The qualitative paradox of non-conglomerability

Abstract

A probability function is non-conglomerable just in case there is some proposition E and partition $\pi$ of the space of possible outcomes such that the probability of E conditional on any member of $\pi$ is bounded by two values yet the unconditional probability of E is not bounded by those values. The paradox of non-conglomerability is the counterintuitive—and controversial—claim that a rational agent’s subjective probability function can be non-conglomerable. In this paper, I present a qualitative analogue of the paradox. I show that, under antecedently plausible assumptions, an analogue of the paradox arises for rational comparative confidence. As I show, the qualitative paradox raises its own distinctive set of philosophical issues.

Appendix

In this Appendix, I show that the non-conglomerable comparative confidence relation $\succeq$ of Sect. 4 is almost representable by de Finetti’s non-conglomerable probability function P of Sect. 2. That is, I show:

Almost Representability If $A|B \succeq C|D$, then $P(A|B) \ge P(C|D)$.

Note that the only constraints on $\succeq$ are Koopman’s axioms, Fairness, and Independence. So, we must show that P satisfies suitable probabilistic analogues of them. For example, since $k|k \succeq a|h$ by Verified Contingency, we must show that $P(k|k) \ge P(a|h)$. Similarly, since $V_i|\Omega \approx V_k|\Omega$ by Fairness, we must show that $P(V_i|\Omega ) = P(V_k|\Omega )$—i.e., that $P(V_i) = P(V_k)$.

I begin with Koopman’s axioms. Because I will not yet appeal to any properties of P beyond its satisfying constraints (1)–(4) from Sect. 2—I will not even appeal to the fact that P is real-valued—it will follow that any conditional probability function satisfying those conditions also satisfies the relevant probabilistic analogues of Koopman’s axioms. Thus, any conditional probability function—real-valued or extended-valued—can almost represent some comparative confidence relation that satisfies Koopman’s axioms.

• 1. Verified Contingency Recall that $P[(A \cap B)|B] = P(A|B)$ (cf. footnote 14). So, $P(a|h) = P[(a \cap h)|h]$ and $P(\lnot a|h) = P[(\lnot a \cap h)|h]$. By finite additivity and constraint (2), $P(a|h) + P(\lnot a|h) = P[(a \cap h)|h] + P[(\lnot a \cap h)|h] = P(h|h) = 1$. Since $P(\lnot a|h) \ge 0$, it follows that $P(a|h) \le 1$. Thus, since $P(k|k) = 1$, $P(k|k) \ge P(a|h)$.

• 2. Implication Suppose $a|h \succeq k|k$, so that $h \subseteq a$. Then $a = (h \cup h')$ for some $h' \in \mathcal {F}$ such that $h \cap h' = \emptyset$. So, by finite additivity, $P(a|h) = P[(h \cup h')|h] = P(h|h) + P(h'|h) = 1 + P(h'|h) \ge 1 = P(k|k)$, as desired.

• 3. Reflexivity Obviously, $P(a|h) \ge P(a|h)$.

• 4. Transitivity Suppose that $P(c|l) \ge P(b|k)$ and $P(b|k) \ge P(a|h)$. Obviously, $P(c|l) \ge P(a|h)$.

• 5. Antisymmetry Suppose that $P(b|k) \ge P(a|h)$. Then, as above, $P(b|k) + P(\lnot b|k) = P[(b \cap k)|k] + P[(\lnot b \cap k)|k] = P(k|k) = 1$. Similarly, $P(a|h) + P(\lnot a|h) = 1$. Thus, $P(\lnot a|h) = 1 - P(a|h) \ge 1 - P(b|k) = P(\lnot b|k)$, as desired.

• 6. Composition

  Suppose:

  • (a) $(a_1 \cap b_1) \cap h_1 \ne \emptyset$ and $(a_2 \cap b_2) \cap h_2 \ne \emptyset$.

  • (b) $P(a_2|h_2) \ge P(a_1|h_1)$.

  • (c) $P[b_2|(a_2 \cap h_2)] \ge P[b_1|(a_1 \cap h_2)]$.

