1
$\begingroup$

given a measurable function $\alpha: (\Omega, \mu) \to \mathbb{R}$, and transformation $\sigma : \Omega \to \Omega $.

I found an example such that $\alpha, \alpha\circ \sigma, \alpha\circ \sigma^2, \alpha\circ \sigma^3 \cdots : \Omega \to \Omega$ are all Independent identical distribution :

Let $(\Omega, \mu) = ([0,1], Leb)$, $\alpha $ takes countable $0-1$ value, and $\sigma$ is piecewise linear map and some $r \in (0,1)$ such that:

$$ \alpha(\omega)=\left\{ \begin{aligned} 0 & & \omega \in [0,r] \\ 1 & & \omega \in [r, 1]\\ \end{aligned} \right. $$

$$ \sigma(\omega)=\left\{ \begin{aligned} \frac{\omega}{r} & & \omega \in [0,r] \\ \frac{\omega-r}{1-r}& & \omega \in [r, 1]\\ \end{aligned} \right. $$

We can also construct similar example such that $\alpha$ takes countable values, i.e. distribution of $\alpha$ has only countable atoms.

Can we construct an example on $\Omega= [0,1] $such that

1, $\alpha, \alpha\circ \sigma, \alpha\circ \sigma^2, \alpha\circ \sigma^3 \cdots $ are Independent identical distribution and

2, distribution of $\alpha$ has no atom and absolutely continuous w.r.t Leb? Thanks in advanced!

$\endgroup$
2
$\begingroup$

The easiest way to do this is to take $\Omega=\{[0,1)\}^{\mathbb N_0}$., $\sigma$ the shift map and $\alpha(\omega)=\omega_0$.

If you don’t like the infinite-dimensional $\Omega$, you can build an example with $\Omega=[0,1)$ at the cost of making the transformation uglier (there is an measure space isomorphism mapping $[0,1)^{\mathbb N_0}$ equipped with the product of Lebesgue measures to $[0,1)$ equipped with Lebesgue, and if you conjugate $\sigma$ by this measure space isomorphism, you obtain a transformation on $[0,1)$ with the desired property.

Let me add: in some sense you should not expect a better answer than this. You seem to have edited the question to require $\Omega=[0,1]$. In this case, the map $\Phi$ from $\Omega$ to $\mathbb R^{\mathbb N_0}$ defined by $\omega\mapsto (\alpha(\sigma^n\omega))_{n\in\mathbb N_0}$ is a factor map. Since its range has infinite entropy, it follows that $\sigma\colon\Omega\to\Omega$ has infinte entropy. This can never be obtained by piecewise smooth maps $\sigma$ like your example.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks, Yes, the one on $[0,1]$ would be uglier via the construction in this way. $\endgroup$ – jason Apr 18 '19 at 1:12
  • $\begingroup$ It’s not super-ugly though: you could write down an explicit transformation. $\endgroup$ – Anthony Quas Apr 18 '19 at 4:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.