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Variety of Evidence

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Abstract

Varied evidence confirms more strongly than less varied evidence, ceteris paribus. This epistemological Variety of Evidence Thesis enjoys widespread intuitive support. We put forward a novel explication of one notion of varied evidence and the Variety of Evidence Thesis within Bayesian models of scientific inference by appealing to measures of entropy. Our explication of the Variety of Evidence Thesis holds in many of our models which also pronounce on disconfirmatory and discordant evidence. We argue that our models pronounce rightly. Against a backdrop of failures of the Variety of Evidence Thesis, the intuitive case for the Variety of Evidence Thesis emerges strengthened. Our models do however not support the general case for the thesis since our explication of it fails to hold in certain cases. The parameter space of this failure is explored and an explanation for the failure is offered.

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Notes

  1. The notion of evidential variety has received a range of labels, the reader is referred to Claveau (2011, p. 249, Footnote 18). Refer to Lloyd (2015) and Vezér (2017) for recent Variety of Evidence reasoning in (the rational reconstruction of) climate science.

  2. This is consistent with the Bovens and Hartmann approach since their analysis “does not apply to unreliable instruments that do not randomize” (Bovens and Hartmann 2003, p. 95).

  3. Variety of Evidence reasoning in climate science has recently been analysed to proceed via different consequences of the hypothesis that temperatures are rising, see Vezér (2017). Consequence considered were patterns in temperature profiles in ice, rock and soil as well as the lengths of mountain glaciers and sizes of tree rings.

  4. Evidence variables which are confirmationally independent regarding the hypothesis variable and their role in confirming the hypothesis with respect to different confirmation measures are investigated in Fitelson (2001). Note that conditional independence with respect to consequence variables (our approach) is not immediately transferable to conditional independence with respect to the hypothesis variable (Fitelson).

  5. A variety of measures of entropy have been put forward. The well of measures of entropy shows no sign of drying up any time soon, see Crupi et al. (2019) and Csiszár (2008) for overviews of entropies and their axiomatic characterisations.

  6. Adopting the standard convention that \(0\cdot \log (0):=0\).

  7. See Colombo et al. (2017) for some of the latest on objectivity in science.

  8. For the record, \({\mathcal {E}}\) has a slightly greater variety score than \({\mathcal {E}}'\): \(V({\mathcal {E}})=1.3086>1.2861=V({\mathcal {E}}')\).

  9. We suspect a slip in Claveau’s in his statement of the VET. Claveau’s proofs in the appendix of his paper only deal with evidence which is in agreement with the predicted value. Furthermore, in Claveau (2011, p. 241) he writes “I take evidence to be always evidence for a specific proposition” [emphasis original]. He might have simply forgotten to state this convention in Claveau (2013). Bovens and Hartman and Earman also only consider confirmatory evidence.

  10. The case for treating two observable consequence on equal epistemological footing in (the reconstruction of) scientific inference has recently been defended in Parkkinen (2016).

  11. A rare meeting of Bayes and Popper, see also Sect. 6.2.2 for disconfirmatory evidence.

  12. There are two equivalent formulations, (1) \(P({\bar{c}}|{\bar{h}})\le P(c|h)\) and (2) \(P(c|h)+P(c|{\bar{h}})\ge 1\).

  13. Let us simply assume for a moment that Poincaré was a Bayesian.

  14. In statistical lingo, P(c|h) describes a rate of true positives (sensitivity) while \(P({\bar{c}}|{\bar{h}})\) describes a rate of true negatives (specificity). \(P(c|{\bar{h}})\ge P({\bar{c}}|h)\) may also be read as saying that an error of Type II is more or equally likely than a Type I error.

  15. It is also conjectured that the two conditions \(P(c_1|h)=P(c_2|h)\) and \(P(c_1|{\bar{h}})=P(c_2|{\bar{h}})\) can be weakened into the single condition \(P(c_1|h)/P(c_1|{\bar{h}})=P(c_2|h)/P(c_2|{\bar{h}})\) and all observations for Scenario B continue to hold.

  16. This also dovetails nicely with, as yet, our unpublished results on VET failure in the original topology of Bovens and Hartmann; Landes and Osimani (2018).

  17. It must be admitted that during writing intuitions were also driven by legal reasoning; the hypothesis that a suspect committed the crime deductively entails that the suspect had a motive, an opportunity and the means.

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Acknowledgements

I would like to thank Stephan Hartmann, Barbara Osimani, Roland Poellinger and Christian Wallmann for very helpful comments and discussions. Thanks are also due to George Pólya for teaching me about reasoning by analogy and the value of limiting cases, both of which were most helpful for devising proofs. This work is supported by the European Research Council (Philosophy of Pharmacology: Safety, Statistical standards and Evidence Amalgamation, grant 639276).

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Correspondence to Jürgen Landes.

Appendices

Appendix 1: Proofs of Main Results

We now give the longer proofs. The propositions to be proved are re-stated for ease of reference.

Proposition 1

For bodies of evidence \({\mathcal {E}},{\mathcal {E}}'\) with \(|{\mathcal {E}}|=|{\mathcal {E}}'|\) it holds that

$$\begin{aligned} V({\mathcal {E}})\ge V({\mathcal {E}}'),\,\,\,{\textit{if}\,\textit{and}\,\textit{only}\,\textit{if }}\,\,\,V_1({\mathcal {E}})\ge V_1({\mathcal {E}}'). \end{aligned}$$

Proof

Using that \(\sum _{C\in {\mathcal {C}}}|C|=|{\mathcal {E}}|\), we find

$$\begin{aligned} V({\mathcal {E}})&=-\sum _{C\in {\mathcal {C}}}\frac{|C|}{|{\mathcal {E}}|}\cdot \log \left( \frac{|C|}{|{\mathcal {E}}|}\right) \\&=-\frac{1}{|{\mathcal {E}}|}\cdot \left( \sum _{C\in {\mathcal {C}}}|C|\cdot \log \left( \frac{|C|}{|{\mathcal {E}}|}\right) \right) \\&=-\frac{1}{|{\mathcal {E}}|}\cdot \left( \sum _{C\in {\mathcal {C}}}|C|\cdot (\log (|C|)-\log (|{\mathcal {E}}|))\right) \\&=-\frac{1}{|{\mathcal {E}}|}\cdot \left( -V_1({\mathcal {E}})+\sum _{C\in {\mathcal {C}}}|C|\cdot (-\log (|{\mathcal {E}}|)\right) \\&=\frac{V_1({\mathcal {E}})}{|{\mathcal {E}}|}+\log (|{\mathcal {E}}|). \end{aligned}$$

