No one can serve two epistemic masters


Consider two epistemic experts—for concreteness, let them be two weather forecasters. Suppose that you aren’t certain that they will issue identical forecasts, and you would like to proportion your degrees of belief to theirs in the following way: first, conditional on either’s forecast of rain being x, you’d like your own degree of belief in rain to be x. Secondly, conditional on them issuing different forecasts of rain, you’d like your own degree of belief in rain to be some weighted average of the forecast of each (perhaps with weights determined by their prior reliability). Finally, you’d like your degrees of belief to be given by an orthodox probability measure. Moderate ambitions, all. But you can’t always get what you want.

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  1. 1.

    There are a wide variety of putative experts and a wide variety of principles of expert deference on offer in the current epistemology literature. However, the reader will note that everything we will say here about Al and Bert goes just as well for chance, your rational future self, and so on and so forth. The reader will also note that if Al and Bert have all your evidence and more, and if they are additionally certain of their own forecasts, then every extant principle of expert deference will entail both (2) and (3). For instance, given an expert \(\mathcal{E}\) who is certain of their own credences and has all your evidence and more, the following ways of treating \(\mathcal{E}\) as an expert are all equivalent: for all pxE (1) \(C(p \mid \mathcal{E}(p)=x) = x\), (2) \(C(p \mid \mathcal{E} = E) = E(p)\), (3) \(C(p \mid \mathcal{E}=E) = E(p \mid \mathcal{E}=E)\), and \(C(p) = \sum _x x \cdot C(\mathcal{E}(p)=x)\). See Gallow (msb) for a proof of this claim and more on the relationship between various principles of expert deference.

  2. 2.

    By an ‘orthodox probability function’, I will mean that C is non-negative, normalized, countably additive, and conglomerable. These assumptions go beyond probabilism in the case where we are considering infinitely many possible values for \(\mathcal{A}\) and \(\mathcal{B}\). However, if we suppose that there are at most finitely many potential values for \(\mathcal{A}\) and \(\mathcal{B}\), then an ‘orthodox probability function’ is just any finitely additive probability. Thanks to an anonymous reviewer for their clarifying comments on this point.

  3. 3.

    Matthew 6:24.

  4. 4.

    See, for starters, Gaifman (1988), Lewis (1980, 1994), Hall (1994), van Fraassen (1984, 1995), Christensen (2010), and Elga (2013).

  5. 5.

    Many of these principles of expert deference look different from (2) and (3); fortunately, in most cases, we can present them in the form of (2) and (3) by simply shifting our attention to a different expert. For instance, Lewis (1994) and Hall (1994) both say that we should not defer to the judgments of chance, but rather the judgments of chance, conditionalized on the proposition that it is chance. In such a case, we could consider the relevant expert to be, not chance itself, but rather chance conditionalized on chance, and we will get back a principle looking like (2) and (3). Cf. Hall and Arntzenius (2003) and Schaffer (2003). (See also footnote 1.)

  6. 6.

    Cf. Gallow (msa).

  7. 7.

    See, for starters, Kelly (2005), Elga (2007), and Christensen (2007, 2010, 2011).

  8. 8.

    See, e.g., Shogenji (ms) and Fitelson and Jehle (2009).

  9. 9.

    See, e.g., Wagner (1985) and Staffel (2015, §6).

  10. 10.

    Cf. Levinstein (2015).


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Thanks to Michael Caie, Daniel Drucker, Harvey Lederman, and an anonymous reviewer for helpful conversations and feedback.

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Correspondence to J. Dmitri Gallow.

Appendix: Proof of Propositions 1 and 2

Appendix: Proof of Propositions 1 and 2


We establish three lemmas, from which the theorems follow immediately. (Note: throughout, I will use ‘C’ indiscriminately for (1) a joint probability density function over the values of \(\mathcal{A}\) and \(\mathcal{B}\), as well as (2) the corresponding marginal densities, and (3) the corresponding probability function. In the event that there are at most finitely many possible values of \(\mathcal{A}\) and \(\mathcal{B}\), ‘C’ will everywhere denote a probability function and integrals may be exchanged for sums throughout.)

Lemma 1

If (2), (3), and (4) hold, then so do (5) and (6).


Since C is a countably additive, conglomerable probability, for all a,

$$\begin{aligned} C(r \mid \mathcal{A}=a)&= \int _0^1 C(r \mid \mathcal{A}= a, \mathcal{B}= b) \cdot C(\mathcal{B}= b \mid \mathcal{A}= a) \cdot db \\&= \int _0^1 \left( \alpha a + \beta b \right) \cdot C(\mathcal{B}= b \mid \mathcal{A}=a) \cdot db \\&= \alpha a \cdot \int _0^1 C(\mathcal{B}= b \mid \mathcal{A}= a) \cdot db \,\,+\,\, \beta \int _0^1 b \cdot C(\mathcal{B}= b \mid \mathcal{A}= a) \cdot db \\&= \alpha a + \beta \mathbb {E}[\mathcal{B}\mid \mathcal{A}=a ] \end{aligned}$$

Then, because \(C(r \mid \mathcal{A}=a)=a\) and \(\beta = 1-\alpha\), we have (6). Following the same procedure, with ‘\(\mathcal{A}\)’ and ‘\(\mathcal{B}\)’ exchanged throughout, establishes (5).

Lemma 2

If (5) and (6) hold, then so does (7).

$$\begin{aligned} \mathbb {E}[\mathcal{A}\mathcal{B}] = \mathbb {E}[\mathcal{A}^2] = \mathbb {E}[\mathcal{B}^2] \end{aligned}$$


$$\begin{aligned} \mathbb {E}[\mathcal{A}\mathcal{B}]&= \int _0^1 \int _0^1 a b \cdot C(\mathcal{A}=a, \mathcal{B}=b) \cdot db \cdot da \\&= \int _0^1 a \cdot C(\mathcal{A}=a) \cdot \left[ \int _0^1 b \cdot C(\mathcal{B}= b \mid \mathcal{A}=a) \cdot db \right] \cdot da \\&= \int _0^1 a \cdot C(\mathcal{A}=a) \cdot \mathbb {E}[\mathcal{B}\mid \mathcal{A}=a] \cdot da \\&= \int _0^1 a^2 \cdot C(\mathcal{A}=a) \cdot da \\&= \mathbb {E}[\mathcal{A}^2] \end{aligned}$$

The same procedure, with ‘\(\mathcal{A}\)’ exchanged for ‘\(\mathcal{B}\)’ throughout, establishes that \(\mathbb {E}[\mathcal{A}\mathcal{B}] = \mathbb {E}[\mathcal{B}^2]\).

Lemma 3

If (7) holds, then so does (8).

$$\begin{aligned} \mathbb {E}[(\mathcal{A}- \mathcal{B})^2] = 0 \end{aligned}$$


$$\begin{aligned} \mathbb {E}[(\mathcal{A}- \mathcal{B})^2]&= \mathbb {E}[\mathcal{A}^2] - 2 \mathbb {E}[\mathcal{A}\mathcal{B}] + \mathbb {E}[\mathcal{B}^2] = 0 \end{aligned}$$

If the expectation of \((\mathcal{A}- \mathcal{B})^2\) is 0, then \(C(\mathcal{A}=\mathcal{B}) = 1\), and (1) is violated.

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Gallow, J.D. No one can serve two epistemic masters. Philos Stud 175, 2389–2398 (2018).

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  • Expert deference
  • Disagreement
  • Linear averaging