t_F} \, (p^*_P \, q^*_F + a \, p^*_P \, \overline{q^*_F} + \overline{a} \, q^*_P \, \overline{q^*_F})\\ P_3(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {F})&-P_3(\mathrm {T}_\mathrm {P}) = \dfrac{\overline{a} \, t_F \, \overline{t_F} \, (p_F - q_F) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)}{p_F \, t_F + q_F \, \overline{t_F}} \end{aligned}$$

\(\square \)

Theorem 4\(E_\mathrm {P}\) confirms \(T_\mathrm {F}\) iff \((p_P - q_P) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)>0\).

Proof

$$\begin{aligned} P_3(\mathrm {T}_\mathrm {F}\mid \mathrm {E}_\mathrm {P})&=\dfrac{P_3(\mathrm {T}_\mathrm {F},\mathrm {E}_\mathrm {P})}{P_3(\mathrm {E}_\mathrm {P})}\\ P_3(\mathrm {T}_\mathrm {F},\mathrm {E}_\mathrm {P})&=P_3(\mathrm {T}_\mathrm {F})\sum _{T^*_P,T^*_F,T_P} P_3(\mathrm {E}_\mathrm {P}\mid T_P)P_3(T_P\mid T^*_P)P_3(T^*_P\mid T^*_F)\\&\quad \cdot P_3(T^*_F\mid \mathrm {T}_\mathrm {F})\\&=t_F \, \left[ (p^*_F + a \, \overline{p^*_F}) \, (p_P \, p^*_P + q_P \, \overline{p^*_P}) + \overline{a} \, \overline{p^*_F} \, (p_P \, q^*_P + q_P \, \overline{q^*_P})\right] \\ P_3(\mathrm {E}_\mathrm {P})&=\sum _{T^*_P,T^*_F,T_P,T_F} P_3(\mathrm {E}_\mathrm {P}\mid T_P)P_3(T_P\mid T^*_P)P_3(T^*_P\mid T^*_F)P_3(T^*_F\mid T_F)\\&\quad \cdot P_3(T_F)\\&=\left( t_F \, (p^*_F + a \, \overline{p^*_F}) + \overline{t_F} \, (q^*_F + a \, \overline{q^*_F})\right) \, (p_P \, p^*_P + q_P \, \overline{p^*_P})\\&\quad + \overline{a} \, (\overline{p^*_F} \, t_F + \overline{q^*_F} \, \overline{t_F}) \, (p_P \, q^*_P + q_P \, \overline{q^*_P})\\ P_3(\mathrm {T}_\mathrm {F})&=t_F\\ P_3(\mathrm {T}_\mathrm {F}\mid \mathrm { E}_\mathrm {P})&-P_3(\mathrm {T}_\mathrm {F})=\dfrac{\overline{a} \, t_F \, \overline{t_F} \, (p_P - q_P) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)}{P_3(\mathrm {E}_\mathrm {P})} \end{aligned}$$

\(\square \)

To show: \(0<P_3(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})<1\)

$$\begin{aligned} P_3(\mathrm {T}^*_\mathrm {P}\mid \mathrm { T}^*_\mathrm {F})&=\dfrac{P_3(\mathrm {T}^*_\mathrm {P},\mathrm {T}^*_\mathrm {F})}{P_3(\mathrm {T}^*_\mathrm {F})}=1\\ P_3(\mathrm {T}^*_\mathrm {P},\mathrm {T}^*_\mathrm {F})&=P_3(\mathrm {T}^*_\mathrm {F})\\ P_3(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})&=\dfrac{P_3(\mathrm {T}^*_\mathrm {P},\mathrm {T}^*_\mathrm {F})}{P_3(\mathrm {T}^*_\mathrm {P)}}=\dfrac{P_3(\mathrm {T}^*_\mathrm {F})}{P_3(\mathrm {T}^*_\mathrm {P})}\\ P_3(\mathrm {T}^*_\mathrm {P})&=\sum _{T^*_F} P_3(\mathrm {T}^*_\mathrm {P}\mid T^*_F)P_3(T^*_F)\\&=P_3(\mathrm {T}^*_\mathrm {F})+aP_3(\lnot \mathrm {T}^*_\mathrm {F})=P_3(\mathrm {T}^*_\mathrm {F}) \, \overline{a}+a\\ P_3(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})&=\dfrac{P_3(\mathrm {T}^*_\mathrm {F})}{P_3(\mathrm {T}^*_\mathrm {F}) \, \overline{a}+a} \end{aligned}$$

Suppose \(P_3(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})=0\), then

$$\begin{aligned} \dfrac{P_3(\mathrm {T}^*_\mathrm {F})}{P_3(\mathrm {T}^*_\mathrm {F}) \, \overline{a}+a}&= 0\\ P_3(\mathrm {T}^*_\mathrm {F})&= 0 \end{aligned}$$

But \(P_3(\mathrm {T}^*_\mathrm {F})\) cannot be equal to 0, since by assumption all probabilities are within the open interval (0, 1) (except for the conditional ones that encode logical consequence).