  By constraint (4), $P[(a_2 \cap b_2)|h_2] = P[b_2|(a_2 \cap h_2)]P(a_2|h_2)$ and $P[(a_1 \cap b_1)|h_1] = P[b_1|(a_1 \cap h_1)]P(a_1|h_1)$. Thus, by (b) and (c), $P[(a_2 \cap b_2)|h_2] \ge P[(a_1 \cap b_1)|h_1]$.

• 7. Decomposition

  Suppose:

  • (a) $(a_1 \cap b_1) \cap h_1 \ne \emptyset$ and $(a_2 \cap b_2) \cap h_2 \ne \emptyset$.

  • (b) $P(a_1|h_1) \ge P(a_2|h_2)$.

  • (c) $P[(a_2 \cap b_2)|h_2] \ge P[(a_1 \cap b_1)|h_1]$.

  As above, $P[(a_2 \cap b_2)|h_2] = P[b_2|(a_2 \cap h_2)]P(a_2|h_2)$ and $P[(a_1 \cap b_1)|h_1] = P[b_1|(a_1 \cap h_1)]P(a_1|h_1)$. Thus, by (b) and (c), it follows that $P[b_2|(a_2 \cap h_2)] \ge P[b_1|(a_1 \cap h_1)]$.

• 8. Alternative Presumption Suppose $P(r|s) \ge P[a|(b \cap h)]$ and $P(r|s) \ge P[a|(\lnot b \cap h)]$. By constraint (4), $P[(a \cap b)|h] = P[a|(b \cap h)]P(b|h)$ and $P[(a \cap \lnot b)|h] = P[a|(\lnot b \cap h)]P(\lnot b|h)$. So, by finite additivity, $P[(a \cap b)|h] + P[(a \cap \lnot b)|h] = P(a|h)$. Using finite additivity again,

  $$\begin{aligned} P(a|h)= & {} P[a|(b \cap h)]P(b|h) + P[a|(\lnot b \cap h)]P(\lnot b|h) \end{aligned}$$

  (32)
  $$\begin{aligned}= & {} P[a|(b \cap h)]P(b|h) + P[a|(\lnot b \cap h)][1-P(b|h)] \end{aligned}$$

  (33)
  $$\begin{aligned}\le & {} P(r|s)P(b|h) + P(r|s)[1 - P(b|h)] \end{aligned}$$

  (34)
  $$\begin{aligned}= & {} P(r|s), \end{aligned}$$

  (35)

  as desired.

• 9. Subdivision

  Suppose the propositions $a_1, \ldots , a_n, b_1, \ldots , b_n$ are such that:

  • (a) $a_i \cap a_j = b_i \cap b_j = \emptyset$ for all $i, j = 1, \ldots , n$ such that $i \ne j$.

  • (b) $a = (a_1 \cup \ldots \cup a_n) \ne \emptyset$ and $b = (b_1 \cup \ldots \cup b_n) \ne \emptyset$.

  • (c) $P(a_n|a) \ge \ldots \ge P(a_1|a)$ and $P(b_n) \ge \ldots \ge P(b_1|b)$.

  Since finite additivity entails that $P(a_1|a) + \ldots + P(a_n|a) = P(a|a) = 1$, it follows from (c) that $P(a_1|a) \le \frac{1}{n}$. For the same reason, $P(b_1|b) \le \frac{1}{n}$. Since $P(b_n|b) \ge P(b_1|b)$, it follows that $P(b_n|b) \ge \frac{1}{n} \ge P(a_1|a)$.

Fairness and Independence place stricter requirements on almost representability than Koopman’s axioms in that not every conditional probability function satisfies their probabilistic analogues. Nonetheless, it is easy to see that P does. In particular, by definition of P:

• Probabilistic Fairness For every $i, j \in \mathbb {N}$: $P(V_i) = P(V_j)$ and $P(H_i) = P(H_j)$.

• Probabilistic Independence For every $i, j \in \mathbb {N}$: $P(H_j|V_i) = P(H_j)$ and $P(V_i|H_j) = P(V_i)$.

Since P satisfies probabilistic analogues of Koopman’s axioms, Fairness, and Independence, it follows that $\succeq$ is almost representable by P.