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:

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