Suppose \(P_3(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})=1\), then

$$\begin{aligned} \dfrac{P_3(\mathrm {T}^*_\mathrm {F})}{P_3(\mathrm {T}^*_\mathrm {F}) \, \overline{a}+a}&= 1\\ P_3(\mathrm {T}^*_\mathrm {F})&= P_3(\mathrm {T}^*_\mathrm {F}) \, \overline{a}+a\\ P_3(\mathrm {T}^*_\mathrm {F}) - P_3(\mathrm {T}^*_\mathrm {F}) \, \overline{a}&= a\\ P_3(\mathrm {T}^*_\mathrm {F}) \, a&= a\\ P_3(\mathrm {T}^*_\mathrm {F})&= 1 \end{aligned}$$

But \(P_3(\mathrm {T}^*_\mathrm {F})\) cannot be equal to 1, for the reason mentioned above.

Hence,

$$\begin{aligned} 0<P_3(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P})<1 \end{aligned}$$

\(\square \)

To show: \(P_3(\mathrm {T}^*_\mathrm {P})>P_3(\mathrm {T}^*_\mathrm {F})\)

$$\begin{aligned} P_3(\mathrm {T}^*_\mathrm {P})&=\sum _{T^*_F} P_3(\mathrm {T}^*_\mathrm {P}\mid T^*_F)P_3(T^*_F)\\&=P_3(\mathrm {T}^*_\mathrm {F})+a \, P_3(\lnot \mathrm {T}^*_\mathrm {F})\\ P_3(\mathrm {T}^*_\mathrm {P})-P_3(\mathrm {T}^*_\mathrm {F})&=P_3(\mathrm {T}^*_\mathrm {F})+a \, P_3(\lnot \mathrm {T}^*_\mathrm {F})-P_3(\mathrm {T}^*_\mathrm {F})\\&=a \, P_3(\lnot \mathrm {T}^*_\mathrm {F})>0 \end{aligned}$$

\(\square \)

Of \(P_5(\mathrm {T}^*_\mathrm {P}\mid \lnot \mathrm { T}^*_\mathrm {F},\mathrm {B})\), \(P_5(\mathrm {T}^*_\mathrm {P}\mid \mathrm {T}^*_\mathrm {F},\lnot \mathrm { B})\), and \(P_5(\mathrm {T}^*_\mathrm {P}\mid \lnot \mathrm {T}^*_\mathrm {F},\lnot \mathrm { B})\), as the most plausible candidate for assigning the value 0 is \(P_5(\mathrm {T}^*_\mathrm {P}\mid \lnot \mathrm { T}^*_\mathrm {F},\mathrm {B})\), since, one could say, given true bridge laws and false \(T^*_F\) (i.e. \(\lnot \mathrm { T}^*_\mathrm {F}\)), theory \(T^*_P\) should not come out as true. As for \(P_5(\mathrm {T}^*_\mathrm {P}\mid \mathrm { T}^*_\mathrm {F},\lnot \mathrm { B})\) and \(P_5(\mathrm {T}^*_\mathrm {P}\mid \lnot \mathrm { T}^*_\mathrm {F}, \lnot \mathrm { B})\), regard them as randomizers and assign them \(a\in (0,1)\) (see Bovens and Hartmann 2003, pp. 57ff.). However, with these probability assignments, a drawback of the original analysis recurs: \(\mathrm {T}^*_\mathrm {P}\) and \(\mathrm {B}\) entail \(\mathrm {T}^*_\mathrm {F}\). In order to show that this entailment holds, observe that in the network in Fig. 5 holds (the only two paths between \(T^*_F\) and B, i.e. \(T^*_F-T^*_P-B\) and \( T^*_F-T_F-E-T_P-T^*_P-B\), are blocked by \(\varnothing \) at \(T^*_P\) and E respectively; so, \(T^*_F\) and B are d-separated by \(\varnothing \)).

To show: \(P_5(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P},\mathrm {B})=1\)

figureb

\(\square \)

To show: \(0<P_4(\mathrm {T}^*_\mathrm {F}\mid \mathrm { T}^*_\mathrm {P},\mathrm {B})<1\)

figurec

\(\square \)

To show: \(P_4(\mathrm {T}^*_\mathrm {P})>P_4(\mathrm {T}^*_\mathrm {F})\) or \(P_4(\mathrm {T}^*_\mathrm {P})<P_4(\mathrm {T}^*_\mathrm {F})\) or \(P_4(\mathrm {T}^*_\mathrm {P})=P_4(\mathrm {T}^*_\mathrm {F})\)

