Non-classical probabilities invariant under symmetries

Abstract

Classical real-valued probabilities come at a philosophical cost: in many infinite situations, they assign the same probability value—namely, zero—to cases that are impossible as well as to cases that are possible. There are three non-classical approaches to probability that can avoid this drawback: full conditional probabilities, qualitative probabilities and hyperreal probabilities. These approaches have been criticized for failing to preserve intuitive symmetries that can be preserved by the classical probability framework, but there has not been a systematic study of the conditions under which these symmetries can and cannot be preserved. This paper fills that gap by giving complete characterizations under which symmetries understood in a certain “strong” way can be preserved by these non-classical probabilities, as well as by offering some results to make it plausible that the strong notion of symmetry here is the right one. Philosophical implications are briefly discussed, but the main purpose of the paper is to offer technical results to help make further philosophical discussion more sophisticated.

Appendix: Construction of highly skewed weakly invariant probabilities

To prove Propositions 2 and 3, we use the methods of West (2020). Let $\mathcal {B}$ be the (real) vector space of all bounded functions from ${\mathbb {Z}}$ to ${\mathbb {R}}$ (i.e., functions f such that there is a real M such that for all x we have $|f(x)| <M$). Let $\mathcal {M}$ be the subset of $\mathcal {B}$ consisting of non-negative functions that are strictly positive at at least one point of ${\mathbb {Z}}$ and that are finitely supported, i.e., are zero except at finitely many points. For two functions f and g in $\mathcal {B}$, define the convolution $f*g$ by:

$$\begin{aligned} (f*g)(x) = \sum _{y=-\infty }^\infty f(y)g(x-y), \end{aligned}$$

whenever this sum is defined. The convolution will always be defined and a member of $\mathcal {B}$ when one of the functions is in $\mathcal {B}$ and the other is finitely supported. It is easy to check that convolution is commutative on $\mathcal {M}$ (this uses the commutativity of $({\mathbb {Z}},+)$), and that we have the associativity property $a*(\phi *\psi )=(a*\phi )*\psi$ whenever $a\in \mathcal {B}$ and $\phi ,\psi \in \mathcal {M}$. Observe that if $\delta _x$ is the function on ${\mathbb {Z}}$ that is zero except at $x\in {\mathbb {Z}}$ where it is equal to one, then $f*\delta _0 = f$.

Define the relation $\sim$ on $\mathcal {B}$ by $a\sim b$ if and only if $a * \phi = b * \psi$ for some $\phi$ and $\psi$ in $\mathcal {B}$. Clearly, $\sim$ is reflexive and symmetric. It is also transitive. For if $a * \phi = b * \psi$ and $b * \zeta = c * \eta$, then

$$\begin{aligned} a*\phi *\zeta = b * \psi * \zeta = b*\zeta *\psi = c*\eta *\psi , \end{aligned}$$

and so $a\sim c$.

Say that a decent cone is a subset C of $\mathcal {B}$ such that:

• (DC1) if $a\sim b$, then $a\in C$ if and only if $b\in C$

• (DC2) if $a,b\in C$ and $\lambda ,\mu \ge 0$, then $\lambda a + \mu b \in C$.

• (DC3) if a is everywhere non-negative, then $a\in C$

• (DC4) if a is everywhere non-positive, and not identically zero, then $a\not \in C$.

Note that it follows from (DC1) and the fact that $\mathcal {M}$ contains a convolutional identity, namely $\delta _0$, that C is closed under convolution with members of $\mathcal {M}$.

Let $-C = \{ -a : a\in C \}$.

Lemma 3

Every decent cone $C_0$ can be extended to a decent cone $C\supseteq C_0$ such that $C\cup -C=\mathcal {B}$ and $C\cap -C = C_0\cap -C_0$.

I will call any such cone C a completion of $C_0$.

Proof of Lemma 3

By Zorn’s Lemma, all we need to do is to suppose that $C_0$ is a decent cone such that $C_0\cup -C_0 \subset \mathcal {B}$, and show there is a decent cone $C_1$ such that $C_0\subset C_1$ and $C_1\cap -C_1 = C_0\cap -C_0$.

