# Existence and uniqueness of a stationary measure

This same question was also posted on MSE https://math.stackexchange.com/questions/3327007/existence-and-uniqueness-of-a-stationary-measure.

Recently I have posted the following question on MO Attractors in random dynamics.

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 $$x_-^*(\sigma) = \text{The unique real root of the polynomial }x^ 3+ \sigma = x,$$ $$x_+^*(\sigma) = \text{The unique real root of the polynomial }x^ 3- \sigma = x,$$ it is easy to see that $$x_+^*(\sigma) = -x_-^*(\sigma)$$.

We can then define the function $$h:\mathbb{N}\times \Delta^ \mathbb{N}\times [x_-^*(\sigma),x_+^*(\sigma)]\to [x_-^*(\sigma),x_+^*(\sigma)],$$ 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 $$\left\{x, \sqrt[3]{x + \sigma \omega_1},\sqrt[3]{\sqrt[3]{x + \sigma \omega_1}+\sigma w_2},\sqrt[3]{\sqrt[3]{\sqrt[3]{x + \sigma \omega_1}+\sigma w_2} + \sigma w_3}, \ldots \right\}.$$

Now, define the following family of Markov kernels $$P_n(x,A) = \nu^{\mathbb{N}}\left(\left\{(\omega)_{n\in \mathbb{N}} \in \Delta^{\mathbb{N}};\ h(n,(\omega_n)_{n\in\mathbb{N}} ,x)\in A \right\}\right).$$

A probability measure $$\mu$$ in $$([x_-^*(\sigma),x_+^*(\sigma)], \mathcal{B}([x_-^*(\sigma),x_+^*(\sigma)])$$ is a called stationary measure if

$$\mu(A) = \int_{[x_-^*(\sigma),x_+^*(\sigma)]} P_1(x,A)\text{d}\mu(x);\ \forall \ A\in \mathcal{B}([x_-^*(\sigma),x_+^*(\sigma)]),$$ where $$\mathcal{B}([x_-^*(\sigma),x_+^*(\sigma)])$$ is the Borel $$\sigma$$-algebra. Moreover, once $$[x_-^*(\sigma),x_+^*(\sigma)]$$ it is easy to prove that there exists at least one stacionary measure.

The answer that I received on MO suggests that there exists only one stationary measure.

Does anyone know if this is true? A reference to such a result is enough for mine purposes.

• Look this book. Chapters 5,6,7.
Aug 22, 2019 at 10:48
• Thx, good reference, I am learning a lot of new techniques Aug 22, 2019 at 22:56

What you’re looking at is an example of an iterated function system: you have $$x_{n+1}$$ is $$h_{\omega_n}(x_n)$$. I haven’t looked in detail, but I assume that there is an interval $$J$$ such that $$h_\omega(x)\in J$$ for all $$\omega \in[-1,1]$$? Assuming this, what remains is to show (ideally) that each $$h_\omega$$ is contracting on all of $$J$$. This would allow one to complete the argument given. Failing this, it would suffice to show that there exists a sequence $$\delta_n\to 0$$ such that $$h_{\omega_1}\circ\ldots h_{\omega_n}(J)$$ is of length at most $$\delta_n$$ for each choice of $$\omega_1,\ldots\omega_n$$. Failing this, I would try for an argument based on negativity of the Lyapunov Exponent.
Assuming that one of these conditions holds, you can find the unique invariant measure by looking at the distribution of $$\lim_{n\to\infty}h_{\omega_{-1}}\circ\ldots\circ h_{\omega_{-n}}(0)$$. The conditions above are to ensure that the limit exists.
• Thx for your answer, but I am facing real issues to solve this question. We can take $J =[x^∗_−(σ),x^∗_+(σ)]$ (where these symbols are explained in the question), it is easy to see that $[x^∗_−(σ),x^∗_+(σ)]$ is the minimal interval with this property. Your suggestion did not work and I will say why. $h_\omega (x) = (x+\omega_0)^{1/3}$, is not a contraction, moreover, it is really hard to prove that $h_{\omega_n}\circ ... \circ h_{\omega_0} (J)$ is an interval. And the calculus of the Lyapunov exponent is not easy once the derivative of $(x)^{1/3}$ is not bounded. Aug 20, 2019 at 12:59
• I have heard that once $[x^∗_−(σ),x^∗_+(σ)]$ a minimal invariant set then can only exist one stationary measure in $[x^∗_−(σ),x^∗_+(σ)]$. Do you know if this is true? But even more so I don't really want to look like I'm not doing anything and just waiting for the answer, I really tried very hard to get your suggestion. Aug 20, 2019 at 13:02
• At least one thing is easy: $h_{\omega_n}\circ\ldots\circ h_{\omega_0}(J)$ is an interval. Each of the $h$"s is a continuous map, which map connected sets to connected sets so this is a composition of $n+1$ continuous maps, which is continuous and maps $J$ to a sub-interval. In fact, each of the $h_\omega$ is monotonic also, which is even better. This means that if one can check that $|h_{\omega_{-1}}\circ\ldots \circ h_{\omega_{-n}}(x_-^*(\sigma))-h_{\omega_{-1}}\circ\ldots \circ h_{\omega_{-n}}(x_+^*(\sigma))|\to 0$, then the position at time 0 is completely determined by $(\omega_n)_{n<0}$. Aug 20, 2019 at 21:07
• Check out en.wikipedia.org/wiki/Harris_chain. You make a very small interval $A$ as above. Then you find you look at the conditional distribution of where you go at the next step under $h_\omega$. This gives you for each $x\in A$ a density on $J$, $\pi_x$ say. All you need to do is find some interval $K$ and some $\delta>0$ where $\pi_x(y)>\delta$ for each $x\in A$ and $y\in K$. This should be easy enough. Now take $\rho$ to be the uniform probability measure on $K$ and check that the second condition of a Harris chain is satisfied. Harris ensures a unique stationary distribution. Aug 24, 2019 at 8:48