Page:Biometrika - Volume 6, Issue 1.djvu/5

The first moment coefficient about the end of the range will therefore be

${\frac {C\int _{0}^{\infty }x^{\frac {n-1}{2}}e^{-{\frac {nx}{2\mu _{2}}}}dx}{I}}={\frac {C\left[{\frac {-2\mu _{2}}{n}}x^{\frac {n-1}{2}}e^{-{\frac {nx}{2\mu _{2}}}}\right]_{x=0}^{x=\infty }}{I}}+{\frac {C\int _{0}^{\infty }{\frac {n-1}{n}}\mu _{2}x^{\frac {n-3}{2}}e^{-{\frac {mx}{2\mu _{2}}}}dx}{I}}$.

The first part vanishes at each limit and the second is equal to

${\frac {{\frac {n-1}{n}}\mu _{2}I}{I}}={\frac {n-1}{n}}\mu _{2}$,

and we see that the higher moment coefficients will be formed by multiplying successively by ${\frac {n+1}{n}}\mu _{2}$, ${\frac {n+3}{n}}\mu _{2}$, etc., just as appeared to be the law of formation of $M'_{2}$, $M'_{3}$, $M'_{4}$, etc.

Hence it is probable that the curve found represents the theoretical distribution of $s^{2}$; so that although we have no actual proof we shall assume it to do so in what follows.

The distribution of $s$ may be found from this, since the frequency of $s$ is equal to that of $s^{2}$ and all that we must do is to compress the base line suitably.

Now if

$y_{1}=\phi (s^{2})$ be the frequency curve of $s^{2}$

and

$y_{2}=\psi (s)$ be„ the„ frequency„ curve„ of„ $s$,

then

$y_{1}d(s^{2})=y_{2}ds$,

or

	$y_{2}ds$	$=2y_{1}sds$,
$\therefore$	$y_{2}$	$=2sy_{1}$.

Hence

$y_{2}=2Cs(s^{2})^{\frac {n-3}{2}}e^{-{\frac {ns^{2}}{2\mu _{2}}}}$

is the distribution of $s$.

This reduces to

$y_{2}=2Cs^{n-2}e^{-{\frac {ns^{2}}{2\mu _{2}}}}$.

Hence $y=Ax^{n-2}e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}$ will give the frequency distribution of standard deviations of samples of $n$, taken out of a population distributed normally with standard deviation $\sigma$. The constant $A$ may be found by equating the area of the curve as follows:—

Area$=A\int _{0}^{\infty }x^{n-2}e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}dx$. (Let $I_{p}$ represent $\int _{0}^{\infty }x^{p}e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}dx$.)

Then

$I_{p}$	$={\frac {\sigma ^{2}}{n}}\int _{0}^{\infty }x^{p-1}{\frac {d}{dx}}\left(-e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}\right)dx$
	$={\frac {\sigma ^{2}}{n}}\left[-x^{p-1}e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}\right]_{x=0}^{x=\infty }+{\frac {\sigma ^{2}}{n}}(p-1)\int _{0}^{\infty }x^{p-2}e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}dx$
	$={\frac {\sigma ^{2}}{n}}(p-1)I_{p-2}$,

since the first part vanishes at both limits.