To that end, fix $c\notin C_0\cup -C_0$. Let $C_1$ be the set of all functions b such that $b*\phi = c*\psi + d$ for some $d \in C_0$, $\phi \in \mathcal {M}$ and $\psi \in \mathcal {M}^* = \mathcal {M}\cup \{0\}$. Since $0\in C_0$ by (DC3) and $c \in C_1$, we have $C_0 \subset C_1$.

For (DC1), suppose $b\in C_1$ is such that $b*\phi = c*\psi + d$, with $d\in C_0$, $\phi \in \mathcal {M}$ and $\psi \in \mathcal {M}^*$, and suppose that $b*\zeta = b'*\zeta '$ for $\zeta ,\zeta '\in \mathcal {M}$. Then $b'*\zeta '*\phi =b*\zeta *\phi =b*\phi *\zeta =c*\psi *\zeta +d*\zeta$, using the commutativity and associativity properties of our convolution. Since $C_0$ is closed under convolutions with members of $\mathcal {M}$, we have $d*\zeta \in C_0$, and $b'\in C_1$.

For (DC2), note that clearly $C_1$ is closed under multiplication by a non-negative scalar, so all we need to do is to show that it is closed under addition. Suppose $b*\phi = c*\psi + d$ and $b'*\phi ' = c*\psi ' + d'$, with $d,d'\in C_0$, $\phi ,\phi '\in \mathcal {M}$ and $\psi ,\psi '\in \mathcal {M}^*$. Then:

$$\begin{aligned} (b+b')*\phi *\phi '&= b*\phi *\phi ' + b'*\phi '*\phi \\&= c*\psi *\phi '+d*\phi ' + c*\psi '*\phi +d'*\phi \\&= c*(\psi *\phi '+\psi '*\phi )+(d*\phi '+d'*\phi ), \end{aligned}$$

and so $b+b'$ is in $C_1$ as $C_0$ is closed under convolution with members of $\mathcal {M}$ and under addition.

Condition (DC3) holds for $C_1$ as it holds for $C_0$.

Next we need to show that $C_1\cap -C_1 \subseteq C_0\cap -C_0$ (the other inclusion is trivial). Suppose that $b \in C_1\cap -C_1$. Thus, $b*\phi = c*\psi + d$ and $-b*\phi ' = c*\psi ' + d'$ for some $d,d'\in C_0$, $\phi ,\phi '\in \mathcal {M}$ and $\psi ,\psi '\in \mathcal {M}^*$. Then

$$\begin{aligned} c*\psi *\phi ' = b*\phi *\phi ' - d*\phi ' \end{aligned}$$

and

$$\begin{aligned} c*\psi '*\phi = -b*\phi '*\phi - d'*\phi . \end{aligned}$$

Since $\phi *\phi '=\phi '*\phi$, adding these two equalities we get

$$\begin{aligned} c*(\psi *\phi '+\psi '*\phi ) = -d*\phi ' - d'*\phi . \end{aligned}$$

If at least one of $\psi$ and $\psi '$ is not identically zero, it follows that $-c\sim d*\phi '+d'*\phi$, and hence that $c \in -C_0$, contrary to our assumptions. If both $\psi$ and $\psi '$ are identically zero, then it follows d and $d'$ are identically zero. In that case $b*\phi = d$ and $-b*\phi ' = d'$, so b and $-b$ are both members of $C_0$, and hence $b\in C_0\cap -C_0$.

Finally, suppose that a is non-positive but not identically zero. Then $-a\in C_1$ by (DC3). Hence if $a\in C_1$, we have $a\in C_1\cap -C_1 = C_0\cap -C_0$, which contradicts the fact that $C_0$ satisfies (DC4). $\square$

Lemma 4

Suppose that C is a subset of $\mathcal {B}$ that is closed under right convolution with members of $\mathcal {M}$ (i.e., if $c\in C$ and $\psi \in \mathcal {M}$, then $c*\psi \in C$), is closed under addition, and satisfies (DC3) and (DC4). Then $C^* = \{ c \in \mathcal {B} : \exists \phi \in \mathcal {M}(c*\phi \in C) \}$ is a decent cone.