$$\begin{aligned} P_4(\mathrm {T}^*_\mathrm {P})&=\sum _{T^*_F,B,T_F} P_4(\mathrm {T}^*_\mathrm {P}\mid T^*_F,B)P_4(B)P_4(T^*_F\mid T_F)P_4(T_F)\\&=t_F \, \left( b \, p^*_F + a \, \overline{b}\, p^*_F + a \, \overline{p^*_F}\right) + \overline{t_F} \, \left( b \, q^*_F + a \, \overline{b} \, q^*_F + a \, \overline{q^*_F}\right) \\ P_4(\mathrm {T}^*_\mathrm {F})&=\sum _{T_F} P_4(\mathrm {T}^*_\mathrm {F}\mid T_F)P_4(T_F)\\&=p^*_F \, t_F + q^*_F \, \overline{t_F}\\ P_4(\mathrm {T}^*_\mathrm {P})-P_4(\mathrm {T}^*_\mathrm {F})&=a-(\overline{a} \, \overline{b} + a) \, (p^*_F \, t_F + q^*_F \, \overline{t_F})\\&=a-(\overline{a} \, \overline{b} + a) \, P_4(\mathrm {T}^*_\mathrm {F}) \end{aligned}$$

So, when \(P_4(\mathrm {T}^*_\mathrm {F})=\frac{a}{\overline{a} \, \overline{b}+a}\), then \(P_4(\mathrm {T}^*_\mathrm {P})=P_4(\mathrm {T}^*_\mathrm {F})\) (note that for \(0<a<1\) and \(0<b<1\), \(0<\frac{a}{\overline{a} \, \overline{b}+a}<1\)). When \(P_4(\mathrm {T}^*_\mathrm {F})<\frac{a}{\overline{a} \, \overline{b}+a}\), then \(P_4(\mathrm {T}^*_\mathrm {P})>P_4(\mathrm {T}^*_\mathrm {F})\). When \(P_4(\mathrm {T}^*_\mathrm {F})>\frac{a}{\overline{a} \, \overline{b}+a}\), then \(P_4(\mathrm {T}^*_\mathrm {P})<P_4(\mathrm {T}^*_\mathrm {F})\).

\(\square \)

Theorem 5\(E_\mathrm {F}\) confirms \(T_\mathrm {P}\) iff \((p_F - q_F) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)>0\).

Proof

$$\begin{aligned} P_4(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {F})&=\dfrac{P_4(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {F})}{P_4(\mathrm {E}_\mathrm {F})}\\ P_4(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {F})&=\sum _{T^*_P,T^*_F,B,T_F} P_4(\mathrm {T}_\mathrm {P}\mid T^*_P)P_4(T^*_P\mid T^*_F,B)P_4(B)P_4(T^*_F\mid T_F)P_4(T_F)\\&\quad \cdot P_4(\mathrm {E}_\mathrm {F}\mid T_F)\\&=p_F \, t_F \, \left( p^*_P \, (b \, p^*_F + a \, \overline{b} + a \, b \, \overline{p^*_F}) + q^*_P \, (\overline{a} \, \overline{b} + \overline{a} \, b \, \overline{p^*_F})\right) \\&\quad + q_F \, \overline{t_F} \, \left( p^*_P \, (b \, q^*_F + a \, \overline{b} + a \, b \, \overline{q^*_F}) + q^*_P \, (\overline{a} \, \overline{b} + \overline{a} \, b \, \overline{q^*_F})\right) \\ P_4(\mathrm {E}_\mathrm {F})&=\sum _{T_F} P_4(\mathrm {E}_\mathrm {F}\mid T_F)P_4(T_F)\\&=p_F \, t_F + q_F \, \overline{t_F}\\ P_4(\mathrm {T}_\mathrm {P})&=\sum _{T^*_P,T^*_F,B,T_F} P_4(\mathrm {T}_\mathrm {P}\mid T^*_P)P_4(T^*_P\mid T^*_F,B)P_4(B)P_4(T^*_F\mid T_F)P(T_F)\\&=p^*_P \, \left( b \, p^*_F \, t_F + b \, q^*_F \, \overline{t_F} + a \, \overline{b} + a \, b \, \overline{p^*_F} \, t_F + a \, b \, \overline{q^*_F} \, \overline{t_F}\right) \\&\quad + q^*_P \, \left( \overline{a} \, \overline{b} + \overline{a} \, b \, \overline{p^*_F} \, t_F + \overline{a} \, b \, \overline{q^*_F} \, \overline{t_F}\right) \\ P_4(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {F})&-P_4(\mathrm {T}_\mathrm {P})=\dfrac{\overline{a} \, b \, t_F \, \overline{t_F} \, (p_F - q_F) \, (p^*_F - q^*_F) \, (p^*_P-q^*_P)}{p_F \, t_F + q_F \, \overline{t_F}} \end{aligned}$$

\(\square \)

Theorem 6\(E_\mathrm {P}\) confirms \(T_\mathrm {F}\) iff \((p_P - q_P) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)>0\).

