2015/03/08

Tinkering around logistic map

I was tinkering around logistic map today, I accidentally discovered a property of it. That is, if you take a statistical view over time and take a look at the histogram of the sequence generated by it, it will look same independent of the input. This immediately hit me that there may be an underlying probability density of this map. In fact, I found out that Ulam and von Neumann studied this! So here we go.

Logistic map is a deterministic recurrence equation. Concisely, it is given as, \begin{align} x_{n+1} = a x_n (1 - x_n) \end{align} where $x_n \in (0,1)$. For $a = 4$, this system exhibits a chaotic behaviour: Its response to a small uncertainty in the input is unpredictable. In other words, if you generate two sequences with very small difference in the initial condition, the sequences will differ vastly.

But as it turns out, it is not totally unpredictable. Take an initial condition $x_0 \in (0,1)$ (we assume we do not choose unstable points for now), and simulate $N$ steps. We have a sequence $(x_n)_{n=1}^N$. Treating these as samples, if you plot the histogram, you will get:


This is true for any $x_0 \in (0,1)$ except for three unstable points $0.25, 0.5, 0.75$. Ulam and von Neumann found out that, if we see this sequence as a sequence of iid samples, the underlying probability density would be: \begin{align} p(x) = \frac{1}{\pi \sqrt{x(1-x)}} \end{align} More interestingly, it is possible to obtain uniform density by applying the following transformation: \begin{align} y_n = (2/\pi) \sin^{-1}(\sqrt{x_n}) \end{align}See the histogram where I used same samples with above after applying this transformation:

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