Help with conditional expectation on the circle

Question

Let $p >1$ a integer, $X = \mathbb{R} / \mathbb{Z}$ and $\mu\colon \mathcal{B}\to [0,1]$ a probability measure on the Borel subsets of $X$ which is $T \colon X \ni x \to (px \text{ mod }1)$ invariant. I need to find a formula for $$E_\mu(f \ | \ T^{-n}\mathcal{B}) $$ for all $f\in L^1(\mu)$.

The only thing I know is that $E_\mu(f \ | \ T^{-n}\mathcal{B}) = g \circ T^n$ for some measurable function $g \colon X \to \mathbb{R}$.

Any help will be appreciated.

Answer 1

For every $x$ in $X$, let $X_n(x)=\{y\in X\mid T^ny=x\}$, then, if $\mu$ is the Lebesgue measure, $$E_\mu(f \ | \ T^{-n}\mathcal{B})(x) =\frac1{\#X_n(x)}\sum_{y\in X_n(x)}f(y),$$ that is, $$E_\mu(f \ | \ T^{-n}\mathcal{B})(x)=\frac1{p^n}\sum_{k=1}^{p^n}f\left(\frac{x+k}{p^n}\right).$$

Answer 2

Such a formula would not exist in any "resonable way" as this expected value is basically the limiting value to which the ergodic averages $\frac{1}{N}\sum_{n=0}^{N-1}f(p^{n}.x)$ converge.

Breaking up the space according to the ergodic decomposition of $\mu$ would yeild a suitable decomposition of the expected value as integral over the ergodic measures. In case of the Lebesgue measure (which may or may not be present in the decomposition) you get the formula that Did wrote (which would converge to the integral a.s.). In other cases, it's hard to know (and even impossible) due to the Bernoullicity of the system $(\mathbb{T}^1,\times p)$, and the plethora of invariant measures which exists (in principle, you can cook up a measure of any given Hausdorff dimension).

Nevertheless, on each ergodic component, the function must be constant as it will be $\times p$-invariant function.

Answer 3

In the general case where we have no further information on $\mu$, one can still say something.

Write $X$ as a finite disjoint union $X = \cup_i X_i$, on which the restriction of $T$ to $X_i$ is a bijective and bimeasurable map $T_i : X_i \to X$, ie $X_i = [\frac{i}{p}, \frac{i+1}{p})$ for $i=0, \ldots, p-1$.

Let $\mu_i$ be the restriction of $\mu$ to $X_i$, and $\nu_i$ the push-forward of $\mu_i$ under $T_i$, that is $\nu_i(A) = \mu(T^{-1}(A) \cap X_i)$.

$\nu_i$ is absolutely continuous with respect to $\mu$, and we denote by $h_i = \frac{ d\nu_i}{d\mu}$ its Radon-Nykodim derivative.

Then, setting $g = \sum_i h_i . (f \circ T_i^{-1})$, we have $E_\mu(f \ | \ T^{-1}\mathcal{B}) = g \circ T$.

When $\mu$ is Lebesgue, we have $\nu_i = \frac 1 p \mu$, whence the formula given in Did's answer, up to the correction I pointed out in the comment.

PS : The same argument works for a larger class of maps, and for measures that are not necessarily invariant, but merely non-singular.