Proof

$$\begin{aligned} P_4(\mathrm {T}_\mathrm {F}\mid \mathrm {E}_\mathrm {P})&=\dfrac{P_4(\mathrm {T}_\mathrm {F},\mathrm {E}_\mathrm {P})}{P_4(\mathrm {E}_\mathrm {P})}\\ P_4(\mathrm {T}_\mathrm {F},\mathrm {E}_\mathrm {P})&=P_4(\mathrm {T}_\mathrm {F})\sum _{T^*_P,T^*_F,B,T_P} P_4(\mathrm {E}_\mathrm {P}\mid T_P)P_3(T_P\mid T^*_P)P_4(T^*_P\mid T^*_F,B)P_4(B)\\&\quad \cdot P_4(T^*_F\mid \mathrm {T}_\mathrm {F})\\&=t_F \, \left[ (b \, p^*_F + a \, \overline{b} + a \, b \, \overline{p^*_F}) \, (p_P \, p^*_P + q_P \, \overline{p^*_P})\right. \\&\quad \left. +\,\, \overline{a} \, (\overline{b} + b \, \overline{p^*_F}) \, (p_P \, q^*_P + q_P \, \overline{q^*_P})\right] \\ P_4(\mathrm {E}_\mathrm {P})&=\sum _{T^*_P,T^*_F,B,T_P,T_F} P_4(\mathrm {E}_\mathrm {P}\mid T_P)P_4(T_P\mid T^*_P)P_4(T^*_P\mid T^*_F,B)P_4(B)\\&\quad \cdot P_4(T^*_F\mid T_F)P_4(T_F)\\&=(b \, p^*_F \, t_F + b \, q^*_F \, \overline{t_F} + a \, \overline{b} + a \, b \, \overline{p^*_F} \, t_F + a \, b \, \overline{q^*_F} \, \overline{t_F}) \, (p_P \, p^*_P + q_P \, \overline{p^*_P})\\&\quad +\,\, \overline{a} \, (\overline{b} + b \, \overline{p^*_F} \, t_F + b \, \overline{q^*_F} \, \overline{t_F}) \, (p_P \, q^*_P + q_P \, \overline{q^*_P})\\ P_4(\mathrm {T}_\mathrm {F})&=t_F\\ P_4(\mathrm {T}_\mathrm {F}\mid \mathrm { E}_\mathrm {P})&-P_4(\mathrm {T}_\mathrm {F})=\dfrac{\overline{a} \, b \, t_F \, \overline{t_F} \, (p_P - q_P) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)}{P_4(\mathrm {E}_\mathrm {P})} \end{aligned}$$

\(\square \)

Theorem 7\(E_\mathrm {F}\) adds to \(E_\mathrm {P}\)’s confirmation of \(T_\mathrm {P}\) iff \((p_F - q_F) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)>0\).

Proof

$$\begin{aligned} P_4(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {P},\mathrm { E}_\mathrm {F})&=\dfrac{P_4(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {P},\mathrm { E}_\mathrm {F})}{P_4(\mathrm {E}_\mathrm {P},\mathrm {E}_\mathrm {F})}\\ P_4(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {P},\mathrm { E}_\mathrm {F})&=P_4({\mathrm {E}_\mathrm {P}\mid \mathrm { T}_\mathrm {P}})\sum _{T^*_P,T^*_F,B,T_F} P_4(\mathrm {T}_\mathrm {P}\mid T^*_P)P_4(T^*_P\mid T^*_F, B)P_4(B)\\&\quad \cdot P_4(T^*_F\mid T_F)P_4(\mathrm {E}_\mathrm {F}\mid T_F)P_4(T_F)\\&=p_P\,\left[ \overline{a}\,b\,(p^*_F\,p_F\,t_F + q^*_F\,q_F\,\overline{t_F})\,(p^*_P - q^*_P)\right. \\&\quad \left. +\,\, (p_F\,t_F + q_F\,\overline{t_F})\,(a\,p^*_P + \overline{a}\,q^*_P)\right] \\ P_4(\mathrm {E}_\mathrm {P},\mathrm {E}_\mathrm {F})&=\sum _{T^*_P,T^*_F,B,T_P,T_F} P_4(\mathrm {E}_\mathrm {F}\mid T_F)P_4(T_F)P_4(\mathrm {E}_\mathrm {P}\mid T_P)P_4(T_P\mid T^*_P)\\&\quad \cdot P_4(T^*_P\mid T^*_F, B)P_4(B)P_4(T^*_F\mid T_F)\\&=p_P\,\left[ \overline{a}\,b\,(p^*_F\,p_F\,t_F + q^*_F\,q_F\,\overline{t_F})\,(p^*_P - q^*_P)\right. \\&\quad \left. +\,\, (p_F\,t_F + q_F\,\overline{t_F})\,(a\,p^*_P + \overline{a}\,q^*_P)\right] \\&\quad +\,\, q_P\,\left[ \overline{a}\,b\,(p^*_F\,p_F\,t_F + q^*_F\,q_F\,\overline{t_F})\,(\overline{p^*_P} - \overline{q^*_P})\right. \\&\quad \left. +\,\, (p_F\,t_F + q_F\,\overline{t_F})\,(a\,\overline{p^*_P} + \overline{a}\,\overline{q^*_P})\right] \\ P_4(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {P})&=\dfrac{P_4(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {P})}{P_4(\mathrm {E}_\mathrm {P})}\\ P_4(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {P})&=P_4({\mathrm {E}_\mathrm {P}\mid \mathrm {T}_\mathrm {P}})\sum _{T^*_P,T^*_F,B,T_F} P_4(\mathrm {T}_\mathrm {P}\mid T^*_P)P_4(T^*_P\mid T^*_F, B)P_4(B)\\&\quad \cdot P_4(T^*_F\mid T_F)P_4(T_F)\\&=p_P\,\left[ \overline{a}\,b\,(p^*_F\,t_F + q^*_F\,\overline{t_F})\,(p^*_P - q^*_P) + a\,p^*_P + \overline{a}\,q^*_P\right] \\ P_4(\mathrm {E}_\mathrm {P})&=p_P\,\left[ \overline{a}\,b\,(p^*_F\,t_F + q^*_F\,\overline{t_F})\,(p^*_P - q^*_P) + a\,p^*_P + \overline{a}\,q^*_P\right] \\&\quad +\,\, q_P\,\left[ \overline{a}\,b\,(p^*_F\,t_F + q^*_F\,\overline{t_F})\,(\overline{p^*_P} - \overline{q^*_P}) + a\,\overline{p^*_P} + \overline{a}\,\overline{q^*_P}\right] \\ (\text {alternative form of }&P_4(\mathrm {E}_\mathrm {P})\text { from the proof of }\mathbf{Theorem~5 })\\ P_4(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {P},\mathrm { E}_\mathrm {F}) - P_4&(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {P})=\dfrac{\overline{a}\,b\,p_P\,q_P\,t_F\,\overline{t_F}\,(p_F - q_F)\,(p^*_F - q^*_F)\,(p^*_P - q^*_P)}{P_4(\mathrm {E}_\mathrm {P},\mathrm {E}_\mathrm {F})\,P_4(\mathrm {E}_\mathrm {P})} \end{aligned}$$