Hence, \(V_1({\mathcal {E}})\) is a positive-slope affine-linear transformation of \(V({\mathcal {E}})\). Positive-slope affine-linear transformation preserve ordinal comparisons. \(\square \)

Proposition 4

For all finite and not-empty sets\(\Omega \)and all probability functionswon\(\Omega \), \( \vec x:=\langle w(\omega ){:}\,\omega \in \Omega ,w(\omega )\in [0,1] \& \sum _{\omega }w(\omega )=1\rangle \), it holds that\(H(\vec x)<H(\frac{|\Omega |}{|\Omega |+1}\vec x,\frac{1}{|\Omega |+1}).\)

Proof

Using that \(\sum _{i=1}^{|\Omega |}x_i=1\) and that \(H(\vec x)\le -\log (\frac{1}{|\Omega |})=\log (|\Omega |)\) we find

$$\begin{aligned}&H\left( \frac{|\Omega |}{|\Omega |+1}\vec x,\frac{1}{|\Omega |+1}\right) -H(\vec x)\\&\quad =-\frac{1}{|\Omega |+1}\left( \log \left( \frac{1}{|\Omega |+1}\right) +\sum _{i=1}^{|\Omega |}|\Omega |x_i\cdot \log \left( \frac{|\Omega |}{|\Omega |+1}x_i\right) \right) -H(\vec x)\\&\quad =-\log \left( \frac{1}{|\Omega |+1}\right) -\frac{|\Omega |}{|\Omega |+1}\cdot \left( \sum _{i=1}^{|\Omega |}x_i\cdot [\log (x_i)+\log (|\Omega |)]\right) -H(\vec x)\\&\quad =\log (|\Omega |+1)-\frac{|\Omega |}{|\Omega |+1}(-H(\vec x)+ \log (|\Omega |))-H(\vec x)\\&\quad =\log (|\Omega |+1)-\frac{|\Omega |}{|\Omega |+1}\cdot \log (|\Omega |)-\frac{1}{|\Omega |+1}H(\vec x)\\&\quad \ge \log (|\Omega |+1)-\frac{|\Omega |}{|\Omega |+1}\log (|\Omega |)-\frac{1}{|\Omega |+1}\log (|\Omega |)\\&\quad =\log (|\Omega |+1)-\log (|\Omega |)\\&\quad >0. \end{aligned}$$

\(\square \)

Theorem 1

In case of Condition A

$$\begin{aligned} {{\mathrm{sign}}}\left( P_{{\mathcal {E}}}(h)-P_{{\mathcal {E}}'}(h)\right) = {{\mathrm{sign}}}\left( P(e_{1}|{\bar{c}}_{1})-P(e_{1}|c_{1})\right) = {{\mathrm{sign}}}(1-Bf_1). \end{aligned}$$

Proof

To simply notation we let \({\vec {f}}\) denote the conjunction of all items evidence pertaining to the \(D_k\) and obtain:

$$\begin{aligned} P_{{\mathcal {E}}}(h)&=\frac{P(h{\vec {f}})}{P({\vec {f}})}=\frac{P({\vec {f}}|h)\cdot P(h)}{\sum _{s=0}^1P({\vec {f}}|h^s)\cdot P(h^s)}\\&=\frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})}{P({\vec {f}}|h)\cdot P(h)}}.\\ P_{{\mathcal {E}}'}(h)&=\frac{P(he_1{\vec {f}})}{P(e_1{\vec {f}})}=\frac{P({\vec {f}}|h)\cdot P(h)\cdot \left( \sum _{i=0}^1P(c_1^{i}|h)P(e_1|c_1^{i})\right) }{\sum _{s=0}^1P({\vec {f}}|h^s)\cdot P(h^s)\cdot \left( \sum _{i=0}^1P(c_1^{i}|h^s)P(e_1|c_1^{i})\right) } \\&=\frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot \left( \sum _{i=0}^1P(c_1^{i}|{\bar{h}})P(e_1|c_1^{i})\right) }{P({\vec {f}}|h)\cdot P(h)\cdot \left( \sum _{i=0}^1P(c_1^{i}|h)P(e_1|c_1^{i})\right) }}. \end{aligned}$$

Hence,

$$\begin{aligned}&{{\mathrm{sign}}}(P_{{\mathcal {E}}}(h)-P_{{\mathcal {E}}'}(h))\\&\quad ={{\mathrm{sign}}}\left( \frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})}{P({\vec {f}}|h)\cdot P(h)}}-\frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot \left( \sum _{i=0}^1P(c_1^{i}|{\bar{h}})P(e_1|c_1^{i})\right) }{P({\vec {f}}|h)\cdot P(h)\cdot \left( \sum _{i=0}^1P(c_1^{i}|h)P(e_1|c_1^{i})\right) }}\right) \\&\quad ={{\mathrm{sign}}}\left( \frac{\sum _{i=0}^1P(c_1^{i}|{\bar{h}})P(e_1|c_1^{i})}{\sum _{i=0}^1P(c_1^{i}|h)P(e_1|c_1^{i})}-1\right) \\&\quad ={{\mathrm{sign}}}\left( \sum _{i=0}^1P(c_1^{i}|{\bar{h}})P(e_1|c_1^{i})-\sum _{i=0}^1P(c_1^{i}|h)P(e_1|c_1^{i})\right) \\&\quad ={{\mathrm{sign}}}\left( P(e_1|c_1)\cdot [P(c_1|{\bar{h}})-P(c_1|h)]+P(e_1|{\bar{c}}_1)\cdot [P({\bar{c}}_1|{\bar{h}})-P({\bar{c}}_1|h)]\right) \\&\quad ={{\mathrm{sign}}}\left( P(e_1|c_1)\cdot [P(c_1|{\bar{h}})-P(c_1|h)]-P(e_1|{\bar{c}}_1)\cdot [P(c_1|{\bar{h}})-P(c_1|h)]\right) \\&\quad ={{\mathrm{sign}}}\left( [P(e_1|c_1)-P(e_1|{\bar{c}}_1)]\cdot [P(c_1|{\bar{h}})-P(c_1|h)]\right) . \end{aligned}$$

