English

On a power series distribution with mean parameterization

General Mathematics 2025-12-05 v2

Abstract

The article examines the distribution of the power series of the function w(y)=(1+1y)12. w(y) = \left( 1 + \sqrt{1 - y} \right)^{-\frac{1}{2}}. The distribution of the considered function into a power series is obtained (1+1y)12=m=0(4m)!16m(2m)!(2m+1)!2ym. \left(1 + \sqrt{1 - y}\right)^{-\frac{1}{2}} = \sum_{m=0}^{\infty} \frac{(4m)! \, 16^{-m}}{(2m)! \, (2m+1)! \, \sqrt{2}} \, y^m. The dispersion function is found ν(x)=x(2x+1)(4x+1),  x>0. \nu(x) = x (2x + 1)(4x + 1), \; x > 0. A distribution with mean parameterization is constructed Pr(ξ=k)=(4k+12k)2kxk(2k+1)k+12(4k+1)2k32,  x>0. \Pr(\xi = k) = \binom{4k + 1}{2k} \, 2^{-k} \, x^k \, (2k + 1)^{k + \frac{1}{2}} \, (4k + 1)^{-2k - \frac{3}{2}}, \; x > 0. It is proved that the raw moments αm\alpha_m, central moments μm\mu_m, cumulants χm,  m=1,2,\chi_m, \; m = 1, 2, \ldots satisfy the following recurrence relations: αm+1=xαm+ν(x)dαmdx,  α0=1,  α1=x;μm+1=mμm1+ν(x)dμmdx,  μ0=1,  μ1=0;χm+1=ν(x)dχmdx,  χ1=x. \alpha_{m+1} = x \alpha_m + \nu(x) \frac{d\alpha_m}{dx}, \; \alpha_0 = 1, \; \alpha_1 = x; \quad \mu_{m+1} = m \mu_{m-1} + \nu(x) \frac{d\mu_m}{dx}, \; \mu_0 = 1, \; \mu_1 = 0; \quad \chi_{m+1} = \nu(x) \frac{d\chi_m}{dx}, \; \chi_1 = x.

Keywords

Cite

@article{arxiv.2511.00069,
  title  = {On a power series distribution with mean parameterization},
  author = {Oleksandr Volkov and Nataliia Voinalovych},
  journal= {arXiv preprint arXiv:2511.00069},
  year   = {2025}
}

Comments

7 pages

R2 v1 2026-07-01T07:16:10.094Z