\(\square \)

Theorem 8\(E_\mathrm {P}\) adds to \(E_\mathrm {F}\)’s confirmation of \(T_\mathrm {F}\) iff \((p_P - q_P) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)>0\).

Proof

$$\begin{aligned} P_4(\mathrm {T}_\mathrm {F}\mid \mathrm { E}_\mathrm {F},\mathrm { E}_\mathrm {P})&=\dfrac{P_4(\mathrm {T}_\mathrm {F},\mathrm {E}_\mathrm {F},\mathrm { E}_\mathrm {F})}{P_4(\mathrm {E}_\mathrm {F},\mathrm {E}_\mathrm {P})}\\ P_4(\mathrm {T}_\mathrm {F},\mathrm {E}_\mathrm {F},\mathrm { E}_\mathrm {P})&=P_4(\mathrm {T}_\mathrm {F})P_4({\mathrm {E}_\mathrm {F}\mid \mathrm { T}_\mathrm {F}})\sum _{T^*_P,T^*_F,B,T_P} P_4(\mathrm {E}_\mathrm {P}\mid T_P)P_4(T_P\mid T^*_P)\\&\quad \cdot P_4(T^*_P\mid T^*_F, B)P_4(B)P_4(T^*_F\mid \mathrm {T}_\mathrm {F})\\&=p_F\,t_F\,\left[ p^*_F\,\left( b\,(p_P\,p^*_P + q_P\,\overline{p^*_P})\right. \right. \\&\quad \left. \left. +\, \overline{b}\,(a\,p_P\,p^*_P + a\,q_P\,\overline{p^*_P} + \overline{a}\,p_P\,q^*_P+\overline{a}\,q_P\,\overline{q^*_P})\right) \right. \\&\quad \left. +\, \overline{p^*_F}\,(a\,p_P\,p^*_P + a\,q_P\,\overline{p^*_P} + \overline{a}\,p_P\,q^*_P+\overline{a}\,q_P\,\overline{q^*_P})\right] \\ P_4(\mathrm {E}_\mathrm {F},\mathrm {E}_\mathrm {P})&=p_F\,t_F\,\left[ p^*_F\,\left( b\,(p_P\,p^*_P + q_P\,\overline{p^*_P})\right. \right. \\&\quad \left. \left. +\, \overline{b}\,(a\,p_P\,p^*_P + a\,q_P\,\overline{p^*_P} + \overline{a}\,p_P\,q^*_P+\overline{a}\,q_P\,\overline{q^*_P})\right) \right. \\&\quad \left. +\, \overline{p^*_F}\,(a\,p_P\,p^*_P + a\,q_P\,\overline{p^*_P} + \overline{a}\,p_P\,q^*_P+\overline{a}\,q_P\,\overline{q^*_P})\right] \\&\quad +q_F\,\overline{t_F}\,\left[ q^*_F\,\left( b\,(p_P\,p^*_P + q_P\,\overline{p^*_P})\right. \right. \\&\quad \left. \left. +\, \overline{b}\,(a\,p_P\,p^*_P + a\,q_P\,\overline{p^*_P} + \overline{a}\,p_P\,q^*_P+\overline{a}\,q_P\,\overline{q^*_P})\right) \right. \\&\quad \left. +\,\, \overline{q^*_F}\,(a\,p_P\,p^*_P + a\,q_P\,\overline{p^*_P} + \overline{a}\,p_P\,q^*_P+\overline{a}\,q_P\,\overline{q^*_P})\right] \\ (\text {alternative form of }&P_4(\mathrm {E}_\mathrm {F},\mathrm {E}_\mathrm {P})\text { from the proof of }\mathbf{Theorem~7 })\\ P_4(\mathrm {T}_\mathrm {F}\mid \mathrm { E}_\mathrm {F})&=\dfrac{P_4(\mathrm {T}_\mathrm {F},\mathrm {E}_\mathrm {F})}{P_4(\mathrm {E}_\mathrm {F})}\\&=\dfrac{P_4(\mathrm {E}_\mathrm {F}\mid \mathrm { T}_\mathrm {F})P_4(\mathrm {T}_\mathrm {F})}{\sum \limits _{T_F} P_4(\mathrm {E}_\mathrm {F}\mid T_F)P_4(T_F)}\\&=\dfrac{p_F\,t_F}{p_F\,t_F + q_F\,\overline{t_F}}\\ P_4(\mathrm {T}_\mathrm {F}\mid \mathrm { E}_\mathrm {F},\mathrm { E}_\mathrm {P}) - P_4&(\mathrm {T}_\mathrm {F}\mid \mathrm { E}_\mathrm {F})=\dfrac{\overline{a}\,b\,p_F\,q_F\,t_F\,\overline{t_F}\,(p_P - q_P)\,(p^*_F - q^*_F)\,(p^*_P - q^*_P)}{P_4(\mathrm {E}_\mathrm {F},\mathrm {E}_\mathrm {P})\,P_4(\mathrm {E}_\mathrm {F})} \end{aligned}$$

