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For any positive integer $N$, consider the Fibonacci sequence $F_n$ of length $N$. Using $F_n$ we can define a Fibonacci discrete probability distribution as follows:

$$p_N(n)=\frac{F_n}{\sum_{k=1}^N F_k}\ \ \forall n=1,2,\ldots,N$$

$$p_N(n)=\frac{F_{N+1-n}}{\sum_{k=1}^N F_k}\ \ \forall n=1,2,\ldots,N$$

With the increasing sequence probability distribution (first equation), and the second equation (associated with decreasing sequence) the question is what is the Entropy of those Distributions. Thank you in advance.

Anthony
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  • For some fixed N or when N goes to infinity? – Did Feb 15 '15 at 19:23
  • I think fixed N as the sum at the denominator goes to infinity for N=infinity. So yes fixed N. I am not sure if one can approximate an answer for N tending to infinity. – Anthony Feb 15 '15 at 19:30
  • Let me rephrase: the distribution $p_N$ is probably unwieldy (and uninteresting) for some fixed $N$, its limit when $N\to\infty$ is more alluring (that is, if one sticks to the second one of the two nonequivalent definitions you propose). – Did Feb 15 '15 at 19:34
  • OK. If that helps... – Anthony Feb 15 '15 at 19:42
  • @Did, For the $2^{nd}$ definition, $\lim_{N\to\infty} p_N(n) = \frac{1}{\phi^{n+1}}$ is a geometric distribution. It isn't anything really special. – achille hui Feb 16 '15 at 05:02
  • You are right for this one. It becomes one. Do you have a solution handy – Anthony Feb 16 '15 at 05:05
  • @achillehui I know (provided one multiplies your phi^(-n-1) by some normalizing factor). No big mystery here (this is probably what you are alluding to in your comment to me), nevertheless this seems the most natural way to get a nontrivial limiting object. – Did Feb 16 '15 at 15:14

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