Proof of Lemma 4

To check condition (DC1), suppose $c*\psi = c'*\psi '$ for $c\in \mathcal {B}$, $c'\in C^*$ and $\psi ,\psi '\in \mathcal {M}$. Fix $\phi \in \mathcal {M}$ such that $c'*\phi \in C$. Then $c*\psi *\phi = c'*\psi '*\phi =c'*\phi *\psi '$. Since C is closed under right convolution with members of $\mathcal {M}$, we have $c'*\phi *\psi '\in C$, and so $c\in C^*$ as desired.

Note that C is closed under multiplication by non-negative scalars since multiplication by a positive scalar $\lambda$ is just convolution with $\lambda \delta _0$, while $0\in C$. It follows that $C^*$ is closed under multiplication by non-negative scalars. To check (DC2), we need only check that $C^*$ is closed under addition. Suppose $a,b\in \mathcal {B}$, $a',b'\in C$ and $\phi ,\phi ',\psi ,\psi '\in \mathcal {M}$ are such that $a * \phi = a' * \phi '$ and $b * \psi = b' * \psi '.$ Then:

$$\begin{aligned} \begin{aligned} (a + b) * \phi *\psi&= a*\phi *\psi + b * \phi *\psi \\&= a'*\phi '*\psi + b * \psi *\phi \\&= a'*\phi '*\psi + b'*\psi '*\phi . \end{aligned} \end{aligned}$$

The right-hand-side is in C, so $a+b$ must be in $C^*$.

That $C^*$ satisfies (DC3) follows from the fact that $C\subseteq C^*$ as $a*\delta _0 = a$. It remains to show that (DC4) is satisfied. Suppose $a \in C^*$ is non-positive but not identically zero. Then $a*\phi \in C$ for some $\phi \in \mathcal {M}$. But $a*\phi$ will also be non-positive but not identically zero, contradicting the fact that C satisfies (DC4). $\square$

Let $1_A$ be the indicator function of a set A, i.e., the function that is 1 on A and 0 outside A.

Lemma 5

Let C be a decent cone such that $C\cup -C=\mathcal {B}$. Stipulate that $A\lessapprox B$ just in case $1_B-1_A \in C$. Then $\lessapprox$ is a total regular qualitative ${\mathbb {Z}}$-invariant probability.

Proof of Lemma 5

Reflexivity of $\lessapprox$ follows from the fact that $0\in C$. Transitivity follows immediately from the fact that a decent cone is closed under addition. Additivity follows from the fact that $1_B-1_A = 1_{B-A}-1_{A-B}$. Since every non-negative function in $\mathcal {B}$ is a member of $C_0\subseteq C$, it follows that $\varnothing \lessapprox A$ for all A.

To prove regularity, suppose A is nonempty. If $A\lessapprox \varnothing$, then $1_\varnothing - 1_A = -1_A$ is in C, contradicting (DC4).

Totality follows from the fact that $C\cup -C = \mathcal {B}$.

Observe that $(1_B - 1_A) * \delta _x = 1_{x+B} - 1_{x+A}$. But a decent cone is invariant under right convolution with members of $\mathcal {M}$ by (DC1), so $x+A \lessapprox x+B$ if and only if $A\lessapprox B$, and thus we have ${\mathbb {Z}}$-invariance. $\square$

West’s result on the existence of a total regular qualitative ${\mathbb {Z}}$-invariant probability then follows by letting $C_0$ be the collection of all functions in $\mathcal {B}$ that are everywhere non-negative, letting C be the completion of $C_0^*$, and applying the above lemmas.

Now say that a function $a\in \mathcal {B}$ has property $X_1$ provided that it is negative only in finitely many places and that $\sum a=\sum _{n=-\infty }^\infty a_n\ge 0$. Say it has property $X_2$ provided that for infinitely many $n>0$ we have $a(n)>0$ and there are only finitely many $n>0$ such that $a(n)<0$. The sum of two functions with property $X_i$ has property $X_i$, for $i=1,2$, and the sum of a function with $X_1$ and a function with property $X_2$ has property $X_2$. Having property $X_i$ is closed under right-convolution with a member of $\mathcal {M}$: in the case of $X_2$, this uses the fact that $\sum (a*\phi )=(\sum a)(\sum \phi )$ if a is negative in only finitely many places and $\phi \in \mathcal {M}$.