By (4) we have \(P(c_1|h)-P(c_1|{\bar{h}})<0\) and hence the proof is completed by noting that

$$\begin{aligned}{{\mathrm{sign}}}(P_{{\mathcal {E}}}(h)-P_{{\mathcal {E}}'}(h))={{\mathrm{sign}}}\left( P(e_1|{\bar{c}}_1)-P(e_1|c_1)\right) . \end{aligned}$$

\(\square \)

Theorem 2

Condition B and the Ceteris Paribus Conditions jointly entail that

$$\begin{aligned}&{{\mathrm{sign}}}(P_{{\mathcal {E}}}(h)-P_{{\mathcal {E}}'}(h)) \\&\quad ={{\mathrm{sign}}}\left( (1-Bf_{C_1})(Bf_{C_1-1}-Bf_{C_2})\left( Bf_{{\mathcal {E}}}-\frac{P({\bar{c}}|h)\cdot P({\bar{c}}|{\bar{h}})}{P(c|h)\cdot P(c|{\bar{h}})}\right) \right) . \end{aligned}$$

Proof

We employ the Ceteris Paribus conditions and simply write P(c|h) and \(P(c|{\bar{h}})\); dropping the subscript of c. To simplify notation we let for \(j,l\in \{0,1\}\)

$$\begin{aligned} \alpha _{jl}:=\prod _{n=1}^{|C_1|-1}P(e_n|c^j_1)\cdot \prod _{g=1}^{|C_2|}P(e_{|C_1|+g}|c^l_2)=\chi _{1j}\cdot \chi _{2l}. \end{aligned}$$

We begin by calculating the posterior probabilities in turn

$$\begin{aligned} P_{{\mathcal {E}}}(h)&=\frac{P(he_1\ldots e_{|C_1|+|C_2|}{\vec {f}})}{P(e_1\ldots e_{|C_1|+|C_2|}{\vec {f}})}\nonumber \\&=\frac{P({\vec {f}}|h)\cdot P(h)\cdot \sum _{j,l=0}^1P(c_1^jc_2^le_1\ldots e_{{|C_1|+|C_2|}}|h)}{\sum _{a=0}^1P({\vec {f}}|h^a)\cdot P(h^a)\cdot \sum _{j,l=0}^1P(c_1^jc_2^le_1\ldots e_{{|C_1|+|C_2|}}|h^a)}\nonumber \\&=\frac{P({\vec {f}}|h)\cdot P(h)\cdot \left( \sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}|c^j_1)\chi _{1j}\cdot \chi _{2l}\right) }{\sum _{a=0}^1P({\vec {f}}|h^a)\cdot P(h^a)\cdot \left( \sum _{j,l=0}^1P(c^j|h^a)P(c^l|h^a)P(e_{|C_1|}|c^j_1)\chi _{1j}\cdot \chi _{2l}\right) } \nonumber \\&=\frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot \left( \sum _{j,l=0}^1P(c^j|{\bar{h}})P(c^l|{\bar{h}})P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) }{P({\vec {f}}|h)\cdot P(h)\cdot \left( \sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) }}, \end{aligned}$$
(16)

recalling that the \(\chi \) were defined in (13). Similarly, we find for \(P_{{\mathcal {E}}'}(h)\) that

$$\begin{aligned} P_{{\mathcal {E}}'}(h)&=\frac{P'(he_1\ldots e_{|C_1|-1} e'_{|C_1|}e_{|C_1|+1}\ldots e_{|C_1|+|C_2|}{\vec {f}})}{P(e_1\ldots e_{|C_1|-1} e'_{|C_1|}e_{|C_1|+1}\ldots e_{|C_1|+|C_2|}{\vec {f}})}\nonumber \\&=\frac{P({\vec {f}}|h)\cdot P(h)\cdot \sum _{j,l=0}^1P'(c_1^jc_2^le_1\ldots e_{|C_1|-1} e'_{|C_1|}e_{|C_1|+1}\ldots e_{|C_1|+|C_2|}|h)}{\sum _{a=0}^1P({\vec {f}}|h^a) P(h^a)\sum _{j,l=0}^1P'(c_1^jc_2^le_1\ldots e_{|C_1|-1} e'_{|C_1|}e_{|C_1|+1}\ldots e_{|C_1|+|C_2|}|h^a)}\nonumber \\&=\frac{P({\vec {f}}|h)\cdot P(h)\cdot \left( \sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}\right) }{\sum _{a=0}^1P({\vec {f}}|h^a)\cdot P(h^a)\cdot \left( \sum _{j,l=0}^1P(c^j|h^a)P(c^l|h^a)P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}\right) } \nonumber \\&=\frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot \left( \sum _{j,l=0}^1P(c^j|{\bar{h}})P(c^l|{\bar{h}})P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}\right) }{P({\vec {f}}|h)\cdot P(h)\cdot \left( \sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}\right) }}. \end{aligned}$$
(17)

Note that the only difference between these two posteriors is that the first posterior contains the term \(P(e_{|C_1|}'|c^j_1)\) while the second posterior contains the term \(P(e_{|C_1|}|c^l_2)\).

The leading factors play no role, the sign of \(P_{{\mathcal {E}}}(h)-P_{{\mathcal {E}}'}(h)\) is thus equal to the sign of

$$\begin{aligned}&\frac{\sum _{j,l=0}^1P(c^j|{\bar{h}})P(c^l|{\bar{h}})P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}}{\sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}}\\&\quad -\,\frac{\sum _{j,l=0}^1P(c^j|{\bar{h}})P(c^l|{\bar{h}})P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}}{\sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}}. \end{aligned}$$

The sign of this expression is equal to the sign of

$$\begin{aligned}&\left( \sum _{j,l=0}^1P(c^j|{\bar{h}})P(c^l|{\bar{h}})P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}\right) \\&\quad \cdot \,\left( \sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) \\&\quad -\,\left( \sum _{j,l=0}^1P(c^j|{\bar{h}})P(c^l|{\bar{h}})P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) \\&\quad \cdot \,\left( \sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}\right) . \end{aligned}$$