\(\square \)

Theorem 7’\(E_\mathrm {F}\) adds to \(E_\mathrm {P}\)’s confirmation of \(T_\mathrm {P}\) iff \((p_F - q_F) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)>0\).

Proof

$$\begin{aligned} P_2(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {P},\mathrm { E}_\mathrm {F})&=\dfrac{P_2(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {P}, \mathrm {E}_\mathrm {F})}{P_2(\mathrm {E}_\mathrm {P},\mathrm {E}_\mathrm {F})}\\ P_2(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {P},\mathrm { E}_\mathrm {F})&=P_2({\mathrm {E}_\mathrm {P}\mid \mathrm { T}_\mathrm {P}})\sum _{T^*_P,T^*_F,T_F} P_2(\mathrm {T}_\mathrm {P}\mid T^*_P)P_2(T^*_P\mid T^*_F)\\&\quad \cdot P_2(T^*_F\mid T_F)P_2(\mathrm {E}_\mathrm {F}\mid T_F)P_2(T_F)\\&=p_P\,\left[ p_F\,t_F\,(p^*_P\,p^*_F + q^*_P\,\overline{p^*_F}) + q_F\,\overline{t_F}\,(p^*_P\,q^*_F + q^*_P\,\overline{q^*_F})\right] \\ P_2(\mathrm {E}_\mathrm {P},\mathrm {E}_\mathrm {F})&=\sum _{T^*_P,T^*_F,T_P,T_F} P_2(\mathrm {E}_\mathrm {F}\mid T_F)P_2(T_F)P_2(\mathrm {E}_\mathrm {P}\mid T_P)P_2(T_P\mid T^*_P)\\&\quad \cdot P_2(T^*_P\mid T^*_F)P_2(T^*_F\mid T_F)\\&=p_P\,\left[ p_F\,t_F\,(p^*_P\,p^*_F + q^*_P\,\overline{p^*_F}) + q_F\,\overline{t_F}\,(p^*_P\,q^*_F + q^*_P\,\overline{q^*_F})\right] \\&\quad + q_P\,\left[ p_F\,t_F\,(\overline{p^*_P}\,p^*_F + \overline{q^*_P}\,\overline{p^*_F}) + q_F\,\overline{t_F}\,(\overline{p^*_P}\,q^*_F + \overline{q^*_P}\,\overline{q^*_F})\right] \\ P_2(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {P})&=\dfrac{P_2(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {P})}{P_2(\mathrm {E}_\mathrm {P})}\\ P_2(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {P})&=P_2(\mathrm {E}_\mathrm {P}\mid \mathrm { T}_\mathrm {P})\sum _{T^*_P,T^*_F,T_F} P_2(\mathrm {T}_\mathrm {P}\mid T^*_P)P_2(T^*_P\mid T^*_F)\\&\quad \cdot P_2(T^*_F\mid T_F)P_2(T_F)\\&=p_P\,\left[ t_F\,(p^*_P\,p^*_F + q^*_P\,\overline{p^*_F}) + \overline{t_F}\,(p^*_P\,q^*_F + q^*_P\,\overline{q^*_F})\right] \\ P_2(\mathrm {E}_\mathrm {P})&=\sum _{T^*_P,T^*_F,T_P,T_F} P_2(\mathrm {E}_\mathrm {P}\mid T_P)P_2(T_P\mid T^*_P)P_2(T^*_P\mid T^*_F)\\&\quad \cdot P_2(T^*_F\mid T_F)P_2(T_F)\\&=p_P\,\left[ t_F\,(p^*_P\,p^*_F + q^*_P\,\overline{p^*_F}) + \overline{t_F}\,(p^*_P\,q^*_F + q^*_P\,\overline{q^*_F})\right] \\&\quad \cdot q_P\,\left[ t_F\,(\overline{p^*_P}\,p^*_F + \overline{q^*_P}\,\overline{p^*_F}) + \overline{t_F}\,(\overline{p^*_P}\,q^*_F + \overline{q^*_P}\,\overline{q^*_F})\right] \\ P_2(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {P},\mathrm { E}_\mathrm {F}) - P_2&(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {P})=\dfrac{p_P\,q_P\,t_F\,\overline{t_F}\,(p_F - q_F)\,(p^*_F - q^*_F)\,(p^*_P - q^*_P)}{P_2(\mathrm {E}_\mathrm {P},\mathrm {E}_\mathrm {F})\,P_2(\mathrm {E}_\mathrm {P})} \end{aligned}$$

