Correction: Non-classical probabilities invariant under symmetries

1 Correction to: Synthese (2021) 199:8507–8532 https://doi.org/10.1007/s11229-021-03173-w

Let $\mathcal F$ be an algebra of subsets of $\Omega$. A full conditional probability on $\mathcal F$ is a real-valued function on $\mathcal F\times (\mathcal F-\{\varnothing \})$ such that:

• (C1) $P(\cdot \mid B)$ is a finitely additive probability function

• (C2) $P(A\cap B\mid C) = P(A\mid C)P(B\mid A\cap C)$.¹

If G is a group (intuitively, a group of symmetries, such as rigid motions on ${\mathbb {R}}^n$) acting on a set $\Omega ^*$ containing $\Omega$, we say that P is G-invariant provided that $P(gA\mid B)=P(A\mid B)$ whenever A, gA and B are all in $\mathcal F$ with B nonempty, $g\in G$, and $A\cup gA\subseteq B$. (There is no assumption here that $\mathcal F$ is itself G-invariant.)

One of the main theorems in Pruss (2021) characterized when exactly a G-invariant full conditional probability on the powerset ${\mathcal {P}}\Omega$ exists. Unfortunately, the proof of Lemma 2 was erroneous. The proof used the claim

$$\begin{aligned} \frac{ \sum _{\mu \in \mathcal B_B} \mu (A) }{ \sum _{\mu \in \mathcal B_B} \mu (B) }\cdot \frac{ \sum _{\mu \in \mathcal B_C} \mu (B) }{ \sum _{\mu \in \mathcal B_C} \mu (C) } = \frac{ \sum _{\mu \in \mathcal B_C} \mu (A) }{ \sum _{\mu \in \mathcal B_C} \mu (C) }, \end{aligned}$$

which it was erroneously said “follows” from the identity $\frac{\alpha }{\beta }\cdot \frac{\beta }{\gamma }=\frac{\alpha }{\gamma }$.

There does not seem to be a simple fix for this, but there is a new proof using the Rényi order in a way inspired by ideas in Armstrong (1989).²

Lemma 1

Let G act on $\Omega ^*\supseteq \Omega$. Suppose that for every nonempty subset E of $\Omega$, there is a G-invariant finitely additive measure $\mu :{\mathcal {P}}\Omega \rightarrow [0,\infty ]$ with $\mu (E)=1$. Let $\mathcal F$ be a finite algebra on $\Omega$. Then there is a G-invariant full conditional probability on $\mathcal F$.

Proof

All the measures in the proof will be finitely additive. If $\mu$ and $\nu$ are measures on the same algebra, say that $\mu \prec \nu$ provided that for all $A\in \mathcal F$, if $\nu (A)>0$, then $\mu (A)=\infty$. Say that a measure $\mu$ is non-degenerate provided that $0<\mu (A)<\infty$ for some A. Then $\prec$ is known as the Rényi order (Armstrong, 1989; Rényi, 1956) and is a strict partial order on non-degenerate measures.

Choose a G-invariant probability measure $\mu _1$ on $\mathcal F$ (there is one on ${\mathcal {P}}\Omega$, so restrict it to $\mathcal F$).

For $n\ge 1$, supposing we have chosen a G-invariant measure $\mu _n$ on ${\mathcal {P}}\Omega$, let

$$\begin{aligned} E_{n+1} = \bigcup \{ B \in \mathcal F : \mu _n(B) = 0 \}. \end{aligned}$$

Note that $\mu _n(E_{n+1})=0$ since $\mathcal F$ is finite, so $E_{n+1}$ is the largest $\mu _n$-null member of $\mathcal F$. If $E_{n+1}=\varnothing$, let $N=n$, and our construction of $\mu _1,\dots ,\mu _N$ is complete.

If $E_{n+1}$ is nonempty, choose a G-invariant measure $\nu$ on ${\mathcal {P}}\Omega$ with $\nu (E_{n+1})=1$. For $A\in \mathcal F$, let $\mu _{n+1}(A) = \nu (A)$ if $A\subseteq E_{n+1}$ and $\mu _{n+1}(A) = \infty$ otherwise.

I claim that $\mu _{n+1}$ is a G-invariant measure on $\mathcal F$. To check finite additivity, suppose A and B are disjoint members of $\mathcal F$. Then if A or B fails to be a subset of $E_{n+1}$, so does $A\cup B$, and so $\mu _{n+1}(A)+\mu _{n+1}(B)=\infty =\mu _{n+1}(A\cup B)$, and if $A\cup B$ fails to be a subset of $E_{n+1}$, so does at least one of A and B. But if A, B and $A\cup B$ are all subsets of $E_{n+1}$, then $\mu _{n+1}$ agrees with $\nu$ as applied to these sets, and $\nu$ is finitely additive.

It remains to check G-invariance. Suppose that $A,gA\in \mathcal F$. If both A and gA are subsets of $E_{n+1}$, the identity $\mu _{n+1}(A)=\mu _{n+1}(gA)$ follows from the G-invariance of $\nu$. If neither is a subset of $E_{n+1}$, then $\mu _{n+1}(A)=\infty =\mu _{n+1}(gA)$. It remains to consider the case where one of A and gA is a subset of $E_{n+1}$ and the other is not. Without loss of generality, suppose that A is a subset of $E_{n+1}$ and gA is not (in the other case, let $A'=gA$ and $g'=g^{-1}$, so $A'$ is a subset of $E_{n+1}$ and $g'A'$ is not). Since $A\subseteq E_{n+1}$, we have $\mu _n(A)=0$. By G-invariance, $\mu _n(gA)=0$, and so $gA\subseteq E_{n+1}$, and thus the case is impossible.