To simplify notation we let for \(j,l\in \{0,1\}\)

$$\begin{aligned} \alpha _{jl}:=\prod _{n=1}^{|C_1|-1}P(e_n|c^j_1)\cdot \prod _{g=1}^{|C_2|}P(e_{|C_1|+g}|c^l_2)=\chi _{1j}\cdot \chi _{2l}. \end{aligned}$$

We obtain the more manageable

$$\begin{aligned}&\left( {\sum _{j,l=0}^1P(c^j|{\bar{h}})P(c^l|{\bar{h}})P(e_{|C_1|}'|c^l_2)\cdot \alpha _{jl}}\right) \cdot \left( {\sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}|c^j_1)\cdot \alpha _{jl}}\right) \\&\quad -\,\left( {\sum _{j,l=0}^1P(c^j|{\bar{h}})P(c^l|{\bar{h}})P(e_{|C_1|}|c^j_1)\cdot \alpha _{jl}}\right) \cdot \left( {\sum _{j,l=0}^1P(c^j|h)P(c^l|h)P(e_{|C_1|}'|c^l_2)\cdot \alpha _{jl}}\right) . \end{aligned}$$

Spelling this out we obtain

$$\begin{aligned}&\left(P({\bar{c}}|{\bar{h}})^2P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{00}+P(c|{\bar{h}})^2P(e_{|C_1|}'|c_2)\alpha _{11}\right. \\&\quad \left. +\,P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})\left[P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{10}+P(e_{|C_1|}'|c_2)\alpha _{01}\right]\right) \\&\quad \cdot \left( P({\bar{c}}|h)^2P(e_{|C_1|}|{\bar{c}}_1)\alpha _{00}+P(c|h)^2P(e_{|C_1|}|c_1)\alpha _{11}\right. \\&\quad \left. +\,P({\bar{c}}|h)P(c|h)\left[P(e_{|C_1|}|c_1)\alpha _{10}+P(e_{|C_1|}|{\bar{c}}_1)\alpha _{01}\right]\right) \\&\quad -\,\left( P({\bar{c}}|{\bar{h}})^2P(e_{|C_1|}|{\bar{c}}_1)\alpha _{00}+P(c|{\bar{h}})^2P(e_{|C_1|}|c_1)\alpha _{11}\right. \\&\quad \left. +\,P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})\left[P(e_{|C_1|}|c_1)\alpha _{10}+P(e_{|C_1|}|{\bar{c}}_1)\alpha _{01}\right]\right) \\&\quad \cdot \left( P({\bar{c}}|h)^2P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{00}+P(c|h)^2P(e_{|C_1|}'|c_2)\alpha _{11}\right. \\&\quad \left. +\,P({\bar{c}}|h)P(c|h)\left[P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{10}+P(e_{|C_1|}'|c_2)\alpha _{01}\right]\right). \end{aligned}$$

Fortunately, all those terms which do not contain \(\alpha _{00}\) nor \(\alpha _{11}\) (these are precisely those terms with \(P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})\)) cancel out. Furthermore, all terms which contain \(\alpha _{00}^2\) and all terms containing \(\alpha _{11}^2\) cancel out.

For the terms containing \(\alpha _{00}\) and \(\alpha _{11}\) we find

$$\begin{aligned}&\alpha _{00}\alpha _{11}\left( P({\bar{c}}|{\bar{h}})^2P(c|h)^2\left[P(e_{|C_1|}'|{\bar{c}}_2)P(e_{|C_1|}|c_1) -P(e_{|C_1|}|{\bar{c}}_1)P(e_{|C_1|}'|c_2)\right]\right. \\&\quad \left. +\,P(c|{\bar{h}})^2P({\bar{c}}|h)^2\left[P(e_{|C_1|}'|c_2)P(e_{|C_1|}|{\bar{c}}_1) -P(e_{|C_1|}|c_1)P(e_{|C_1|}'|{\bar{c}}_2)\right]\right) . \end{aligned}$$

Under the standing assumption that

$$\begin{aligned} Bf_{C_1}=\frac{P(e_{|C_1|}|c_1)}{P(e_{|C_1|}|{\bar{c}}_1)}=\frac{P(e_{|C_1|}'|c_2)}{P(e_{|C_1|}'|{\bar{c}}_2)} \end{aligned}$$

we note that also these terms cancel out.

