On a power series distribution with mean parameterization

Abstract

The article examines the distribution of the power series of the function $w(y)={\left(1+\sqrt{1-y}\right)}^{-\frac{1}{2}}.$ The distribution of the considered function into a power series is obtained ${\left(1+\sqrt{1-y}\right)}^{-\frac{1}{2}}={\sum }_{m=0}^{\infty }\frac{(4m)!16^{-m}}{(2m)!(2m+1)!\sqrt{2}}{y}^m.$ The dispersion function is found $\hat{I}½(x)=x(2x+1)(4x+1),x>0.$ A distribution with mean parameterization is constructed $\mathrm{Pr}(\hat{I}¾=k)=\left(\genfrac{}{}{0ex}{}{4k+1}{2k}\right)2^{-k}{x}^k(2k+1{)}^{k+\frac{1}{2}}(4k+1{)}^{-2k-\frac{3}{2}},x>0.$ It is proved that the raw moments $\hat{I}{\pm }_m$, central moments $\hat{I}{¼}_m$, cumulants $\ddot{I}{‡}_m,m=1,2,\dots$ satisfy the following recurrence relations: $\hat{I}{\pm }_{m+1}=x\hat{I}{\pm }_m+\hat{I}½(x)\frac{d\hat{I}{\pm }_m}{dx},\hat{I}{\pm }_0=1,\hat{I}{\pm }_1=x;\hat{I}{¼}_{m+1}=m\hat{I}{¼}_{m-1}+\hat{I}½(x)\frac{d\hat{I}{¼}_m}{dx},\hat{I}{¼}_0=1,\hat{I}{¼}_1=0;\ddot{I}{‡}_{m+1}=\hat{I}½(x)\frac{d\ddot{I}{‡}_m}{dx},\ddot{I}{‡}_1=x.$

7 pages