Say that a function has property X provided it has $X_1$ or $X_2$. Having property X is thus closed under addition and right-convolution with $\mathcal {M}$.

Proof of Proposition 2

Let $C_0$ be the set of functions $a\in \mathcal {B}$ that have property X. Note that any non-negative function has property $X_1$ and hence is in $C_0$, and the only non-positive function that can have X is zero.

Let C be a completion of $C_0^*$. Define $\lessapprox$ as in Lemma 5. This will be a regular total G-invariant qualitative probability.

Let’s examine $C_0^* \cap -C_0^*$. Suppose $a \in C_0^* \cap -C_0^*$. Then $a*\phi$ and $-a*\psi$ have property X for some $\phi ,\psi \in \mathcal {M}$. Hence, so do $a*\phi *\psi$ and $-a*\psi *\phi$ since property X is closed under right $\mathcal {M}$ convolution. Note that $\phi *\psi =\psi *\phi$. Let $b=a*\phi *\psi$. Then b and $-b$ both have property X. There are three cases to consider: both functions have $X_1$, both have $X_2$, and one has $X_1$ while the other has $X_2$. It is clearly impossible that both b and $-b$ have $X_2$. And if a function has $X_1$, then its negative is positive in only finitely many places, and so that negative cannot have $X_2$. Thus, the remaining case is where both functions have $X_1$. The only way this can be is if both functions sum to zero and are finitely supported. But if $a*\phi *\psi$ is finitely supported, so is a, and if one sums to zero, so does the other since $\sum a = (\sum a)(\sum \phi )(\sum \psi )$. Thus, any function in $C_0^*\cap -C_0^*$ is finitely supported and sums to zero. Conversely, any finitely supported function that sums to zero has $X_1$ and so does its negative.

So, $C_0^*\cap -C_0^*=C\cap -C$ is the set of all finitely supported functions that sum to zero.

Suppose B has infinitely many positive members and A does not. Then $1_B-1_A$ has property $X_2$, and hence is in C, and so $A\lessapprox B$. If we also had $B\lessapprox A$, we would have $1_B-1_A$ in $C\cap -C$, which would require $1_B-1_A$ to be finitely supported, which it’s not. Thus, $A<B$.

Finally, suppose A and B are finite and of the same cardinality. Then $1_B-1_A$ and $1_A-1_B$ both have $X_1$, and so $A\approx B$. $\square$

Now suppose C is a decent cone such that $C\cup -C=\mathcal {B}$. For any $a,b\in \mathcal {B}$, define $a\lessapprox b$ provided that $b-a\in C$. This is a total vector space preorder on $\mathcal {B}$ (i.e., it’s a total preorder such that if $a\lessapprox b$, $\lambda \ge 0$ and c in the vector space, then $\lambda a + c \le \lambda b + c$).

We now need a useful lemma about total vector space preorders. Let

$$\begin{aligned} \gamma (a,b) = \inf \{ \alpha \in {\mathbb {R}}: a \lessapprox \alpha b \} \end{aligned}$$

if $b\not \approx 0$. If $b\approx 0$ and $a\not \approx 0$, let $\gamma (a,b)=\infty$. And make $\gamma (a,b)$ be undefined if $a\approx 0$ and $b\approx 0$.

Lemma 6

If $c\not \approx 0$, then $\gamma (a+b,c)=\gamma (a,c)+\gamma (b,c)$. Moreover, if $0\lessapprox a,b,c$, then

$$\begin{aligned} \gamma (a,c) = \gamma (a,b)\gamma (b,c) \end{aligned}$$

whenever the right-hand-side is well-defined. If $b\not \approx 0$, then $\gamma (b,b)=1$. Finally, if $0\lessapprox a$ and $0<b$, then $0\le \gamma (a,b)$.