What remains is the much more manageable

$$\begin{aligned}&\left( P({\bar{c}}|{\bar{h}})^2P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{00}+P(c|{\bar{h}})^2P(e_{|C_1|}'|c_2)\alpha _{11}\right) \\&\quad \cdot \left( P({\bar{c}}|h)P(c|h)\left[ P(e_{|C_1|}|c_1)\alpha _{10}+P(e_{|C_1|}|{\bar{c}}_1)\alpha _{01}\right] \right) \\&\quad -\,\left( P({\bar{c}}|{\bar{h}})^2P(e_{|C_1|}|{\bar{c}}_1)\alpha _{00}+P(c|{\bar{h}})^2P(e_{|C_1|}|c_1)\alpha _{11}\right) \\&\quad \cdot \,\left( P({\bar{c}}|h)P(c|h)\left[ P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{10}+P(e_{|C_1|}'|c_2)\alpha _{01}\right] \right) \\&\quad +\,\left( P({\bar{c}}|h)^2P(e_{|C_1|}|{\bar{c}}_1)\alpha _{00}+P(c|h)^2P(e_{|C_1|}|c_1)\alpha _{11}\right) \\&\quad \cdot \,\left( P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})\left[ P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{10}+P(e_{|C_1|}'|c_2)\alpha _{01}\right] \right) \\&\quad -\,\left( P({\bar{c}}|{\bar{h}})^2P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{00}+P(c|{\bar{h}})^2P(e_{|C_1|}'|c_2)\alpha _{11}\right) \\&\quad \cdot \,\left( P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})\left[ P(e_{|C_1|}|c_1)\alpha _{10}+P(e_{|C_1|}|{\bar{c}}_1)\alpha _{01}\right] \right) \\&\quad =P({\bar{c}}|h)P(c|h)\left( \left[ P({\bar{c}}|{\bar{h}})^2\alpha _{00}+P(c|{\bar{h}})^2Bf_{C_1}\alpha _{11}\right] P(e_{|C_1|}|{\bar{c}}_2)\left[ P(e_{|C_1|}|c_1)\alpha _{10}+P(e_{|C_1|}|{\bar{c}}_1)\alpha _{01}\right] \right. \\&\quad \left. -\,\left[ P({\bar{c}}|{\bar{h}})^2\alpha _{00}+P(c|{\bar{h}})^2Bf_{C_1}\alpha _{11}\right] P(e_{|C_1|}|{\bar{c}}_1)\left[ P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{10}+P(e_{|C_1|}'|c_2)\alpha _{01}\right] \right) \\&\quad +\,P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})\left( \left[ P({\bar{c}}|h)^2\alpha _{00}+P(c|h)^2Bf_{C_1}\alpha _{11}\right] P(e_{|C_1|}'|{\bar{c}}_1)\left[ P(e_{|C_1|}'|{\bar{c}}_2)\alpha_{10}+P(e_{|C_1|}'|c_2) \alpha _{01}\right] \right. \\&\quad\left. -\,\left[ P({\bar{c}}|{\bar{h}})^2\alpha_{00}+P(c|{\bar{h}})^2Bf_{C_1}\alpha _{11}\right]P(e_{|C_1|}'|{\bar{c}}_2)\left[ P(e_{|C_1|}|c_1)\alpha_{10}+P(e_{|C_1|}|{\bar{c}}_1)\alpha _{01}\right] \right) \\&\quad =P({\bar{c}}|h)P(c|h)\left[ P({\bar{c}}|{\bar{h}})^2\alpha _{00}+P(c|{\bar{h}})^2Bf_{C_1}\alpha _{11}\right] \\&\quad \left( P(e_{|C_1|}'|{\bar{c}}_2)\left[ P(e_{|C_1|}|c_1)\alpha _{10}+P(e_{|C_1|}|{\bar{c}}_1)\alpha _{01}\right] -P(e_{|C_1|}|{\bar{c}}_1)\left[ P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{10}+P(e_{|C_1|}'|c_2)\alpha _{01}\right] \right) \\&\quad +\,P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})\left[ P({\bar{c}}|h)^2\alpha _{00}+P(c|h)^2Bf_{C_1}\alpha _{11}\right] \\&\quad \left( P(e_{|C_1|}|{\bar{c}}_1)\left[ P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{10}+P(e_{|C_1|}'|c_2)\alpha _{01}\right] -P(e_{|C_1|}'|{\bar{c}}_2)\left[ P(e_{|C_1|}|c_1)\alpha _{10}+P(e_{|C_1|}|{\bar{c}}_1)\alpha _{01}\right] \right) \\&\quad =\,\left( P({\bar{c}}|h)P(c|h)\left[ P({\bar{c}}|{\bar{h}})^2\alpha _{00}+P(c|{\bar{h}})^2Bf_{C_1}\alpha _{11}\right]\right.\\ &\left.\quad-\,P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})\left[ P({\bar{c}}|h)^2\alpha _{00}+P(c|h)^2Bf_{C_1}\alpha _{11}\right] \right) \\ &\quad \cdot \,\left( P(e_{|C_1|}'|{\bar{c}}_2)\left[ P(e_{|C_1|}|c_1)\alpha _{10}+P(e_{|C_1|}|{\bar{c}}_1)\alpha _{01}\right] -P(e_{|C_1|}|{\bar{c}}_1)\left[ P(e_{|C_1|}'|{\bar{c}}_2)\alpha _{10}+P(e_{|C_1|}'|c_2)\alpha _{01}\right] \right) . \end{aligned}$$

We now investigate both factors in turn. We begin with the second and find that it is equal to

$$\begin{aligned}&P(e_{|C_1|}'|{\bar{c}}_2)P(e_{|C_1|}|{\bar{c}}_1)\left(\left[Bf_{C_1}\alpha _{10}+\alpha _{01}\right]-\left[\alpha _{10}+Bf_{C_1}\alpha _{01}\right]\right) \\&\quad=P(e_{|C_1|}'|{\bar{c}}_2)P(e_{|C_1|}|{\bar{c}}_1)(Bf_{C_1}-1)(\alpha_{10}-\alpha _{01}). \end{aligned}$$

We note that

$$\begin{aligned} \alpha _{10}-\alpha _{01}&=\prod _{n=1}^{|C_1|-1}P(e_n|c_1)\cdot \prod _{g=1}^{|C_2|}P(e_{|C_1|+g}|{\bar{c}}_2)-\prod _{n=1}^{|C_1|-1}P(e_n|{\bar{c}}_1)\cdot \prod _{g=1}^{|C_2|}P(e_{|C_1|+g}|c_2)\\&=\left( \prod _{n=1}^{|C_1|-1}P(e_n|{\bar{c}}_1)\prod _{g=1}^{|C_2|}P(e_{|C_1|+g}|{\bar{c}}_2)\right) \cdot \left( \prod _{n=1}^{|C_1|-1}\frac{P(e_n|c_1)}{P(e_n|{\bar{c}}_1)}-\prod _{g=1}^{|C_2|}\frac{P(e_{|C_1|+g}|c_2)}{P(e_{|C_1|+g}|{\bar{c}}_2)}\right) \\&=\left( \prod _{n=1}^{|C_1|-1}P(e_n|{\bar{c}}_1)\prod _{g=1}^{|C_2|}P(e_{|C_1|+g}|{\bar{c}}_2)\right) \cdot \left( Bf_{C_1-1}-Bf_{C_2}\right) . \end{aligned}$$

