Page:Biometrika - Volume 6, Issue 1.djvu/4

In a similar tedious way I find:

$M'_{3}=\mu _{2}^{3}{\frac {(n-1)(n+1)(n+3)}{n^{3}}}$,

and

$M'_{4}=\mu _{2}^{4}{\frac {(n-1)(n+1)(n+3)(n+5)}{n^{4}}}$.

The law of formation of these moment coefficients appears to be a simple one, but I have not seen my way to a general proof.

If now $M_{R}$ be the $R$^th moment coefficient of $s^{2}$ about its mean, we have

$M_{2}$	$=\mu _{2}^{2}{\frac {(n-1)}{n^{2}}}\{(n+1)-(n-1)\}=2\mu _{2}^{2}{\frac {(n-1)}{n^{2}}}$,
$M_{3}$	$=\mu _{2}^{3}\left\{{\frac {(n-1)(n+1)(n+3)}{n^{3}}}-{\frac {3(n-1)}{n}}\cdot {\frac {2(n-1)}{n^{2}}}-{\frac {(n-1)^{2}}{n^{3}}}\right\}$
	$=\mu _{2}^{3}{\frac {(n-1)}{n^{3}}}\{n^{2}+4n+3-6n+6-n^{2}+2n-1\}=8\mu _{2}^{3}{\frac {(n-1)}{n^{3}}}$,
$M_{4}$	$={\frac {\mu _{2}^{4}}{n^{4}}}\{(n-1)(n+1)(n+3)(n+5)-32(n-1)^{2}-12(n-1)^{3}-(n-1)^{4}\}$
	$={\frac {\mu _{2}^{4}(n-1)}{n^{4}}}\{n^{3}+9n^{2}+23n+15-32n+32-12n^{2}+24n-12-n^{3}+3n^{2}-3n+1\}$
	$={\frac {12\mu _{2}^{4}(n-1)(n+3)}{n^{4}}}$.

Hence

$\beta _{1}={\frac {M_{3}^{2}}{M_{2}^{3}}}={\frac {8}{n-1}}$, $\beta _{2}={\frac {M_{4}}{M_{2}^{2}}}={\frac {3(n+3)}{n-1}}$,
$\therefore 2\beta _{2}-3\beta _{1}-6={\frac {1}{n-1}}\{6(n+3)-24-6(n-1)\}=0$.

Consequently a curve of Professor Pearson’s type III. may be expected to fit the distribution of $s^{2}$.

The equation referred to an origin at the zero end of the curve will be

$y=Cx^{p}e^{-\gamma x}$,

where

$\gamma =2{\frac {M_{2}}{M_{3}}}={\frac {4\mu _{2}^{2}(n-1)n^{3}}{8n^{2}\mu _{2}^{3}(n-1)}}={\frac {n}{2\mu _{2}}}$,

and

$p={\frac {4}{\beta _{1}}}-1={\frac {n-1}{2}}-1={\frac {n-3}{2}}$.

Consequently the equation becomes

$y=Cx^{\frac {n-3}{2}}e^{-{\frac {nx}{2\mu _{2}}}}$,

which will give the distribution of $s^{2}$.

The area of this curve is $C\int _{0}^{\infty }x^{\frac {n-3}{2}}e^{-{\frac {nx}{2\mu _{2}}}}dx=I$ (say).