De Finetti's theorem says that an exchangeable sequence of random variables $X_i$ is a mixture of i.i.d. random variables. In other words, if $\mu$ is a measure on $\mathbb{R}^\infty$ that is invariant under exchanging finitely many coordinates (a symmetric measure), then there is some probability measure $\eta$ on probability measures such that $\mu = \int \nu^\infty \, d \eta(\nu)$.

Further, I know the following.

  1. The product measures of the form $\nu^{\infty}$ are the extreme points for the convex set of symmetric measures. They are also ergodic with respect to the group of transformations which exchange finitely many coordinates. So $\mu = \int \nu^\infty \, d \eta(\nu)$ is an ergodic decomposition.

  2. For $\mu$-a.e. $x=\{x_i\}_{i\in\mathbb{N}}\in \mathbb{R}^\infty$, there is some probability measure $\nu_x$ on $\mathbb{R}$ such that for all measurable sets open balls $A \subseteq \mathbb{R}$,

    (A) $\quad$ ${\displaystyle \lim_{k\rightarrow\infty} \frac{1}{k} \sum_{i<k} \mathbf{1}_A(x_i) = \nu_x(A) }$.

    Moreover, if $P^n_k$ is the set of all injective functions $\pi \colon [n] \rightarrow [k]$, then for all bounded continuous functions $f\colon \mathbb{R}^n \rightarrow \mathbb{R}$,

    (B) $\quad$ ${\displaystyle \lim_{k\rightarrow\infty} \frac{1}{|P^n_k|} \sum_{\pi \in P^n_k} f(x_{\pi(0)},\ldots ,x_{\pi(n-1)}) = \int_{\mathbb{R}^n} f\, d \nu_x^n}$.

  3. Equations (A) and (B) and de Finetti's theorem can all be proved using reverse martingales. Indeed, $M_{-k}(x) = \frac{1}{|P^n_k|} \sum_{\pi \in P^n_k} f(x_{\pi(0)},\ldots ,x_{\pi(n-1)})$ is a reverse martingale.

My questions are as follows.

  • To what extent are equations (A) and (B) instances of some variant of the pointwise ergodic theorem? (I guess (A) is just Birkoff's pointwise ergodic theorem with the shift map---although I am not sure why the shift map comes in. But (B) is not so clear to me.)

  • When may an ergodic average be represented as a reverse martingale?

  • Similarly, for which types of pointwise ergodic theorems and ergodic decompositions is there a proof using reverse martingales?

Pointers to any relevant references would be helpful.

  • $\begingroup$ It seems there's something wrong with the order of the quantifiers in 2. For instance, if the marginal distribution of $X_1$ is non-atomic, then the limit in (A) won't exist for all $A$ (if $A$ is allowed to depend on $x$, then $A$ could for example pick out certain individual values of the $x_i$ but not others in an inconvenient way). Do you mean: for all $A$, then for $\mu$-a.e. $x$, ...? $\endgroup$ Apr 10, 2012 at 20:08
  • $\begingroup$ @James Martin: I think it is correct. This isn't true for all x, but for a.e. x. In other words, there are $x$'s so random that the $x_i$ are "nicely" distributed in a way that gives the measure $\nu_x$. For example, if the $x_i$ were uniformly distributed, then (A) would hold for the $\nu_x$ equal to the Lebesgue measure. This is similar to the concept of being a generic in ergodic theory (see terrytao.wordpress.com/2008/02/04/254a-lecture-9-ergodicity). Let me know if you still disagree. $\endgroup$
    – Jason Rute
    Apr 10, 2012 at 20:29
  • $\begingroup$ @James Martin: Ok, I see your point. I could take $A={x_0,x_1, \ldots}$. I got this from a paper, which in turn, got it from Kallenberg, Probabilistic symmetries and invariance principles, Proposition 1.4: books.google.com/… I must be reading the a.s. in that statement incorrectly. I think it works if I assume $A$ ranges over all open balls. (I could also use continuous functions like equation (B)). $\endgroup$
    – Jason Rute
    Apr 10, 2012 at 21:40
  • $\begingroup$ Agreed - works fine with open balls. $\endgroup$ Apr 10, 2012 at 23:36

2 Answers 2


It was recently pointed out by Bill Johnson in a comment to a question concerning "Fully exchangeable random sequences" that if an infinite sequence of random variables is exchangeable, then it is "fully exchangeable", that is to say its distribution is in fact invariant by any permutation (even if the permutation has no fixed point). The same argument shows that the distribution of an infinite exchangeable sequence of random variables is invariant under the shift map. Hence you can view (A) as the pointwise ergodic theorem for the shift map.


