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k-Statistic

An Unbiased Estimator of the Cumulants $\kappa_i$ of a Distribution. The expectation values of the $k$-statistics are therefore given by the corresponding Cumulants

$\displaystyle \left\langle{k_1}\right\rangle{}$ $\textstyle =$ $\displaystyle \kappa_1$ (1)
$\displaystyle \left\langle{k_2}\right\rangle{}$ $\textstyle =$ $\displaystyle \kappa_2$ (2)
$\displaystyle \left\langle{k_3}\right\rangle{}$ $\textstyle =$ $\displaystyle \kappa_3$ (3)
$\displaystyle \left\langle{k_4}\right\rangle{}$ $\textstyle =$ $\displaystyle \kappa_4$ (4)

(Kenney and Keeping 1951, p. 189). For a sample of size, $N$, the first few $k$-statistics are given by
$\displaystyle k_1$ $\textstyle =$ $\displaystyle m_1$ (5)
$\displaystyle k_2$ $\textstyle =$ $\displaystyle {N\over N-1}m_2$ (6)
$\displaystyle k_3$ $\textstyle =$ $\displaystyle {N^2\over (N-1)(N-2)}m_3$ (7)
$\displaystyle k_4$ $\textstyle =$ $\displaystyle {N^2[(N+1)m_4-3(N-1){m_2}^2]\over(N-1)(N-2)(N-3)},$ (8)

where $m_1$ is the sample Mean, $m_2$ is the sample Variance, and $m_i$ is the sample $i$th Moment about the Mean (Kenney and Keeping 1951, pp. 109-110, 163-165, and 189; Kenney and Keeping 1962). These statistics are obtained from inverting the relationships


$\displaystyle \left\langle{m_1}\right\rangle{}$ $\textstyle =$ $\displaystyle \mu$ (9)
$\displaystyle \left\langle{m_2}\right\rangle{}$ $\textstyle =$ $\displaystyle {N-1\over N} \mu_2$ (10)
$\displaystyle \left\langle{{m_2}^2}\right\rangle{}$ $\textstyle =$ $\displaystyle {(N-1)[(N-1)\mu_4+(N^2-2N+3){\mu_2}^2]\over N^3}$ (11)
$\displaystyle \left\langle{m_3}\right\rangle{}$ $\textstyle =$ $\displaystyle {(N-1)(N-2)\over N^2}\mu_3$ (12)
$\displaystyle \left\langle{m_4}\right\rangle{}$ $\textstyle =$ $\displaystyle {(N-1)[(N^2-3N+3)\mu_4+3(2N-3){\mu_2}^2]\over N^3}.$ (13)


The first moment (sample Mean) is

\begin{displaymath}
m_1 \equiv \left\langle{x}\right\rangle{}= {1\over N} \sum_{i=1}^N x_i,
\end{displaymath} (14)

and the expectation value is
\begin{displaymath}
\left\langle{m_1}\right\rangle{} = \left\langle{{1\over N}\sum_{i=1}^N x_i}\right\rangle{} =\mu.
\end{displaymath} (15)

The second Moment (sample Standard Deviation) is
$\displaystyle m_2$ $\textstyle \equiv$ $\displaystyle \left\langle{(x-\mu)^2}\right\rangle{} = \left\langle{x^2}\right\...
...mu\left\langle{x}\right\rangle{}+\mu^2 = \left\langle{x^2}\right\rangle{}-\mu^2$  
  $\textstyle =$ $\displaystyle {1\over N}\sum_{i=1}^N {x_i}^2-\left({{1\over N} \sum_{i=1}^N {x_i}}\right)^2$  
  $\textstyle =$ $\displaystyle {1\over N} \sum_{i=1}^N {x_i}^2 -{1\over N^2} \left({\sum_{i=1}^N {x_i}^2+\sum_{\scriptstyle{i,j=1}\atop \scriptstyle i\not=j}^N x_ix_j}\right)$  
  $\textstyle =$ $\displaystyle {N-1\over N^2} \sum_{i=1}^N {x_i}^2-{1\over N^2} \sum_{\scriptstyle {i,j=1}\atop \scriptstyle i\not= j}^N x_ix_j,$ (16)