\(\square \)

Theorem 8’\(E_\mathrm {P}\) adds to \(E_\mathrm {F}\)’s confirmation of \(T_\mathrm {F}\) iff \((p_P - q_P) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P){>}0\).

Proof

\(\square \)

Theorem 9 Given a, \(p_F\), \(q_F\), \(p^*_F\), \(q^*_F\), \(p^*_P\), \(q^*_P\), and \(t_F\) are constant and \(p_F>q_F\), \(p^*_F>q^*_F\), and \(p^*_P>q^*_P\), if b increases (decreases), then \(d(T_\mathrm {P},E_\mathrm {F})\) increases (decreases).

Proof

From the proof of the Theorem 5 above, we have that:

$$\begin{aligned} P_4(\mathrm {T}_\mathrm {P}\mid \mathrm { E}_\mathrm {F})-P_4(\mathrm {T}_\mathrm {P})&=\dfrac{\overline{a} \, b \, t_F \, \overline{t_F} \, (p_F - q_F) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)}{p_F \, t_F + q_F \, \overline{t_F}} \end{aligned}$$

Observe that, given \(t_F\), a, \(p_F\), \(q_F\), \(p^*_F\), \(q^*_F\), \(p^*_P\), and \(q^*_P\) are constant and \(p_F>q_F\), \(p^*_F>q^*_F\), and \(p^*_P>q^*_P\), if b increases, then \(\overline{a} \, b \, t_F \, \overline{t_F} \, (p_F - q_F) \, (p^*_F - q^*_F) \, (p^*_P-q^*_P)\) increases. As the denominator, i.e. \(p_F \, t_F + q_F \, \overline{t_F}\), does not dependent on b, then if b increases, \(\frac{\overline{a} \, b \, t_F \, \overline{t_F} \, (p_F - q_F) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)}{p_F \, t_F + q_F \, \overline{t_F}}\) increases, i.e. \(d(\mathrm {T}_\mathrm {P},\mathrm {E}_\mathrm {F})\) increases.

\(\square \)

Theorem 10 Given a, \(p_P\), \(q_P\), \(p^*_F\), \(q^*_F\), \(p^*_P\), \(q^*_P\), and \(t_F\) are constant and \(p_P>q_P\), \(p^*_F>q^*_F\), and \(p^*_P>q^*_P\), if b increases (decreases), then \(d(T_\mathrm {F},E_\mathrm {P})\) increases (decreases).

Proof

From the proof of the Theorem 6 above, we have that:

$$\begin{aligned} P_4(\mathrm {T}_\mathrm {F}\mid \mathrm { E}_\mathrm {P})-P_4(\mathrm {T}_\mathrm {F})&=\dfrac{\overline{a} \, b \, t_F \, \overline{t_F} \, (p_P - q_P) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)}{P_4(\mathrm {E}_\mathrm {P})} \end{aligned}$$

Notice that, in contrast to the previous proof, the denominator, i.e. \(P_4(\mathrm {E}_\mathrm {P})\), is dependent on b; so, changing the value of b would also change the value of \(P_4(\mathrm {E}_\mathrm {P})\). Alternative way of writing \(P_4(\mathrm {E}_\mathrm {P})\) so that it better serves the purpose of this proof is:

$$\begin{aligned} P_4(\mathrm {E}_\mathrm {P})&=\overline{a} \, b \, (p^*_F \, t_F + q^*_F \, \overline{t_F}) \, (p_P - q_P) \, (p^*_P - q^*_P) + a \, (p_P \, p^*_P + q_P \, \overline{p^*_P})\\&\quad + \overline{a} \, (p_P \, q^*_P + q_P \, \overline{q^*_P}) \end{aligned}$$

