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Briefly, the way symbols are generated is:

Consider a one-dimensional chaotic map $T: [0,1]→[0,1]$ and a time series $\{x_n\}_{n=1}^N$ generated with this map. Define a threshold $A$ and a thresholding function $π$:

$$\pi(x) := \begin{cases} 1 &\text{if} \quad x>A \\ 0 & \text{else}\end{cases}$$

Consider a symbolic sequence $\{s_n\}_{n=1}^N$ obtained by applying $π$ to the elements of $\{x_n\}_{n=1}^N$, i.e., $s_i := π(x_i)$

For example, let $A =0.5$ be the threshold and the time series $x = [0.1,\, 0.56,\, 0.6,\, ...]$. Then

$\pi(x_1) = 0$;

$\pi(x_1) = 1$;

$\pi(x_2) = 1$.

Thus, the symbolic representation is $s = [0,1,1]$

I am facing technical difficulties in following the paper. My question is: Such binary sequences are just i.i.d. random variables of some distribution. How can we say that these sequences $s$ are chaotic?

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  • $\begingroup$ This is more of a math question. I don't think this differs fundamentally from any other pseudo-random generation schemes (most of which are probably based on chaotic maps even though that word wasn't in use when they were invented). In general, though, the quality of a pseudo-random number generator always has to be tested based on the relevant criteria, starting with a chaotic system is absolutely no guarantee for pseudo-randomness. $\endgroup$ – CuriousOne Jun 1 '15 at 18:53
  • $\begingroup$ So, a map $\pi: X \rightarrow \Sigma_2$. – I fail to understand this: You are mapping from a single sequence to the set of all sequences? Why? What is the connection to the rest of the question? $\endgroup$ – Wrzlprmft Jun 1 '15 at 20:29
  • $\begingroup$ I am mapping each real number $x \in X$ to its corresponding symbol from $\Sigma_2 = {0,1}$ depending on which partition $x$ falls within. The resulting symbolic orbit is called the symbolic dynamics of the real orbit. $\endgroup$ – SKM Jun 1 '15 at 20:36
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When such 0/1 sequences are generated, they are just i.i.d random variables of some distribution.

No, they aren’t.

As for identically distributed, consider any sequence when changing $A$. The higher $A$, the higher the probability that the symbol is $1$.

As for independently, consider the sequence generated by a tent map and then transform all values via $$f(x) := \left(x-\tfrac{1}{5}\right) \bmod 1.$$

The corresponding sequence could as well have been generated by a somewhat shifted tent map, which looks like this:

shifted tent map

If you select $A=\tfrac{7}{10}$ (see plot), it should become obvious that the probability that a $1$ is followed by another $1$ is higher than the probability that that a $0$ is followed by a $1$, so the symbol sequence has some memory.


How can we say that these sequences S are chaotic?

It all boils down to your definition of chaoticity at the end, but consider the following: Let $T$ be the classical tent map (forget the above shift) and let $A:=\tfrac{1}{2}$. Now, if we represent numbers in binary, the effect of the map on a number can be understood as removing its first digit after the decimal point, e.g., $$T(0.10100101011101) = 0.0100101011101,$$ $$T(0.0100101011101) = 0.100101011101,$$ and so on. Due to our threshold selection, we also have that $s_i$ is the digit removed from $x_i$ by $T$. Thus your sequence $s$ is the sequence of digits of $x_0$.

Thus, the more precisely, we know $x_0$, the longer can we accurately predict the sequence $s$, but any finite precision will make it impossible for us to predict $s$ forever. This is very reminiscent of the Butterfly Effect.

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  • $\begingroup$ Thank you for your reply with the graphs. Somethings are still unclear: The paper that I linked to and many others mention that the sequences are iid random variables. Hence, they are used as pseudo random number generators. I am not sure why you say they are not. $\endgroup$ – SKM Jun 1 '15 at 21:32
  • $\begingroup$ Next, the shift map is said to be chaotic according to literature but are the symbols so generated chaotic? What I mean is if the iterates generated from an initial condition $x_0$ Is chaotic, then for next iterate $x_1$, the binary representation of rational number $x_1 = 0.6$ (say) would be chaotic as well? In essence, by symbolic dynamics do we mean converting real number to its binary? Is the output $0.01001010...$ chaotic? $\endgroup$ – SKM Jun 1 '15 at 21:33
  • $\begingroup$ Then, I think the shift is to the left of the decimal point and dropping off the symbol, thus shifting to the left. So, $T(0.101) = 1.01 =, T(1.01) = 0.1..$ Lastly, is there a way to model/ represent the symbolic representation by an equation? Literature says that A Bernoulli Map acts as the Shift Map. So, can we say that Bernoulli Map is used to represent the symbolic dynamics? Thank you for your valuable insights and effort. $\endgroup$ – SKM Jun 1 '15 at 21:38
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    $\begingroup$ Well, the sequences are generated by a deterministic process and exhibit no repetiton – if you choose the right starting value $x_1$, e.g., an irrational number (“real number” in my previous comment was not exactly precise) for the tent map. E.g., for $x_1 = π-3$, you will never see any repetion, yet you can exactly predict the sequence. $\endgroup$ – Wrzlprmft Jun 1 '15 at 21:58
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    $\begingroup$ What would be those properties (hallmarks of chaos) applicable to symbolic dynamics such that they are chaotic – I mentioned those already: No regularity, generated by a deterministic process and dependence on initial conditions. Note that the dynamics is value-continuous and time-discrete in this case, not symbolic. $\endgroup$ – Wrzlprmft Jun 2 '15 at 6:57

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