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The Consistency of Probabilistic Regresses: Some Implications for Epistemological Infinitism

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Abstract

This note employs the recently established consistency theorem for infinite regresses of probabilistic justification (Herzberg in Stud Log 94(3):331–345, 2010) to address some of the better-known objections to epistemological infinitism. In addition, another proof for that consistency theorem is given; the new derivation no longer employs nonstandard analysis, but utilises the Daniell–Kolmogorov theorem.

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Notes

  1. An event U is called contingent given a probability measure P if and only if 0 < P(U) < 1.

  2. The symmetric difference of two events AB is defined as \(A\Updelta B:=(A\setminus B)\cup (B\setminus A). \)

  3. We have nevertheless decided to stick with the term ‘justification’ for much of this paper simply because it seems to be the more common terminological choice in the literature.

  4. In fact, the property of being a finite ordinal, i.e. a natural number, can even be defined in set-theoretically “robust” terms: It is definable through a \(\Updelta_0\) and thus through an absolute formula. In other words, whether a given set is a natural number does not depend on the interpretation in a particular transitive model of set theory (cf. Jech 2000, Lemma 12.10).

  5. It should be mentioned that the Daniell–Kolmogorov theorem admits a particularly simple proof by means of nonstandard analysis, cf. Herzberg (2011).

  6. To see this, apply the Transfer Principle to the sentence \({\forall n\in{\bf N}}\left( S(n+1)\Rightarrow \forall k\leq n\left(S(k)\right)\right), \) which in turn follows from the hypothesis \({\forall n\in{\bf N}}\left(S(n+1)\Rightarrow S(n)\right)\) by mathematical induction.

  7. In the special case where \(\varepsilon=\frac{1}{2}, \) we have \(Q^{j,\infty}_{i}=\frac{1}{2}\) for all \(j\in{\bf N}\) and each \(i\in\{0,1\}. \)

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Correspondence to Frederik Herzberg.

Appendix: A “Standard” Proof of the Consistency Theorem for Probabilistic Regresses

Appendix: A “Standard” Proof of the Consistency Theorem for Probabilistic Regresses

For all of this section, we fix two sequences \(\langle\alpha_k\rangle_{k\in{\bf N}},\langle\beta_k\rangle_{k\in{\bf N}}\in[0,1]^{\bf N};\) as in all of this paper, N denotes the set of all nonnegative integers. In this section, we shall prove the following result, which is a slight generalisation of the consistency theorem in Herzberg (2010):

Theorem A.1

There exists some probability measure P and some sequence of P-measurable events \(\langle S_k\rangle_{k\in{\bf N}}\) such that \(P\left(\left.S_k\right|S_{k+1}\right)=\alpha_k\) as well as \(P\left(\left.S_k\right|\complement S_{k+1}\right)=\beta_k\) for all \(k\in{\bf N}\).

Essentially, the proof is based on the construction of a reversed non-homogeneous Markov chain with state space {0,1} that was started an infinitely long time ago. We use the notation \(\prod\nolimits_{\leftarrow}\) to refer to products of matrices whose order of multiplication has been reversed, so that \({\prod}_{\mathop{\leftarrow}\limits_{k=m}}^{n}T^k=T^n\cdots T^m\) for all \(n\geq m\in{\bf N}. \) Likewise, \({\prod}_{\mathop{\leftarrow}\limits_{k=m}}^{\infty} T^k=\lim\nolimits_{n\rightarrow\infty} T^n\cdots T^m. \)

For all \(k\in{\bf N}, \) put

$$ T^k=\left(\begin{array}{cc} \alpha_k &1-\alpha_k\\ \beta_k &1-\beta_k \end{array}\right). $$
(3)

