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Let $\Delta$ be the interval $[-1,1]$, then we can consider the probability space $(\Delta , \mathcal{B}(\Delta),\nu)$, where $\mathcal{B}(\Delta)$ is the Borel $\sigma$-algebra and $\nu$ is equal half of the Lebesgue measure.

Then we can endow the space $\Delta^{\mathbb{N}}:= \{ (\omega_n)_{n\in \mathbb{N}};\ \omega_n \in \Delta, \ \forall \ n\in \mathbb{N}\}$ with the $\sigma$-algebra $\mathcal{B}(\Delta^{\mathbb{N}})$ (Borel $\sigma$-algebra of $\Delta^{\mathbb{N}}$ induced by the product topology) and the probability measuare $\nu^{\mathbb{N}}$ in the measurable space$(\Delta^{\mathbb{N}},\mathcal{B}(\Delta^{\mathbb{N}}))$, such that $$\nu^{\mathbb{N}} \left(A_1\times A_2\times \ldots \times A_n \times \prod_{i=n+1}^{\infty} \Delta\right)=\nu(A_1) \cdot \ldots\cdot \nu(A_n). $$

Now, let $\sigma>2/(3\sqrt{3})$ be a real number, and define $$x_-^* = \text{The unique real root of the polynomial }x^ 3+ \sigma = x, $$ $$x_+^* = \text{The unique real root of the polynomial }x^ 3- \sigma = x, $$ it is easy to see that $x_+^* = -x_-^*$.

We can then define the function $$h:\mathbb{N}\times \Delta^ \mathbb{N}\times \mathbb{R}\to \mathbb{R} $$ in the following recursive way,

  • $h(0,(\omega_n)_{n},x) = x$, $\forall\ (\omega_n)_n\in \mathbb{N}$ and $\forall\ x\in \mathbb{R}$;
  • $h(i+1,(\omega_n)_{n},x) = \sqrt[3]{h(i,(\omega_n)_{n},x) + \sigma \omega_i}.$

This way we are for, every $x \in \mathbb{R}$ and $(\omega_n)_n\in\Delta^\mathbb {N}$, defining the following sequence $$\left\{x, \sqrt[3]{x + \sigma \omega_1},\sqrt[3]{\sqrt[3]{x + \sigma \omega_1}+\sigma w_2},\sqrt[3]{\sqrt[3]{\sqrt[3]{x + \sigma \omega_1}+\sigma w_2} + \sigma w_3}, \ldots \right\}.$$

Using Mathematica I did some numerical experiments and seems like that $$\lim_{n\to\infty}\left| h(n,(\omega_n)_n, x^*_+) - h(n,(\omega_n)_n, x^*_-) \right| = 0,\ \text{for $\nu^{\mathbb{N}}$-almost surely $(\omega_n)_n$} \in \Delta^{\mathbb{N}}. $$

Can anyone tell me, or present me a good reason why this should happen?


Numerical evidences

Below this paragraph you can find my Mathematica code, imputing a value of $\sigma$ and a number of iterations ($b$) this program has as output two discrete plot, the first one is the discrete plot of the sequences $h(\cdot, (\omega_n)_n, x_+^ *)$ and $h(\cdot, (\omega_n)_n, x_-^ *)$ in the same plot, the second one is a discrete plot of the difference $h(\cdot , (\omega_n)_n, x_+^ *) - h(\cdot , (\omega_n)_n, x_-^ *)$, for some random sequence $(\omega_n)_n$.

\[Sigma] = 2/(3*Sqrt[3]) + 0.15; *Insert here your value of sigma*
b = 10000; *Insert here your the number of iterations*
A = Solve[x^3 + \[Alpha] == x , x, Reals];
B = Solve[x^3 - \[Alpha] == x , x, Reals];
y = x /. A[[1]];
z = x /. B[[1]];
W = Table[RandomReal[{-\[Sigma], \[Sigma]}], {i, 1, b}]; 
P = RecurrenceTable[{x[n + 1] == CubeRoot[x[n] + W[[n]]], x[1] == y}, 
x, {n, 1, b} ];
Q = RecurrenceTable[{x[n + 1] == CubeRoot[x[n] + W[[n]]], x[1] == z}, 
x, {n, 1, b} ];
DiscretePlot[{P[[i]], Q[[i]]}, {i, 1, b}, PlotRange -> {{1, b}, {-1.5, 1.5}}]
DiscretePlot[Q[[i]] - P[[i]], {i, 1, b},PlotRange -> {{1, b}, {-2, 2}}, PlotStyle -> Brown]

Example 1: $\sigma = 1$ and $b= 50$ enter image description here

Example 2: $\sigma = \frac{2}{3 \sqrt{3}} +0.15$ and $b= 100000.$

enter image description here

for some reason when $\sigma$ is big, the convergence of $\left| h(n,(\omega_n)_n, x^*_+) - h(n,(\omega_n)_n, x^*_-) \right|$ seems to occur "faster", and I do not have idea why this happens.

Can anyone help me?

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1 Answer 1

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Since this is a monotone continuous random dynamical system such that the underlying Markov process is uniquely ergodic, it admits a unique random fixed point. The standard proof goes as follows: by monotonicity and continuity, the bounded set attractor $A$ is such that $A_+ = \sup A$ and $A_- = \inf A$ are random fixed points. Since the law of $A_\pm$ is a (and therefore the by unique ergodicity) invariant measure, one must have $A_+ = A_-$ almost surely.

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  • $\begingroup$ Thank you so much for your answer. I was looking exactly for a result as you described "Since this is a monotone continuous random dynamical system such that the underlying Markov process is uniquely ergodic, it admits a unique random fixed point" Could you recommend me some reference so I can learn about it? $\endgroup$ Aug 4, 2019 at 13:18
  • $\begingroup$ See for example the list of references in Section 1.1 of arxiv.org/pdf/1411.1340.pdf. $\endgroup$ Aug 4, 2019 at 13:39
  • $\begingroup$ Thx you so much. $\endgroup$ Aug 4, 2019 at 13:46
  • $\begingroup$ Can you please me definite the "underlying Markov process" I am not sure about you mean by that? $\endgroup$ Aug 13, 2019 at 21:40
  • $\begingroup$ I think that is the family of process $$\{\Phi_{n}^{x}(\omega) := h(n,x,\omega)\}, $$ with transition probability $$P_{n}(x,\Gamma) = \int_{\Omega} 1_{\Gamma}\circ h(n,\omega,x)\ \text{d} \nu^{\mathbb{N}}(\omega), $$ but I am not sure. $\endgroup$ Aug 13, 2019 at 21:55

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