# Independent identical distribution sequence

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!

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.

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