Given a field $F$ of subsets of $\Omega$, we can define full conditional probabilities to be a function $P:F\times (F-\{ \varnothing \}) \to [0,1]$ such that:

  1. $P(-|B)$ is a finitely-additive probability function for each $B\in F-\{\varnothing\}$
  2. $P(A|B)P(B|C)=P(A|C)$ if $A\subseteq B\subseteq C$.

Suppose $G$ is a group acting on $\Omega$ and $F$ is $G$-invariant. Then there are two natural ways of defining $G$-invariance for $P$. Strong $G$-invariance is defined by $P(gA|B)=P(A|B)$ whenever $g\in G$ and $A\cup gA \subseteq B$. Weak $G$-invariance is defined by $P(gA|gB)=P(A|B)$ for all $A,B,g$. It's not hard to prove that strong $G$-invariance implies weak $G$-invariance. (The easiest way for me is to go through exchange rates. More on those in a sec.)

Main question: Does weak $G$-invariance imply strong $G$-invariance?

Armstrong in Prop. 1.3 in his paper in this volume advertises a positive answer, but as far as I can see--though maybe someone can show me otherwise--the proof offered just plain doesn't work. (It only shows that if $P$ is weakly $G$-invariant, then the associated exchange rate $r(A,B)=P(A|A\cup B)/P(B|A\cup B)$ (where $1/0=\infty$ and $1/\infty=0$) satisfies $r(gA,gB)=r(A,B)$. But what we need to show is that $r(gA,A)=1$.)

The answer is positive in the special case where $G$ is generated by elements of finite order. (It's easiest to work with the associated exchange rate $r$. This satisfies $r(A,B)r(B,C)=r(A,C)$ whenever the left-side is defined. Then weak and strong invariance for $P$ are equivalent to the universal conditions $r(A,B)=r(gA,gB)$ and $1=r(gA,A)$, respectively. If $g^n=e$ then $1=r(A,A)=r(A,gA)r(gA,g^2 A)\dots r(g^{n-1}A,g^n A)$ if the rhs is defined, and by weak invariance all the factors are equal and hence must all equal $1$.) So there won't be any counterexamples for finite $\Omega$.

Here are two cases I've thought a fair amount about that might lead to a counterexample, but I haven't succeeding in getting a counterexample. The cases are of independent interest to me.

Case 1: Let $\Omega = \{0,1\}^{\mathbb Z}$ and let $F$ be usual product $\sigma$-field. Let $\mathbb Z$ act (additively) on $\Omega$ by shifting the sequences: $(x+\omega)(y) = \omega(y-x)$. Let $H_n = \{ \omega : \forall k\ge n(\omega(k)=1) \}$. It's easy to see that there is no strongly $\mathbb Z$-invariant full conditional probability $P$ on $F$ such that $P(H_n | H_{n+1}) < 1$. But is there a weakly $\mathbb Z$-invariant one? A positive answer implies a negative answer to my main question.

(If in addition to shift-invariance we require reflection invariance, then we have a group that's generated by element of finite order and so there is neither a weak nor a strongly invariant $P$.)

Case 2: Let $\Omega = \mathbb R^2$ and let $F$ be all subsets. If $G$ is all rigid motions (combinations of translations and rotations), there is no strongly $G$-invariant $P$ because of the Sierpinski-Mazurkiewicz paradox. Is there a weakly $G$-invariant $P$? Again, a positive answer implies a negative answer to the main question.

(Again, if we replace $G$ with all isometries, the answer is negative, because all isometries are generated by reflections.)


1 Answer 1


The answer is negative and the counterexample is a lot easier than the two cases I considered in the question. (It would still be nice to have an answer to the two cases.)

Let $\Omega = \mathbb Z$, acting on itself by translation. Let $F$ be the algebra of finite unions of intervals $(a,b) \cap \mathbb Z$ for $a,b \in \mathbb Z \cup \{ \pm \infty \}$. Set: $$ P(A|B) = \begin{cases} 0&\text{ if }\sup B > \sup A\cap B\\ 1&\text{ if }\sup B = \sup A\cap B.\\ \end{cases} $$

Clearly $P$ is weakly invariant. But it's not strongly invariant. For let $B = (-\infty,1] \cap \mathbb Z$ and let $A=(-\infty,0] \cap \mathbb Z$. Then $P(1+A|B)=1$ but $P(A|B)=0$.

It remains to show that $P$ is a full conditional probability. Fix a non-empty $B\in F$. Clearly $P(B|B)=1$. If $\sup B = \infty$, then $P(A|B)$ is 1 iff $A$ contains an interval going off to infinity on the right and otherwise $P(A|B)$ is zero. No two sets $A$ with $P(A|B)=1$ are disjoint, so $P(-|B)$ is additive on $F$. If $\sup B < \infty$, then $P(-|B)$ is a probability measure with all its weight at $\sup B$. So in either case we have condition 1. And 2 follows from the fact that if $\sup A < \sup C$ then $\sup A < \sup B$ or $\sup B < \sup C$.

More generally, the following conditions are equivalent:

  1. Weak $G$-invariance and strong $G$-invariance are equivalent for all full conditional probability spaces $(\Omega,F,P)$ where $G$ acts on $\Omega$ and $F$ is $G$-invariant.
  2. $G$ has no non-trivial order preserving action on a totally-ordered set ($gx \le gy$ iff $x\le y$)
  3. for every order-preserving action of $G$ on a totally preordered set $(X,\le)$ we have $x \le gx \le x$ for all $x,g$
  4. $G$ has no non-trivial left-invariant total preorder

The equivalence of 2 and 4 is proved by Yves Cornullier here in response to a question of mine.

Trivially, not-2 implies not-3. And not-3 implies not-2 since if we have an order-preserving action on a totally preordered set $X$ where for some $x$ and $g$ we have $x \not \le gx$ (the failure of $gx \le x$ comes to the same thing--just replace $g$ with $g^{-1}$), then we will have a non-trivial order-preserving action on the largest totally ordered quotient of $X$.

Suppose not-2. Then there is an order-preserving non-trivial action of $G$ on a totally ordered set $X$. Let $F$ be an algebra generated by intervals as before. Let $$ P(A|B) = \begin{cases} 0&\text{ if }\exists x\in B\text{ with }x > A\cap B\\ 1&\text{ otherwise,}\\ \end{cases} $$ where $x>A$ just means that $x>y$ for all $y\in A$. As before, $P$ is weakly $G$-invariant but not strongly so, contrary to 1.

On the other hand, suppose $P$ is a counterexample to 1, i.e., weakly but not strongly $G$-invariant. Let $r$ be the associated exchange rate. Define the total preorder $\le$ on $F$ by $A \le B$ iff $r(A,B) \le 1$. (This is equivalent to $P(A|A\cup B) \le P(B|A\cup B)$.) Then $G$ acts on $F$ in the obvious way, and the action is order-preserving by weak invariance. Suppose we don't have strong invariance. Then for some $A$ we have $r(A,gA) \ne 1$. But then either $A < gA$ or $gA < A$, and so we have not-3.


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