For all \(j,j^{\prime}\in {\bf N}\) with j′ ≥ j and for all \(i,i'\in\{0,1\}, \) let

$$ Q^{j,j'}_{i,i'} = {^\top \!\! e_{i'}}\left(\prod_{\mathop{\leftarrow}\limits_{k=j}}^{j'-1} T^k\right)e_{i} $$

wherein \(e_0=\left(\begin{array}{c}0\\ 1\end{array}\right)\) and \(e_1=\left(\begin{array}{c}1\\ 0\end{array}\right), \) and let

$$ Q^{j,\infty}_{i} = {^\top \!\! c}\left(\prod_{\mathop{\leftarrow}\limits_{Pk=j}}^{\infty} T^k\right)e_{i} $$

for some \(c=\left(\begin{array}{c}1-\varepsilon\\ \varepsilon\end{array}\right)\) with \(\varepsilon\in (0,1). \) Footnote 7 Define for all finite \(J\subseteq {\bf N}\) with \(\{j_1,\ldots,j_{\# J}\}, \) a measure P J on {0,1}J by means of the equation

$$ P_J\{\vec i\}= \left(\prod_{\ell=1}^{\#J-1} Q^{j_{\ell},j_{\ell+1}}_{i_{j_{\ell}},i_{j_{\ell}+1}} \right) Q^{j_{\#J},\infty}_{i_{j_{\#J}}} $$

for all \(\vec i\in \{0,1\}^J. \)

Lemma A.2

P J is a probability measure on {0,1}J for all finite \(J\subseteq {\bf N}. \)

Proof

Note that all the T k are transition probability matrices, that the set of transition probability matrices is closed under multiplication and that the set of transition probability matrices is a closed subset of the set of matrices of that dimension. Therefore, \({\prod_{\mathop{\leftarrow} \limits_{k}}}T^k\) is a transition probability matrix. By definition, this means that each entry is nonnegative and the sum of each row vector equals 1. Formally,

$$ \left({\prod_{\mathop{\leftarrow}\limits_{k}}T^k}\right) \underbrace{\left(\begin{array}{c}1\\ 1\end{array}\right)}_{=e_0+e_1}=\left(\begin{array}{c}1\\ 1\end{array}\right), $$

thus \({^\top \!\! e_{i}}\left({\prod_{\mathop{\leftarrow} \limits_{k}}T^k}\right)\left(e_0+e_1\right)=1\) for each \(i\in\{0,1\}. \) This immediately entails that

$$ Q^{j,j'}_{0,i'} +Q^{j,j'}_{1,i'}=1 $$

for all \(j,j'\in{\bf N}\) and each \(i'\in\{0,1\}. \) However, it also means that

$$ {^\top \!\!c}\left({\prod_{\mathop{\leftarrow} \limits_{k}}T^k}\right)\left(e_0+e_1\right)=\left((1-\varepsilon){^\top \!\! e_{0}}+\varepsilon {^\top \!\! e_{1}}\right)\left({\prod_{\mathop{\leftarrow} \limits_{k}}T^k}\right)\left(e_0+e_1\right)=1 $$

which implies that

$$ Q^{j,\infty}_{0}+Q^{j,\infty}_{1}=1 $$

for all \(j\in{\bf N}. \) So,

$$ Q^{j,j'}_{0,i'} +Q^{j,j'}_{1,i'}=1,\quad Q^{j,\infty}_{0}+Q^{j,\infty}_{1}=1 $$

for all \(j,j'\in{\bf N}\) and each \(i'\in{\bf N}. \) This can be used to prove inductively in #J that \(\sum_{\vec i\in\{0,1\}^J}P_J\{\vec i\}=1. \) Since clearly \(P_J\{\vec i\}\geq 0\) for all \( \vec i\in\{0,1\}^J, \) we conclude that P J must be a probability measure on {0,1}J. \(\square\)

Let \(\mathcal{H}\) be the set of all finite subsets of N.

Lemma A.3

\(\left\langle P_J\right\rangle_{J\in\mathcal{H}}\) is a projective family of probability measures.