After some algebra we find for the negative of the first factor that

$$\begin{aligned}&\left[ P({\bar{c}}|h)^2\alpha _{00}+P(c|h)^2Bf_{C_1}\alpha _{11}\right] \cdot P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})\nonumber \\&\qquad -\,\left[ P({\bar{c}}|{\bar{h}})^2\alpha _{00}+P(c|{\bar{h}})^2Bf_{C_1}\alpha _{11}\right] \cdot P({\bar{c}}|h)P(c|h)\nonumber \\&\quad =\alpha _{00}\cdot \left[ P({\bar{c}}|h)^2\cdot P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})- P({\bar{c}}|{\bar{h}})^2\cdot P({\bar{c}}|h)P(c|h)\right] \nonumber \\&\qquad +\,Bf_{C_1}\alpha _{11}\cdot \left[ P(c|h)^2\cdot P({\bar{c}}|{\bar{h}})P(c|{\bar{h}})- P(c|{\bar{h}})^2\cdot P({\bar{c}}|h)P(c|h)\right] \nonumber \\&\quad =\alpha _{00}\cdot [P({\bar{c}}|h)P({\bar{c}}|{\bar{h}})]\cdot [P({\bar{c}}|h)P(c|{\bar{h}})- P({\bar{c}}|{\bar{h}})P(c|h)]\nonumber \\&\qquad +\,Bf_{C_1}\alpha _{11}\cdot [P(c|h)P(c|{\bar{h}})]\cdot [ P({\bar{c}}|{\bar{h}})P(c|h)- P({\bar{c}}|h)P(c|{\bar{h}})]\nonumber \\&\quad =\left[ P({\bar{c}}|{\bar{h}})P(c|h)- P({\bar{c}}|h)P(c|{\bar{h}})\right] \nonumber \\&\qquad \cdot \,\left[ Bf_{C_1}\cdot \alpha _{11}\cdot P(c|h)\cdot P(c|{\bar{h}}) - \alpha _{00}\cdot P({\bar{c}}|h)\cdot P({\bar{c}}|{\bar{h}})\right] . \end{aligned}$$
(18)

Since \(P(c|h)>P(c|{\bar{h}})\), we have \(P({\bar{c}}|{\bar{h}})=1-P(c|{\bar{h}})>1-P(c|h)=P({\bar{c}}|h)\). Thus, \(P({\bar{c}}|{\bar{h}})P(c|h)- P({\bar{c}}|h)P(c|{\bar{h}})>0\).

Using that \(Bf_{C_1}\alpha _{11}/\alpha _{00}=Bf_{{\mathcal {E}}}\), i.e., (10) holds, we find that

$$\begin{aligned}&{{\mathrm{sign}}}(P_{{\mathcal {E}}}(h)-P_{{\mathcal {E}}'}(h))\\&\quad ={{\mathrm{sign}}}\left( (Bf_{C_1}-1)(Bf_{C_1-1}-Bf_{C_2})\right. \\&\qquad \left. \cdot \left[ -Bf_{C_1}\cdot \alpha _{11}\cdot P(c|h)\cdot P(c|{\bar{h}}) + \alpha _{00}\cdot P({\bar{c}}|h)\cdot P({\bar{c}}|{\bar{h}})\right] \right) \\&\quad ={{\mathrm{sign}}}\left( (1-Bf_{C_1})(Bf_{C_1-1}-Bf_{C_2})\left( Bf_{{\mathcal {E}}}-\frac{P({\bar{c}}|h)\cdot P({\bar{c}}|{\bar{h}})}{P(c|h)\cdot P(c|{\bar{h}})}\right) \right. . \end{aligned}$$

\(\square \)

Corollary 3

If\(E_1,\ldots ,E_{|C_1|}\)are the children of\(C_1\), then for all possible measurements\(E_1=e_1,\ldots ,E_{|C_1|}=e_{|C_1|}\)

$$\begin{aligned} P(h|e_1\ldots e_{|C_1|}{\vec {f}})<P(h|c_1 {\vec {f}}). \end{aligned}$$

Proof

The proof is a relatively simple exercise in Bayesian network calculations

$$\begin{aligned} P(h|e_1\ldots e_{|C_1|}{\vec {f}})&=\frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot \left( \sum _{j=0}^1P(c^j|{\bar{h}})\prod _{n=1}^{|C_1|}P(e_n|c^j)\right) }{P({\vec {f}}| h)\cdot P(h) \cdot \left( \sum _{j=0}^1P(c^j|h)\prod _{n=1}^{|C_1|}P(e_n|c^j)\right) }}.\\ P(h|c_1 {\vec {f}})&=\frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot (P(c|{\bar{h}}))}{P({\vec {f}}|h)\cdot P(h)\cdot P(c|h)}}. \end{aligned}$$

Hence,

$$\begin{aligned}&{{\mathrm{sign}}}(P(h|e_1\ldots e_{|C_1|}{\vec {f}})-P(h|c_1 {\vec {f}}))\\&\quad ={{\mathrm{sign}}}\left( \frac{P(c|{\bar{h}})}{P(c|h)}-\frac{\sum _{j=0}^1P(c^j|{\bar{h}})\prod _{n=1}^{|C_1|}P(e_n|c^j)}{\sum _{j=0}^1P(c^j|h)\prod _{n=1}^{|C_1|}P(e_n|c^j)}\right) \\&\quad ={{\mathrm{sign}}}\left( P(c|{\bar{h}})\sum _{j=0}^1P(c^j|h)\prod _{n=1}^{|C_1|}P(e_n|c^j)-P(c|h)\cdot \sum _{j=0}^1P(c^j|{\bar{h}})\prod _{n=1}^{|C_1|}P(e_n|c^j)\right) \\&\quad ={{\mathrm{sign}}}\left( P(c|{\bar{h}})[P(c|h)\prod _{n=1}^{|C_1|}P(e_n|c)+P({\bar{c}}|h)\prod _{n=1}^{|C_1|}P(e_n|{\bar{c}})]\right. \\&\qquad \left. -\,P(c|h)\cdot \left[ P(c|{\bar{h}})\prod _{n=1}^{|C_1|}P(e_n|c)+P({\bar{c}}|{\bar{h}})\prod _{n=1}^{|C_1|}P(e_n|{\bar{c}})\right] \right) \\&\quad ={{\mathrm{sign}}}\left( P(c|{\bar{h}})P({\bar{c}}|h)\prod _{n=1}^{|C_1|}P(e_n|{\bar{c}})-P(c|h)\cdot P({\bar{c}}|{\bar{h}})\prod _{n=1}^{|C_1|}P(e_n|{\bar{c}})\right) \\&\quad ={{\mathrm{sign}}}\left( P(c|{\bar{h}})P({\bar{c}}|h)-P(c|h)\cdot P({\bar{c}}|{\bar{h}})\right) . \end{aligned}$$

By (4) we have that \(P(c|{\bar{h}})<P(c|h)\) and that \(P({\bar{c}}|h)<P({\bar{c}}|{\bar{h}})\). Hence, the bracket is negative and it follows that

$$\begin{aligned} P\left( h|e_1\ldots e_{|C_1|}{\vec {f}}\right) <P\left( h|c_1 {\vec {f}}\right) . \end{aligned}$$