Proof of Proposition 3

Let C be as in the proof of Proposition 2. By abuse of notation, use $\lessapprox$ both for the qualitative probability in that proof and for our vector space preorder. We then have $A\lessapprox B$ if and only if $1_A\lessapprox 1_B$. It follows from the non-negativity and regularity of the qualitative probability that $0\lessapprox 1_A$ for every A, with the inequality being strict if A is nonempty.

Define $c(A,B)=\gamma (1_A,1_B)$. It follows from Lemma 6 that this is a coherent exchange rate. Then $P_c$ will be a full conditional probability. Moreover, $\gamma (a*\phi ,b*\phi )=\gamma (a,b)$ for all $\phi \in \mathcal {M}$ by (DC1). Letting $\phi =\delta _x$, we see that c satisfies the weak ${\mathbb {Z}}$-invariance condition $c(x+A,x+B)=c(A,B)$. It follows that $P=P_c$ is weakly ${\mathbb {Z}}$-invariant by Lemma 1.

Next, suppose m and n are distinct integers (the case $m=n$ is trivial). Let $a_\alpha = 1_{\{m\}} - \alpha \cdot 1_{\{m,n\}}$. If $\alpha >1/2$, then $-a_\alpha$ has property $X_1$. It follows that $-a_\alpha \in C$, so $1_{\{m\}} \lessapprox \alpha \cdot 1_{\{m,n\}}$. Thus, $\gamma (1_{\{m\}},1_{\{m,n\}}) \le 1/2$. On the other hand, if $\alpha <1/2$, then $a_\alpha$ has property $X_1$. It follows that $a_\alpha \in C$. The only way we could have $-a_\alpha$ in C as well is if $a_\alpha \in C\cap -C$, which according to the proof of Proposition 2 would require that $a_\alpha$ sum to zero, which it does not for $\alpha <1/2$. So, we do not have $1_{\{m\}} \lessapprox \alpha \cdot 1_{\{m,n\}}$, and hence $\gamma (1_{\{m\}},1_{\{m,n\}})\ge 1/2$. Thus, $\gamma (1_{\{m\}},1_{\{m,n\}})=1/2$, and it follows that $P(\{m\}\mid \{m,n\})=1/2$. Swapping m and n we get that $P(\{m\}\mid \{m,n\})=P(\{n\}\mid \{m,n\})$.

Now, suppose that A has only finitely many positive integers and B has infinitely many. Then $P(A\mid A\cup B)=c(A,A\cup B)=\gamma (1_A,1_{A\cup B})$. Fix any $\alpha >0$. Then $\alpha \cdot 1_{A\cup B}-1_A$ has property $X_2$, and hence is in C, so $1_A \lessapprox \alpha \cdot 1_{A\cup B}$. Thus, $\gamma (1_A,1_{A\cup B}) \le 0$, and by Lemma 6 we have $\gamma (1_A,1_{A\cup B})=0$. Thus, $P(A\mid A\cup B)=0$, and so we must have $P(B\mid A\cup B)=1$ by finite additivity. $\square$

Proof of Lemma 6

For convenience, write:

$$\begin{aligned} \gamma ^+(a,b) = \gamma (a,b) \end{aligned}$$

and

$$\begin{aligned} \gamma ^-(a,b) = \sup \{ \alpha \in R : \alpha b \lessapprox a \}, \end{aligned}$$

whenever $b\not \approx 0$, with the expressions undefined otherwise. Observe that $a\lessapprox \alpha b$ if and only if $(-\alpha ) b\lessapprox -a$, so we have the duality

$$\begin{aligned} \gamma ^+(a,b) = -\gamma ^-(-a,b). \end{aligned}$$

It is very easy to see that $\gamma ^+(\cdot ,b)$ is subadditive:

$$\begin{aligned} \gamma ^+(a+a',b)\le \gamma ^+(a,b)+\gamma ^+(a',b) \end{aligned}$$

and that $\gamma ^-(\cdot ,b)$ is superadditive:

$$\begin{aligned} \gamma ^-(a+a',b)\ge \gamma ^-(a,b)+\gamma ^-(a',b). \end{aligned}$$