Proof

Let H = J ∪ {h} with \(h\not\in J. \) Without loss of generality, let h > max J = j #J . Let p H J be the projection of H onto J. Then

$$ \begin{aligned} & P_H\left({p_J^H}^{-1}\{\vec i\}\right)\\ &\quad=P_H\left(\{\vec i\}\times\{0,1\}\right)= P_H\left(\{\vec i\}\times\{0\}\right)+ P_H\left(\{\vec i\}\times\{1\}\right)\\ &\quad=\left(\prod_{\ell=1}^{\#J-1} Q^{j_{\ell},j_{\ell+1}}_{i_{j_{\ell}},i_{j_{\ell}+1}} \right) Q^{j_{\#J},h}_{i_{j_{\#J}},0} Q^{h,\infty}_{0} + \left(\prod_{\ell=1}^{\#J-1} Q^{j_{\ell},j_{\ell+1}}_{i_{j_{\ell}},i_{j_{\ell}+1}} \right) Q^{j_{\#J},h}_{i_{j_{\#J}},1} Q^{h,\infty}_{1}\\ &\quad=\left(\prod_{\ell=1}^{\#J-1} Q^{j_{\ell},j_{\ell+1}}_{i_{j_{\ell}},i_{j_{\ell}+1} }\right) \left(Q^{j_{\#J},h}_{i_{j_{\#J}},0}Q^{h,\infty}_{0} + Q^{j_{\#J},h}_{i_{j_{\#J}},1}Q^{h,\infty}_{1} \right) \\ \end{aligned} $$
(4)

Now, it is not difficult to verify that

$$ Q^{j_{\#J},h}_{i_{j_{\#J}},0}Q^{h,\infty}_{0} + Q^{j_{\#J},h}_{i_{j_{\#J}},1}Q^{h,\infty}_{1} = Q^{j_{\#J},\infty}_{i_{j_{\#J}}}. $$
(5)

Indeed, \(e_0 {^\top \!\! e_0} + e_1 {^\top \!\! e_1}= \left(\begin{array}{cc}1 & 0\\ 0 & 1 \end{array}\right), \) hence

$$ \begin{aligned} &Q^{h,\infty}_{0}Q^{j_{\#J},h}_{i_{j_{\#J}},0} + Q^{h,\infty}_{1}Q^{j_{\#J},h}_{i_{j_{\#J}},1}\\ &\quad= {^\top \!\! c}\left(\prod_{\mathop{\leftarrow} \limits_{k=h}}^{\infty} T^k\right)e_{0} {^\top \!\! e_{0}}\left(\prod_{\leftarrow \atop k=j_{\#J}}^{h-1} T^k\right)e_{i_{j_{\#J}}} \\ &\quad + {^\top \!\! c}\left(\prod_{\mathop{\leftarrow} \limits_{k=h}}^{\infty} T^k\right)e_{1} {^\top \!\! e_{1}}\left(\prod_{\leftarrow \atop k=j_{\#J}}^{h-1} T^k\right)e_{i_{j_{\#J}}}\\ &\quad= {^\top \!\! c}\left(\prod_{\mathop{\leftarrow} \limits_{k=j_{\#J}}}^{h-1} T^k\right)\left(\prod_{\mathop{\leftarrow} \limits_{k=h}}^{\infty} T^k\right)e_{i_{j_{\#J}}} ={^\top \!\! c}\left(\prod_{\leftarrow \atop k=j_{\#J}}^{\infty} T^k\right)e_{i_{j_{\#J}}} = Q^{j_{\#J},\infty}_{i_{j_{\#J}}}. \end{aligned} $$

