Angular processes related to Cauchy random walks (Q2882284)
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scientific article; zbMATH DE number 6030155
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Angular processes related to Cauchy random walks |
scientific article; zbMATH DE number 6030155 |
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4 May 2012
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angular random walk
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Cauchy distribution
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hyperbolic trigonometry
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arcsine law
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continued fractions
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Fibonacci numbers
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nonlinear transformation of random variables
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Angular processes related to Cauchy random walks (English)
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Motivated by studies on angular random walks either on \(\mathbb{R}^{2}\) or the Poincaré space \(\mathbb{H}_{+}^{2}\), describing the rotation of a moving particle around the starting point, the authors are led to analyze the random variable \(U=(\gamma C_{1}+\delta C_{2})/(\alpha -\beta C_{1}C_{2}),\) where \(C_{1}\) and \(C_{2}\) are independent Cauchy random variables, and \(\beta ,\gamma ,\delta >0\) and \(\alpha +\beta \neq 0.\) By means of cumbersome computations, they arrive at the conclusion that \(U\) is still Cauchy distributed with scale parameter \(a_{U}\) and location parameter \(b_{U}\) defined in terms of the corresponding parameters of \(C_{1}\) and \(C_{2}.\) In the last part of the paper, the authors examine the sequences \((V_{n})_{n\geq 1}\) and \((U_{n})_{n\geq 1}\) given by \(V_{1}=1/(1+C),\) \(V_{n+1}=1/(1+V_{n})\) and \(U_{1}=1/(1+C^{2}),\) \(U_{n+1}=1/(1+U_{n})\), where \(C\) is a standard Cauchy random variable. They prove that \(V_{n}\) has a Cauchy distribution with scale parameter \(a_{n}=1/F_{2n+1}\) and location parameter \(b_{n}=F_{2n}/F_{2n+1}\), where \((F_{n})_{n\geq 0}\) is the Fibonacci sequence with \(F_{0}=0\) and \(F_{1}=1\). Observing that \(a_{n}\rightarrow 0\) and \(b_{n}\rightarrow \varphi -1,\) where \(\varphi =(1+\sqrt{5})/2\) is the golden ratio, they conclude that \((V_{n})_{n\geq 1}\) converges in distribution to \(\varphi -1\). While the random variable \(U_{1}\) possesses the arcsine law, the sequence \((U_{n})_{n\geq 1}\) converges in distribution to \(\varphi -1\) as well.
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