(My understanding of this material has significantly gone up in the months since I asked it, and I will attempt to answer my own question.)

In general, if $(\Omega,\mathcal{B},\mathbb{P},\{T_g\})$ is a measure preserving system where $\{T_g\}$ is a commuting (semi-)group action, then one can consider the limits of averages over any chain of finite subsets $H_1 \subseteq H_2 \subseteq … \subseteq G$, that is averages of the form $$A_n (f) = \frac{1}{|H_n|} \sum_{g \in H_n} f \circ T_g.$$

In this particular case of de Finetti's theorem, we can take $G$ to be finite permutations of $\mathbb{N}$, and $T_g$ to be the transformations on $\Omega = \mathbb{R}^\infty$ associated with that permutation (by permuting the coordinates of $x=(x_n) \in R^\infty$. Then $H_n$ is the set of permutations which only permute the first $n$ coordinates.

Notice, in this particular case, that each $H_n$ is a subgroup and therefore $$\frac{1}{|H_n|} \sum_{g \in H_n} f \circ T_g = \mathbb{E}[f \mid \mathcal{I}nv(H_n) ]$$ where $\mathcal{I}nv(H_n)$ is the sigma-algebra of sets which are $T_g$-invariant for all $g \in H_n$.

Now one has a reverse martingale $\mathbb{E}[f \mid \mathcal{I}nv(H_n) ]$, which converges to $\mathbb{E}[f \mid \mathcal{I}nv(G) ]$, since $\mathcal{I}nv(G) = \bigcap_n \mathcal{I}nv(H_n)$. Hence, the ergodic decomposition $\mu_x$ comes from the conditional probability $\mu_x = \mathbb{P}[ \cdot \mid \mathcal{I}nv(G)] (x)$.

In the de Finetti case, since each $\mu_x$ must be invariant under the permutations of coordinates $\{T_g\}$, it is a product measure of the form $\mu_x = \nu^\infty_x$ for some $\nu_x$.

  1. To answer the first question is seems that yes, equations (A) and (B) are examples of a pointwise ergodic theorem.

  2. To answer the second question, it seems that such ergodic averages can be represented as reverse martingales when the $H_n$ are finite subgroups of $G$. (This leads to an observation that even when $H_n$ are not finite, if they are subgroups, we could define an ergodic average of $f$ over $H_n$ as $\mathbb{E}[ f \mid \mathcal{I}nv(H_n)]$.

  3. As for the third question, many pointwise ergodic theorems can be proved using reverse martingale techniques. The idea is to first work on the group $G$ itself rather than a probability space. In this setting, the ergodic averages become $$A_n (f) (x) = \frac{1}{|H_n|} \sum_{g \in H_n} f(gx)$$ for $x \in G$ and $f \in \ell_1 (G)$. This is reminiscent of the averages in the Lebesgue differentiation theorem---think of $\{gx: g \in H_n\}$ as a ball around $x$.

    These averages can be approximated by reverse martingales on $\ell_1 (G)$: rather than average over a ball around $x$, partition $G$ and average over the part that $x$ is in (if the $H_n$ are subgroups then the parts can be the cosets and the two types of averages are the same!).

    Then on can use the geometry of $G$ to turn a theorem about reverse martingales (for example a maximal theorem) into a theorem about ergodic averages. (This is similar to how one can prove the Lebesgue differentiation theorem using (forward) martingales, with some additional lemmas to handle the geometry.) Finally, one can use the Calderon transfer principle to transfer the result on $\ell_1(G)$ to a result on any measure preserving system with a $G$-action.


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