Asymptotic mean in an one-dimensional chaotic system I have the following one-dimensional dynamical system
$$x_{t+1}=\begin{cases}
ax_t,\quad\mbox{if }x_t\leq 1,\\
ax_t^{c},\quad\mbox{if }x_t\geq 1,\\
\end{cases}$$
where $x_0>0$ is an arbitrary initial value, $a>1$ and $c<-1$.
I know that the fixed point is $x^*=a^\frac{1}{1-c}$ and it is unstable. My understanding is that the system is chaotic. However, I was wondering whether I can analytically compute the asymptotic mean of $x_t$, i.e., $\lim_{t\rightarrow\infty}\mathbb{E}[x_t]$? Do I need to derive first the invariant distribution of $x$? If so, what is the procedure?
Edit:
Moreover, I have a variable $y_t$ related to $x_t$ in the following way
$$
y_t=\begin{cases}1,\quad\mbox{ if }x_t\leq 1,\\
x_t,\quad\mbox{if }x_t\geq 1.\\\end{cases}
$$
I simulated the trajectory of $x_t$ for arbitrary $x_0$ and found that
$$
\lim_{t\rightarrow\infty}\left(\prod_{s=0}^ty_s\right)^\frac{1}{t+1}=x^*.
$$
Is it possible to show it analytically?
 A: I've spotted that taking the logarithm of the difference equation might make things simpler. We get the new difference equation
$$g_{t+1}=\begin{cases}
a' + g_t,\quad\mbox{if }g_t\leq 0,\\
a' + c g_t,\quad\mbox{if }g_t\geq 0,\\
\end{cases}$$
where $g_t=\log x_t$ and $a' = \log a$. Further, we can rewrite this equation into a more comfortable form by using the heaviside theta function
$$g_{t+1} = a' + g_t + (c-1)g_t\Theta(g_t)$$
To get some non-trivial dynamics I have chosen $a'=1$, $c=-3.5$. The velocity plot is as follows

The point where the velocity 0 zero is an attractor, because the the velocity points towards it from both sides. However, the step size is very big, so the system never really lands onto the attractor, just bounces around it. The equilibrium value obtained from setting velocity to zero is $g^* = \frac{a'}{1-c}$
Just to check, I have attempted to simulate $g_t$. Your $y_t$ we will replace with $h_t = \Theta(g_t) g_t$. The logarithm of the desired geometric mean will convert to the arithmetic mean.
$$\frac{1}{T}\sum_t h_t =^? g^*$$
It indeed appears that for reasonable values of $a$ and $c$ that $g_t$ oscillates in some complicated way around its equilibrium value $g^*$, and that $h_t$ converges to that equilibrium value over time.

Edit: Ok, here's a trick that works for the general problem, but it is indeed quite lucky that it does. Notice that in the original equation, the last term is simply $h_t$
$$g_{t+1} = a' + g_t + (c-1)h_t$$
then we can take the sum of both sides
$$\sum_{\tau=0}^{t} g_{\tau+1} = \sum_{\tau=0}^{t} a' + g_\tau + (c-1)h_\tau$$
Let $s^g_t = \sum_{\tau=0}^{t} g_\tau$ and $s^h_t = \sum_{\tau=0}^{t} h_\tau$
$$s^g_{t+1} - g_0 = a't + s^g_t + (c-1)s^h_t$$
Let $\mu^g_t = \frac{s^g_t}{t}$ and $\mu^h_t = \frac{s^h_t}{t}$
$$\mu^g_{t+1} - \frac{g_0}{t+1} = \frac{t}{t+1}(a' + \mu^g_t + (c-1)\mu^h_t)$$
In the limit of $t \rightarrow \infty$
$$\mu^g_{\infty} = a' + \mu^g_{\infty} + (c-1)\mu^h_\infty$$
This is where we get even more lucky, and $\mu^g$ simply cancels out from both sides, and
$$\mu_h^{\infty} = \frac{a'}{1-c} = g^*$$
