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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?

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  • $\begingroup$ do you mean to take $x_0$ as a random variable? if so, what is its law? $\endgroup$
    – Albert
    Dec 16, 2020 at 12:55
  • $\begingroup$ I just edited $x_1$ to $x_0$. $x_0$ is just an arbitrary initial value of the system. My understanding of ergodic theory is that the distribution of $x$ should converge to some kind of invariant distribution regardless of the initial value, right? $\endgroup$
    – K_I
    Dec 16, 2020 at 13:03
  • $\begingroup$ @K_l not quite. if you pick an invariant ergodic measure $\mu$, then $\mu$-almost every $x$ will have its orbit equidistributing according to $\mu$. but there are in general many ergodic measures. it may happen in some cases that there is a preferred measure, for instance if there is a unique measure absolutely continuous with respect to the Lebesgue measure. this will be the case for some specific values of $a$ and $c$ but not in general $\endgroup$
    – Albert
    Dec 16, 2020 at 13:05
  • $\begingroup$ In any case it is certainly not true that the asymptotic distribution is the same for every $x_0$ (think of the case where $x_0$ is a fixed point) $\endgroup$
    – Albert
    Dec 16, 2020 at 13:08
  • $\begingroup$ Thanks. I just added a second part of the question. In the simulation, I pick arbitrary initial values for $x$, the product in the question always converge to the same value. That made me thought that the asymptotic distribution is the same regardless of the initial value. $\endgroup$
    – K_I
    Dec 16, 2020 at 13:29

1 Answer 1

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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

enter image description here

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.

enter image description here

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

$$\mu_h^{\infty} = \frac{a'}{1-c} = g^*$$

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  • $\begingroup$ Thanks for the time! I did a similar simulation. What puzzling me is that while $g_t$ oscillates around its equilibrium value, but the geometric mean converges to that value over time. $\endgroup$
    – K_I
    Dec 16, 2020 at 15:56
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    $\begingroup$ There's actually nothing too surprising in the fact that it converges. I think that, except for some really bizarre functions, most functions with values bound to a finite interval have a converging mean. It is indeed nontrivial why this mean would coincide with the equilibrium of the upper half of the function. I'll see if I can figure out how to derive the distribution, but with non-linear functions its always a hit or miss $\endgroup$ Dec 16, 2020 at 16:16
  • $\begingroup$ Thanks! The step of relating $g_{t+1}$, $g_t$ and $h_t$ is crucial! $\endgroup$
    – K_I
    Dec 17, 2020 at 14:40

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