2
$\begingroup$

According to my simulations, it looks like the number of times that the $N$ first iterates $u_0$, $\ldots$, $u_{N-1}$ of the sequence $(u_n)$ defined here meets an interval $I$ is close to $N|I|$ for any initial value $u_0=x$.

This observation is just the motivation of the following question. I don't need to recall the sequence $(u_n)$ in order to state the question.

Because of this observation, one could expect that for almost all $x$,
$$ \lim_{N \to \infty} \sum_{k=0}^{\infty} a_{N,k} {\boldsymbol 1}_{T^kx \in I} = |I| $$ where $T$ is an irrational rotation, and $a_{N,k} = \dfrac{\#\{i | 0 \leq i \leq N-1, K_i = k\}}{N}$ where $K_N=\sum_{i = 0}^{N-1} \epsilon_i2^i$ and ${(\epsilon_n)}_{n \geq 0}$ is a sequence of independent $0$-$1$ symmetric Bernoulli random variables (thus $a_{N,k}=0$ for $k > 2^N-1$ and $\sum_{k=0}^{2^N-1} a_{N,k}=1)$.

Observe that the $a_{N,k}$ are random, so I should also say that the above limit holds for almost all sequences ${(\epsilon_n)}_{n \geq 0}$.

Thus the expectation is that the following limit holds for almost every realisation of the $a_{N,k}$: $$ \sum_{k=0}^{\infty} a_{N,k} f(T^k x) \overset{a.s.}{\to} E(f) $$ for every $f \in L^1$. Is it true ?

I suspect this is true when $T$ has discrete spectrum and $T^{2^n}$ is ergodic for every $n \geq 0$ (i.e. the product of $T$ with the dyadic odometer is ergodic). By the way my simulations yield the same observation when $T$ is an ergodic product of two rotations.

Let me rephrase the problem in terms of an algorithm. The input is a trajectory y(0), y(1), ... of an ergodic stationary process and the output is x(0), x(1), ...:

SET x(0)=y(0)
SET i=0
REPEAT
  SET k=2^i
  SIMULATE epsilon = 0 or 1
  IF epsilon=0 SET x(i+1)=x(i) ELSE SET x(i+1)=y(k)
  SET i=i+1

Then the claim is that the empirical distribution given by x(0), x(1), ... is the invariant measure of the stationary process.

$\endgroup$
8
  • $\begingroup$ I'm pretty sure that what you're looking for is false. If I understand correctly, you are averaging along an extremely sparse sequence of times. You're taking a random sequence of bits $\ldots\epsilon_n \ldots\epsilon_1 \epsilon_0$, and turning them into a sequence of integers: $m_n$ is the integer with the binary expansion $\epsilon_{n-1}\ldots\epsilon_0$. What you're then asking is whether the averages along the random sequence $(m_n)$ is "good for $L^1$-convergence". Theorems of Wierdl and co-authors should give a negative answer since the $(m_n)$ grow exponentially. $\endgroup$ Feb 18, 2016 at 20:26
  • $\begingroup$ @AnthonyQuas The sequence is sparse except there are some blocks $m_n=m_{n+1}=\ldots=m_{n+k}$. I'd like the reference of the theorems you mention. $\endgroup$ Feb 18, 2016 at 23:02
  • 1
    $\begingroup$ So the existence of constant blocks (of length with bounded expectation) shouldn't change things. See Wierdl and Rosenblatt's article in Petersen and Salama: Ergodic Theory and its connections with harmonic analysis; see also Bellow Bellow, Alexandra(1-NW) On "bad universal'' sequences in ergodic theory. II. Measure theory and its applications (Sherbrooke, Que., 1982), 74–78, Lecture Notes in Math., 1033, Springer, Berlin, 1983. $\endgroup$ Feb 18, 2016 at 23:17
  • $\begingroup$ Thank you @AnthonyQuas. These references perfectly fit the problem. I see, the claim does not seem to hold for $\alpha=\sum\frac{1}{2^{n!}}$. However I think the sequence meets the interval $[3/4, 3/4+\epsilon]$ infinitely often in this case, and such a property is enough for my real problem (while this is still not exactly my real problem). But well, the claim as it is stated in my post seems to be wrong. $\endgroup$ Feb 19, 2016 at 10:18
  • $\begingroup$ So it occurs to me, another reference for why things go wrong if you have unbounded functions is my paper with Máté Wierdl, math.uvic.ca/faculty/aquas/papers/paper24.pdf. It sounds as though you're mainly interested in positive results (for which you need to average "nice" functions). $\endgroup$ Feb 19, 2016 at 18:05

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.