\(\square \)

Proposition 2

If Condition B, the last two ceteris paribus condition and if \(P(c_1|h)=1=P(c_2|h)\) hold, then

$$\begin{aligned} {{\mathrm{sign}}}\left( P_{{\mathcal {E}}}(h)-P_{{\mathcal {E}}'}(h)\right)&= {{\mathrm{sign}}}\left( \frac{\sum _{j,l=0}^1P(c^j_1|{\bar{h}})P(c^l_2|{\bar{h}})P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}}{P(e_{|C_1|}'|c_2)}\right. \\&\quad \left. -\frac{\sum _{j,l=0}^1P(c^j_1|{\bar{h}})P(c^l_2|{\bar{h}})P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}}{P(e_{|C_1|}|c_1)}\right) . \end{aligned}$$

Proof

The proof is a simple exercise in Bayesian network calculations. First, we use (16) and (17) to obtain

$$\begin{aligned} P_{{\mathcal {E}}}(h)&=\frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot \left( \sum _{j,l=0}^1P(c^j_1|{\bar{h}})P(c^l_2|{\bar{h}})P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) }{P({\vec {f}}|h)\cdot P(h)\cdot \left( \sum _{j,l=0}^1P(c^j_1|h)P(c^l_2|h)P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) }}.\\ P_{{\mathcal {E}}'}(h)&= \frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot \left( \sum _{j,l=0}^1P(c^j_1|{\bar{h}})P(c^l_2|{\bar{h}})P(e_{|C_1|}'|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) }{P({\vec {f}}|h)\cdot P(h)\cdot \left( \sum _{j,l=0}^1P(c^j_1|h)P(c^l_2|h)P(e_{|C_1|}'|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) }}. \end{aligned}$$

Using that \(P(c_1|h)=1=P(c_2|h)\) and hence \(P({\bar{c}}_1|h)=0=P({\bar{c}}_2|h)\) the expressions simplify to

$$\begin{aligned} P_{{\mathcal {E}}}(h)= \frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot \left( \sum _{j,l=0}^1P(c^j_1|{\bar{h}})P(c^l_2|{\bar{h}})P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) }{P({\vec {f}}|h)\cdot P(h)\cdot \left( P(e_{|C_1|}|c_1)\cdot \chi _{11}\cdot \chi _{21}\right) }} \end{aligned}$$

and

$$\begin{aligned} P_{{\mathcal {E}}'}(h)= \frac{1}{1+\frac{P({\vec {f}}|{\bar{h}})\cdot P({\bar{h}})\cdot \left( \sum _{j,l=0}^1P(c^j_1|{\bar{h}})P(c^l_2|{\bar{h}})P(e_{|C_1|}'|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) }{P({\vec {f}}|h)\cdot P(h)\cdot \left( P(e_{|C_1|}'|c_1)\cdot \chi _{11}\cdot \chi _{21}\right) }}. \end{aligned}$$

The claimed result follows by re-substituting the definitions of \(\chi _{1j}\) and \(\chi _{2l}\). \(\square \)

Proposition 3

If Condition B, the last two ceteris paribus condition,\(P(c_1|h)=1=P(c_2|h)\) and if \(P(e_{|C_1|}|c_1)= P(e_{|C_1|}'|c_2)>P(e_{|C_1|}'|{\bar{c}}_2)\)hold, then

$$\begin{aligned} {{\mathrm{sign}}}(P_{{\mathcal {E}}}(h)-P_{{\mathcal {E}}'}(h)) =-{{\mathrm{sign}}}\left( \frac{\alpha _{0,0}\cdot \beta _{0,0}+\alpha _{0,1}\cdot \beta _{0,1}}{\alpha _{1,0}\cdot \beta _{1,0}}+\prod _{n=1}^{|C_1|-1}\frac{P(e_n|c_1)}{P(e_n|{\bar{c}}_1)}\right) , \end{aligned}$$

where the\(\alpha \)and\(\beta \)are parameters independent of\(|C_1|\)which are defined in (19).

Proof

Under the assumption that \(P(e_{|C_1|}|c_1)= P(e_{|C_1|}'|c_2)\) the denominators are equal. Furthermore, the terms with \(j=l=1\) cancel out. We hence find

$$\begin{aligned}&{{\mathrm{sign}}}\left( P(h|e_1\ldots e_Le_{|C_1|}e_{{|C_1|}+1}\ldots e_{{|C_1|}+{|C_2|}}{\vec {f}})\right. \\&\quad \left. -\,P(h|e_1\ldots e_{|C_1|-1} e_{|C_1|}'e_{{|C_1|}+1}\ldots e_{{|C_1|}+{|C_2|}}{\vec {f}})\right) \\&\quad ={{\mathrm{sign}}}\left( \sum _{\begin{array}{c} j,l=0\\ (j,l)\ne (1,1) \end{array}}^1P(c^j_1|{\bar{h}})P(c^l_2|{\bar{h}})P(e_{|C_1|}'|c^l_2)\cdot \chi _{1j}\cdot \chi _{2l}\right. \\&\qquad \left. -\,\sum _{\begin{array}{c} j,l=0\\ (j,l)\ne (1,1) \end{array}}^1P(c^j_1|{\bar{h}})P(c^l_2|{\bar{h}})P(e_{|C_1|}|c^j_1)\cdot \chi _{1j}\cdot \chi _{2l}\right) . \end{aligned}$$

