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Pearson from Gaussian

Consider changing $\rho (\sigma )$, the distribution of classes, to

\begin{displaymath}\rho _{n}(\sigma )=N\left( \frac{w^{n-1}}{\sigma ^{n}}\right) \exp \left( -%
\frac{w^{2}}{2\sigma ^{2}}\right) ,\qquad n>1.

The normalization N is

\begin{displaymath}N=\frac{1}{2^{\left( n-3\right) /2}\Gamma \left( \frac{n-1}{2}\right) }.

Convolution of this with the Gaussian distribution produces Pearson distributions[2]

\begin{displaymath}P_{P,b}(B_{i})=\frac{(norm)}{w\left( \frac{B_{i}^{2}}{w^{2}}+1\right) ^{b}}%
,\qquad b=\frac{n}{2}>\frac{1}{2},

\begin{displaymath}P_{P,b}(\vert{\bf B}\vert)=(norm)\frac{B^{2}}{w^{3}}\left( \frac{1}{\left( \frac{%
B^{2}}{w^{2}}+1\right) ^{(b+1)}}\right) .

This includes the Lorentzian as the b=1 special case, and the Gaussian as the $b\rightarrow \infty $ limit. While it is thus expected that the relaxation function GP,b(t) will evolve smoothly from Lorentzian KT to Gaussian KT as b increases, b=2 (n=4) is the only nontrivial case for which closed form G(t) is known:

\begin{displaymath}G_{P,2}(t)=\frac{1}{3}+\frac{2}{3}\left( 1+at-a^{2}t^{2}\right) e^{-at}.

Jess H. Brewer