No one can serve two epistemic masters

Abstract

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.

Appendix: Proof of Propositions 1 and 2

Proof

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).

Proof

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).

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

(7)

Proof

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

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).

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

(8)

Proof

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

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