With (subscripts represent the values for j and l)

$$\begin{aligned} \alpha _{0,0}&:=P({\bar{c}}_1|{\bar{h}})P({\bar{c}}_2|{\bar{h}})\cdot \prod _{g=1}^{|C_2|}P(e_{{|C_1|}+g}|{\bar{c}}_2)\nonumber \\ \beta _{0,0}&:=P(e_{|C_1|}'|{\bar{c}}_2)-P(e_{|C_1|}|{\bar{c}}_1)\nonumber \\ \alpha _{1,0}&:=P(c_1|{\bar{h}})P({\bar{c}}_2|{\bar{h}})\cdot \prod _{g=1}^{|C_2|}P(e_{{|C_1|}+g}|{\bar{c}}_2)\nonumber \\ \beta _{1,0}&:=P(e_{|C_1|}'|{\bar{c}}_2)-P(e_{|C_1|}|c_1)\nonumber \\ \alpha _{0,1}&:=P({\bar{c}}_1|{\bar{h}})P({\bar{c}}_2|{\bar{h}})\cdot \prod _{g=1}^{|C_2|}P(e_{{|C_1|}+g}|c_2)\nonumber \\ \beta _{0,1}&:=P(e_{|C_1|}'|c_2)-P(e_{|C_1|}|{\bar{c}}_1) \end{aligned}$$
(19)

this becomes equal to

$$\begin{aligned}&{{\mathrm{sign}}}\left( \alpha _{0,0}\cdot \beta _{0,0}\cdot \prod _{n=1}^{|C_1|-1}P(e_n|{\bar{c}}_1)+\alpha _{0,1}\cdot \beta _{0,1}\cdot \prod _{n=1}^{|C_1|-1}P(e_n|{\bar{c}}_1) \right. \\&\qquad \left. +\,\alpha _{1,0}\cdot \beta _{1,0}\cdot \prod _{n=1}^{|C_1|-1}P(e_n|c_1)\right) \\&\quad ={{\mathrm{sign}}}\left( [\alpha _{0,0}\cdot \beta _{0,0}+\alpha _{0,1}\cdot \beta _{0,1}]\cdot \prod _{n=1}^{|C_1|-1}P(e_n|{\bar{c}}_1)+\alpha _{1,0}\cdot \beta _{1,0}\cdot \prod _{n=1}^{|C_1|-1}P(e_n|c_1)\right) . \end{aligned}$$

Note that the \(\alpha \) and the \(\beta \) parameters do not change value with varying \(|C_1|\) and are hence treated as constants.

Using that \(\alpha _{1,0}\cdot \beta _{1,0}<0\) and \(\prod _{n=1}^{|C_1|-1}P(e_n|{\bar{c}}_1)>0\) we find that

$$\begin{aligned}&{{\mathrm{sign}}}\left( P\left( h|e_1\ldots e_{|C_1|-1}e_{|C_1|}e_{{|C_1|}+1}\ldots e_{{|C_1|}+{|C_2|}}{\vec {f}}\right) \right) \\&\quad -P\left( h|e_1\ldots e_{|C_1|-1} e_{|C_1|}'e_{{|C_1|}+1}\ldots e_{{|C_1|}+{|C_2|}}{\vec {f}}\right) \\&\quad =-{{\mathrm{sign}}}\left( \frac{\alpha _{0,0}\cdot \beta _{0,0}+\alpha _{0,1}\cdot \beta _{0,1}}{\alpha _{1,0}\cdot \beta _{1,0}}+\frac{\prod _{n=1}^{|C_1|-1}P(e_n|c_1)}{\prod _{n=1}^{|C_1|-1}P(e_n|{\bar{c}}_1)}\right) . \end{aligned}$$

\(\square \)

Appendix 2: High Arity Variables

We now address the claim that the so-far established technical results, also hold for models which employ higher arity hypothesis and/or consequence variables.

Denote by \(h^2,h^3,\ldots \) the values of H different from h and by \(c^2,c^3,\ldots \) the values of consequence variable C different from c. \(h^0\) is the (possibly infinite) disjunction of the \(h^i\) with \(i\ge 2\). \(c^0\) is the (possibly infinite) disjunction of the \(c^k\) with \(k\ge 2\). To simplify notation we let \(P(c_j^0|h^i):=1-P(c_j|h^i)\).

The ceteris paribus conditions are that for every consequence variables \(C_j\) and every evidence variable E pertaining to the consequence variable \(C_j\) it holds that

$$\begin{aligned} P(c_j^0|h^i)=P(c_j^0|h^2)=P({\bar{c}}_j|{\bar{h}}) \end{aligned}$$
(20)
$$\begin{aligned} P(e|c^k_j)=P(e|c^2_j)=P(e|{\bar{c}}_j). \end{aligned}$$
(21)

This formalises the thought that all values of the hypothesis variable H different from h are equal. Furthermore, the conditional probability of the evidence does not depend on the particular value \(c^k\). We can hence define our Bayes factors in the usual way unequivocally as \(P(e|c)/P(e|{\bar{c}})\).

Proposition 5

If (20) and (21) hold, then Corollarys 1and 2hold for higher arities, too.

Proof

Using the ceteris paribus condition for the second and third equality, we find

$$\begin{aligned} \sum _{i\ge 2}\sum _{k\ge 2}P(h^i)\cdot P(c^k_j|h^i)\cdot P(e|c^k_j)&=\sum _{i\ge 2}P(h^i)\cdot \sum _{k\ge 2} P(c^k_j|h^i)\cdot P(e|c^k_j)\\&=P(e|c^2_j)\cdot \sum _{i\ge 2}P(h^i)\cdot \sum _{k\ge 2} P(c^k_j|h^i)\\&=P(e|{\bar{c}}_j)\cdot \sum _{i\ge 2}P(h^i)\cdot P(c^0_j|h^i)\\&=P(e|{\bar{c}}_j)\cdot P({\bar{c}}_j|{\bar{h}})\cdot P({\bar{h}}). \end{aligned}$$

To complete the proof it suffices make the following observations:

  1. 1.

    Whenever the term ‘\(P(e|{\bar{c}}_j)\cdot P({\bar{c}}_j|{\bar{h}})\cdot P({\bar{h}})\)’ appears in a proof for the basic model it is replaced by the term ‘\(\sum _{i=2}\sum _{k=2}P(h^i)\cdot P(c^k_j|h^i)\cdot P(e|c^k_j)\)’. Since both terms are equal, no new difficulties arise.

  2. 2.

    The terms of the form ‘\(P({\bar{c}}_j|{\bar{h}})\)’ for \(i\ge 2\) are replaced by terms of the form \(\sum _{k\ge 2}P(c_j^k|h^i)\)’. The latter is equal to \(P(c_j^0|h^i)\). By the first ceteris paribus assumption for greater arities, these terms are equal. \(\square \)

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Landes, J. Variety of Evidence. Erkenn 85, 183–223 (2020). https://doi.org/10.1007/s10670-018-0024-6

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