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 $b= 50$

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

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?