In particular, if $b\not \approx 0$, we have $0=\gamma ^+(0,b)\le \gamma ^+(a,b)+\gamma ^+(-a,b)$ for all a. Hence, $-\gamma ^+(-a,b)\le \gamma ^+(a,b)$. Suppose $-\gamma ^+(-a,b)<\gamma ^+(a,b)$. Choose $\alpha \in (-\gamma ^+(-a,b),\gamma ^+(a,b))$. Since $\alpha <\gamma ^+(a,b)$, we do not have $a\lessapprox \alpha b$, and hence we have $\alpha b < a$ by totality. Thus, we have $-a < -\alpha b$, so $\gamma ^+(-a,b)\le -\alpha$, or $\alpha \le -\gamma ^+(-a,b)$, a contradiction to the choice of $\alpha$. Hence,

$$\begin{aligned} \gamma ^+(-a,b)=-\gamma ^+(a,b), \end{aligned}$$

as long as $b\not \approx 0$. The same is true for $\gamma ^-$ by the earlier proved duality between $\gamma ^+$ and $\gamma ^-$. Therefore,

$$\begin{aligned} \gamma ^+(a,b)=-\gamma ^+(-a,b)=\gamma ^-(a,b), \end{aligned}$$

if $b\not \approx 0$. It follows that if $b\not \approx 0$, then $\gamma (\cdot ,b)$ is both subadditive and superadditive, and hence it is additive.

Now, if $0\lessapprox a$ and $0<b$, then $\gamma (a,b)=\gamma ^+(a,b)=\gamma ^-(a,b)$. If $0<a$ and $0\approx b$, then $\gamma (a,b)=\infty$. And if $0\approx a$ and $0\approx b$, then $\gamma (a,b)$ is undefined.

I claim that if $0\lessapprox a,b,c$, and $\gamma (a,b)\gamma (b,c)$ is defined, then $\gamma (a,c)=\gamma (a,b)\gamma (b,c)$.

To see this, consider first the case where $0<b$ and $0<c$. Then if $a \lessapprox \alpha b$ and $b\lessapprox \beta c$, we have $a\lessapprox \alpha \beta c$. It follows that $\gamma ^+(a,c) \le \gamma ^+(a,b)\gamma ^+(b,c)$. Moreover, if $\alpha b \lessapprox a$ and $\beta c\lessapprox b$, then $\alpha \beta c \lessapprox a$, and so $\gamma ^-(a,c) \ge \gamma ^-(a,b)\gamma ^-(b,c)$. But since $\gamma ^+(u,v)=\gamma ^-(u,v)$ whenever $0<v$, we thus have $\gamma (a,c)=\gamma (a,b)\gamma (b,c)$.

Next, consider the case where $0\approx c$. Then $\gamma (b,c)$ is either undefined or equal to infinity. If it is undefined, we don’t need to prove anything. So suppose it’s equal to infinity. For $\gamma (a,b)\gamma (b,c)$ to be defined, we must have $\gamma (a,b)$ defined and strictly positive. This will happen only if $0<a$. But in that case $\gamma (a,c)=\infty$, and so we have $\gamma (a,c)=\infty =\gamma (a,b)\gamma (b,c)$.

For the last case, suppose that $0<c$ but $0\approx b$. For $\gamma (a,b)$ to be defined, we must have $0<a$. In that case $\gamma (a,b)=\infty$. For the product $\gamma (a,b)\gamma (b,c)$ to be defined, we must have $0<\gamma (b,c)$, which is impossible if $b\approx 0$.

Next, suppose $0\not \approx b$. Then $\gamma ^+(b,b) \le 1$ and $\gamma ^-(b,b) \ge 1$ by reflexivity of $\lessapprox$, hence $\gamma (b,b)=\gamma ^+(b,b)=\gamma ^-(b,b)=1$.

Finally, if $0\lessapprox a$ and $0<b$, then $0\le \gamma ^-(a,b)=\gamma (a,b)$. $\square$