Further, let us introduce the following abbreviations:

$$\begin{aligned} \mathrm {C}&:=\overline{a} \, b \, t_F \, \overline{t_F} \, (p_P - q_P) \, (p^*_F - q^*_F) \, (p^*_P - q^*_P)\\ \mathrm {D}&:=\overline{a} \, b \, (p^*_F \, t_F + q^*_F \, \overline{t_F}) \, (p_P - q_P) \, (p^*_P - q^*_P) \end{aligned}$$

Observe that both \(\mathrm {C}\) and \(P_4(\mathrm {E}_\mathrm {P})\) increase as b increases (other values remaining constant). To see which of the two, \(\mathrm {C}\) or \(P_4(\mathrm {E}_\mathrm {P})\), increases faster with the increase of b, we calculate \(\mathrm {C}-\mathrm {D}\).

$$\begin{aligned} \mathrm {C}-\mathrm {D}=-\overline{a} \, b \, (p_P - q_P) \, (p^*_P - q^*_P) \, (t_F^2 \, p^*_F + t_F \, \overline{t_F} \, q^*_F + \overline{t_F} \, q^*_F) \end{aligned}$$

So, given \(p_P>q_P\) and \(p^*_P>q^*_P\), \(\mathrm {C}<\mathrm {D}\) (as a consequence \(\mathrm {C}<P_4(\mathrm {E}_\mathrm {P})\); so \(\frac{\mathrm {C}}{P_4(\mathrm {E}_\mathrm {P})}<1\)). Hence, with the increase of b, \(P_4(\mathrm {E}_\mathrm {P})\) increases faster than \(\mathrm {C}\), that is, the slope of \(P_4(\mathrm {E}_\mathrm {P})\) is greater than the slope of \(\mathrm {C}\). Nevertheless, even with a very large slope of \(P_4(\mathrm {E}_\mathrm {P})\) and a very small slope of \(\mathrm {C}\), \(\frac{\mathrm {C}}{P_4(\mathrm {E}_\mathrm {P})}\) still increases, as shown in Fig. 7. So, if b increases, \(d(\mathrm {T}_\mathrm {F},\mathrm {E}_\mathrm {P})\) increases. \(\square \)

Fig. 7
figure7

Dependence of \(P_4(\mathrm {E}_\mathrm {P})\) (left figure) and \(\mathrm {C}\) and \(d(\mathrm {T}_\mathrm {F},\mathrm {E}_\mathrm {P})\) (right figure) on b

Theorem 11\(\varDelta _0=0\) iff (\(p^*_F=q^*_F\)) or (\(p^*_P=q^*_P\)). And \(\varDelta _0>0\) if (\(p^*_F>q^*_F\)) and if (\(p^*_P>q^*_P\)).

Proof

$$\begin{aligned} P_1(\mathrm {T}_\mathrm {F},\mathrm {T}_\mathrm {P})&=t_F \, t_P\\&=t_F \, \left[ p^*_P \, \left( b \, p^*_F \, t_F + b \, q^*_F \, \overline{t_F} + a \, \overline{b} + a \, b \, \overline{p^*_F} \, t_F + a \, b \, \overline{q^*_F} \, \overline{t_F}\right) \right. \\&\quad \left. +\,\, q^*_P \, \left( \overline{a} \, \overline{b} + \overline{a} \, b \, \overline{p^*_F} \, t_F + \overline{a} \, b \, \overline{q^*_F} \, \overline{t_F}\right) \right] (P_4(\mathrm {T}_\mathrm {P})\hbox { instead of }P_1(\mathrm {T}_\mathrm {P})\\ P_4(\mathrm {T}_\mathrm {F},\mathrm {T}_\mathrm {P})&=P_4(\mathrm {T}_\mathrm {F})\sum _{T^*_P,T^*_F,B} P_4(\mathrm {T}_\mathrm {P}\mid T^*_P)P_4(T^*_P\mid T^*_F,B)P_4(B)P_4(T^*_F\mid \mathrm {T}_\mathrm {F})\\&=t_F \, \left[ p^*_P \, (b \, p^*_F + a \, \overline{b} + a \, b \, \overline{p^*_F}) + q^*_P \, (\overline{a} \, \overline{b} + \overline{a} \, b \, \overline{p^*_F})\right] \\ \varDelta _0&:=P_4(\mathrm {T}_\mathrm {F},\mathrm {T}_\mathrm {P})-P_1(\mathrm {T}_\mathrm {F},\mathrm {T}_\mathrm {P})\\ \varDelta _0&=\overline{a} \, b \, t_F \, \overline{t_F} \, (p^*_F - q^*_F) \, (p^*_P - q^*_P) \end{aligned}$$

\(\square \)

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Tešić, M. Confirmation and the generalized Nagel–Schaffner model of reduction: a Bayesian analysis. Synthese 196, 1097–1129 (2019). https://doi.org/10.1007/s11229-017-1501-1

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Keywords

  • Confirmation
  • Nagelian reduction
  • Thermodynamics and statistical mechanics
  • Bayesian network models