Next note that that $\mu _{n+1} \prec \mu _n$. For if $\mu _n(A)>0$, then A is not a subset of $E_{n+1}$ and so $\mu _{n+1}(A)=\infty$.

The finiteness of $\mathcal F$ guarantees that the construction must terminate in a finite number N of steps, since we cannot have an infinite sequence of non-degenerate measures on a finite algebra $\mathcal F$ that are totally ordered by $\prec$.

We have thus constructed a sequence of G-invariant measures $\mu _1,\ldots ,\mu _N$ such that $\mu _N\prec \dots \prec \mu _1$. I claim that for any nonempty $A\in \mathcal F$, there is a unique $n=n_A$ such that $0<\mu _n(A)<\infty$. Uniqueness follows immediately from the ordering $\mu _N\prec \dots \prec \mu _1$, so only existence needs to be shown. By our construction, the only $\mu _N$-null set is $\varnothing$, so $\mu _N(A)>0$. Let n be the smallest index such that $\mu _n(A)>0$. If $\mu _n(A)<\infty$, we are done. So suppose $\mu _n(A)=\infty$. We cannot have $n=1$, since $\mu _1$ is a probability measure on $\mathcal F$. Thus, $n>1$. By minimality of n, we must have $\mu _{n-1}(A)=0$. Thus, $A\subseteq E_n$, and so $\mu _n(A)\le \mu _n(E_n)=1$, a contradiction.

Now, for any $(A,B)\in \mathcal F\times (\mathcal F-\{\varnothing \})$, let $P(A\mid B)=\mu _{n(B)}(A\cap B)/\mu _{n(B)}(B)$. Then $P(\cdot \mid B)$ is finitely additive since $\mu _{n(B)}$ is.

Next, suppose we have A, B and C with $A\cap C$ nonempty. If $n(A\cap C)=n(C)$, then let $\mu =\mu _{n(C)}=\mu _{n(A\cap C)}$, so we have

$$\begin{aligned} P(A\mid C)P(B\mid A\cap C)&= \frac{\mu (A\cap C)}{\mu (C)} \cdot \frac{\mu (B\cap A\cap C)}{\mu (A\cap C)} \\&= \frac{\mu (A\cap B\cap C)}{\mu (C)} = P(A\cap B\mid C). \end{aligned}$$

Now suppose that $n(A\cap C)\ne n(C)$ so $\mu _{n(C)}(A\cap C)\notin (0,\infty )$. Since $\mu _{n(C)}(A\cap C)\le \mu _{n(C)}(C) <\infty$, we must have $\mu _{n(C)}(A\cap C)=0$. But then $P(A\mid C)=\mu _{n(C)}(A\cap C)/\mu _{n(C)}(C)=0$ and $P(A\cap B\mid C)=\mu _{n(C)}(A\cap B\cap C)/\mu _{n(C)}(C)=0$, and so both sides of (C2) are zero.

Finally, G-invariance of P follows immediately from G-invariance of the $\mu _n$. $\square$

We then get the following which is the same as the Lemma 2 in Pruss (2021) whose proof was flawed.

Corollary 1

Let G act on $\Omega ^*\supseteq \Omega$. There is a G-invariant full conditional probability on ${\mathcal {P}}\Omega$ if and only if for every nonempty subset E of $\Omega$ there is a G-invariant finitely additive measure $\mu :{\mathcal {P}}\Omega \rightarrow [0,\infty ]$ with $\mu (E)=1$.

Proof

First suppose there is a G-invariant full conditional probability P on ${\mathcal {P}}\Omega$. Then if E were a nonempty paradoxical subset of $\Omega ^*$, we could partition E into disjoint subsets A and B that could be decomposed under the action of G to form all of E, so that $1=P(E\mid E)=P(A\mid E)+P(B\mid E)=P(E\mid E)+P(E\mid E)=2$ by the finite additivity and G-invariance of $P(\cdot \mid E)$. But if E is not a paradoxical subset, then by Tarski’s Theorem (Tomkowicz and Wagon 2016, Cor 11.2) there is a G-invariant finitely additive measure $\mu$ on ${\mathcal {P}}\Omega ^*$ with $\mu (E)=1$, and we can then restrict $\mu$ to ${\mathcal {P}}\Omega$.

Conversely, suppose for every nonempty E there is a $\mu$ as in the statement of the Corollary. For a finite algebra $\mathcal F$ on $\Omega$, let P be a G-invariant full conditional probability on $\mathcal F$ by Lemma 1. Let $P_{\mathcal F}(A\mid B) = P(A\mid B)$ for $(A,B) \in \mathcal F\times (\mathcal F-\{\varnothing \})$ and $P_{\mathcal F}(A\mid B)=0$ for all other $(A,B)\in {\mathcal {P}}\Omega \times ({\mathcal {P}}\Omega -\{\varnothing \})$. The set F of all finite algebras $\mathcal F$ on $\Omega$, ordered by inclusion, is a directed set. Since $[0,1]^{{\mathcal {P}}\Omega \times ({\mathcal {P}}\Omega -\{\varnothing \})}$ is a compact set by the Tychonoff Theorem, there will be a convergent subnet of the net $(P_{\mathcal F})_{\mathcal F \in F}$, and the limit of that subnet then satisfies the conditions for a G-invariant full conditional probability. $\square$