Inserting Equations (5) into (4) yields

$$ \begin{aligned} P_H\left({p_J^H}^{-1}\{\vec i\}\right) = & \left(\prod_{\ell=1}^{\#J-1} Q^{j_{\ell},j_{\ell+1}}_{i_{j_{\ell}},i_{j_{\ell}+1} }\right) Q^{j_{\#J},\infty}_{i_{j_{\#J}}} \end{aligned} $$

whence by definition \(P_H\left({p_J^H}^{-1}\{\vec i\}\right)= P_J\{\vec i\}\) for arbitrary \(\vec i\in \{0,1\}^J\) and all finite \(H,J\subseteq {\bf N}\) with \(J\subseteq H\) and \(\#\left(H\setminus J\right)=1. \) This implies (cf. Bauer (2002, p. 307, Bemerkung 2)) that \(\left\langle P_J\right\rangle_{J\in\mathcal{H}}\) is a projective family of probability measures.\(\square\)

In the following theorem, \(\mathcal{P}(\{0,1\})^{\otimes {\bf N}}\) denotes the infinite product of countably many identical copies of the power-set of {0,1}; it is the smallest σ-algebra with respect to which all projections are measurable. Given any \(J\in\mathcal{H}, \) the map π J is the projection onto {0,1}J defined as the map \(\pi_J:\{0,1\}^{\bf N}\rightarrow\{0,1\}^J,\quad \omega\mapsto \langle\omega(k)\rangle_{k\in J}\)

Theorem A.4

There exists a probability measure \(P:\mathcal{P}(\{0,1\})^{\otimes {\bf N}}\rightarrow[0,1]\) such that \(P_J=P\left(\pi_J^{-1}(\cdot)\right)\) for all \(J\in \mathcal{H}\).

Proof

By the Daniell-Kolmogorov theorem, there must be a probability measure P on \(\Upomega=\{0,1\}^{\bf N}\) such that the finite-dimensional distributions of the process X defined by \(X_k:\Upomega\rightarrow\{0,1\},\quad\omega\mapsto \omega(k)\) (for all \(k\in{\bf N}\)) coincide with the probability measures P J \(J\in\mathcal{H}. \) \(\square\)

Now we can, at last, prove the consistency theorem stated at the beginning as a corollary to Theorem A.4:

Proof of Theorem A.1

Let P and X be as in the proof of Theorem A.4. For all \(k\in{\bf N}, \) set \(S_k=\left\{X_k=1\right\}, \) so that \(\complement S_k=\left\{X_k=0\right\}, \)

Then, for all \(n>k\in{\bf N}, \)

$$ P\left\{X_{k}=1\right\}= T^k\cdots T^n\left(\begin{array}{c} P\left\{X_{n+1}=1\right\}\\ P\left\{X_{n+1}=0\right\}\end{array}\right), $$

hence

$$P(S_k)= T^k\cdots T^n\left(\begin{array}{c} P(S_{n+1})\\ 1-P(S_{n+1})\end{array}\right) $$

as well as

$$ T^k=\left(\begin{array}{cc}P\left(\left.\left\{X_{k}=1\right\}\right|\left\{X_{k+1}=1\right\}\right)& P\left(\left.\left\{X_{k}=0\right\}\right|\left\{X_{k+1}=1\right\}\right)\\ P\left(\left.\left\{X_{k}=1\right\}\right|\left\{X_{k+1}=0\right\}\right) & P\left(\left.\left\{X_{k}=0\right\}\right|\left\{X_{k+1}=0\right\}\right) \end{array}\right), $$

whence

$$ T^k=\left(\begin{array}{cc}P\left(\left.S_k\right|S_{k+1}\right)& P\left(\left.\complement S_k\right|S_{k+1}\right)\\ P\left(\left.S_k\right|\complement S_{k+1}\right) & P\left(\left.\complement S_k\right|\complement S_{k+1}\right) \end{array}\right). $$

Comparing this with the definition of T k in Equation (3) leads us to conclude

$$ P\left(\left.S_k\right|S_{k+1}\right)=\alpha_k,\quad P\left(\left.S_k\right|\complement S_{k+1}\right)=\beta_k $$

for all \(k\in{\bf N}. \square \)

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Herzberg, F. The Consistency of Probabilistic Regresses: Some Implications for Epistemological Infinitism. Erkenn 78, 371–382 (2013). https://doi.org/10.1007/s10670-011-9358-z

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