# Gauss–Kuzmin distribution

Parameters (none) ${\displaystyle k\in \{1,2,\ldots \}}$ ${\displaystyle -\log _{2}\left[1-{\frac {1}{(k+1)^{2}}}\right]}$ ${\displaystyle 1-\log _{2}\left({\frac {k+2}{k+1}}\right)}$ ${\displaystyle +\infty }$ ${\displaystyle 2\,}$ ${\displaystyle 1\,}$ ${\displaystyle +\infty }$ (not defined) (not defined) 3.432527514776...[1][2][3]

In mathematics, the Gauss–Kuzmin distribution is a discrete probability distribution that arises as the limit probability distribution of the coefficients in the continued fraction expansion of a random variable uniformly distributed in (0, 1).[4] The distribution is named after Carl Friedrich Gauss, who derived it around 1800,[5] and Rodion Kuzmin, who gave a bound on the rate of convergence in 1929.[6][7] It is given by the probability mass function

${\displaystyle p(k)=-\log _{2}\left(1-{\frac {1}{(1+k)^{2}}}\right)~.}$

## Gauss–Kuzmin theorem

Let

${\displaystyle x={\frac {1}{k_{1}+{\frac {1}{k_{2}+\cdots }}}}}$

be the continued fraction expansion of a random number x uniformly distributed in (0, 1). Then

${\displaystyle \lim _{n\to \infty }\mathbb {P} \left\{k_{n}=k\right\}=-\log _{2}\left(1-{\frac {1}{(k+1)^{2}}}\right)~.}$

Equivalently, let

${\displaystyle x_{n}={\frac {1}{k_{n+1}+{\frac {1}{k_{n+2}+\cdots }}}}~;}$

then

${\displaystyle \Delta _{n}(s)=\mathbb {P} \left\{x_{n}\leq s\right\}-\log _{2}(1+s)}$

tends to zero as n tends to infinity.

## Rate of convergence

In 1928, Kuzmin gave the bound

${\displaystyle |\Delta _{n}(s)|\leq C\exp(-\alpha {\sqrt {n}})~.}$

In 1929, Paul Lévy[8] improved it to

${\displaystyle |\Delta _{n}(s)|\leq C\,0.7^{n}~.}$

Later, Eduard Wirsing showed[9] that, for λ=0.30366... (the Gauss-Kuzmin-Wirsing constant), the limit

${\displaystyle \Psi (s)=\lim _{n\to \infty }{\frac {\Delta _{n}(s)}{(-\lambda )^{n}}}}$

exists for every s in [0, 1], and the function Ψ(s) is analytic and satisfies Ψ(0)=Ψ(1)=0. Further bounds were proved by K.I.Babenko.[10]