and the expectation value is
$\displaystyle \left\langle{m_2}\right\rangle{}$ $\textstyle =$ $\displaystyle {N-1\over N} \left\langle{{1\over N} \sum_{i=1}^N {x_i}^2}\right\...
...{\sum_{\scriptstyle{i,j=1}\atop \scriptstyle i\not= j}^N x_ix_j}\right\rangle{}$  
  $\textstyle =$ $\displaystyle {N-1\over N} \mu'_2-{N(N-1)\over N^2}\mu^2,$ (17)

since there are $N(N-1)$ terms $x_ix_j$, using
\begin{displaymath}
\left\langle{x_ix_j}\right\rangle{}=\left\langle{x_i}\right\...
...langle{x_j}\right\rangle{}=\left\langle{x_i}\right\rangle{}^2,
\end{displaymath} (18)

and where $\mu'_2$ is the Moment about 0. Using the identity
\begin{displaymath}
\mu'_2=\mu_2+\mu^2
\end{displaymath} (19)

to convert to the Moment $\mu_2$ about the Mean and simplifying then gives
\begin{displaymath}
\left\langle{m_2}\right\rangle{} = {N-1\over N} \mu_2.
\end{displaymath} (20)

The factor $(N-1)/N$ is known as Bessel's Correction.


The third Moment is


$\displaystyle m_3$ $\textstyle \equiv$ $\displaystyle \left\langle{(x-\mu)^3}\right\rangle{} = \left\langle{x^3-3\mu x^2+3\mu^2x-\mu^3}\right\rangle{}$  
  $\textstyle =$ $\displaystyle \left\langle{x^3}\right\rangle{}-3\mu\left\langle{x^2}\right\rangle{}+3\mu^2\left\langle{x}\right\rangle{}-\mu^3$  
  $\textstyle =$ $\displaystyle \left\langle{x^3}\right\rangle{}-3\mu\left\langle{x^2}\right\rangle{}+3\mu^3-\mu^3$  
  $\textstyle =$ $\displaystyle \left\langle{x^3}\right\rangle{}-3\mu\left\langle{x^2}\right\rangle{}+2\mu^3$  
  $\textstyle =$ $\displaystyle {1\over N} \sum {x_i}^3 -3\left({{1\over N}\sum x_i}\right)\left({{1\over N} \sum {x_j}^2}\right)+2\left({{1\over N} \sum x_i}\right)^3$  
  $\textstyle =$ $\displaystyle {1\over N}\sum {x_i}^3-{3\over N^2}\left({\sum x_i}\right)\left({\sum {x_j}^2}\right)+{2\over N^3} \left({\sum x_i}\right)^3.$ (21)

Now use the identities
$\quad \left({\sum {x_i}^2}\right)\left({\sum x_j}\right)=\sum{x_i}^3+\sum{x_i}^2x_j$ (22)
$\quad \left({\sum x_i}\right)^3=\sum{x_i}^3+3\sum{x_i}^2x_j+6\sum x_ix_jx_k,$ (23)
where it is understood that sums over products of variables exclude equal indices. Plugging in


\begin{displaymath}
m_3=\left({{1\over N}-{3\over N^2}+{2\over N^3}}\right)\sum{...
...r N^3}}\right)\sum{x_i}^2x_j+6\cdot{2\over N^3}\sum x_ix_jx_k.
\end{displaymath} (24)

The expectation value is then given by


\begin{displaymath}
\left\langle{m_3}\right\rangle{}= \left({{1\over N}-{3\over ...
...\mu_2'\mu+{12\over N^3} {\textstyle{1\over 6}}N(N-1)(N-2)\mu^3
\end{displaymath} (25)

where $\mu_2'$ is the Moment about 0. Plugging in the identities
$\displaystyle \mu'_2$ $\textstyle =$ $\displaystyle \mu_2+\mu^2$ (26)
$\displaystyle \mu'_3$ $\textstyle =$ $\displaystyle \mu_3+3\mu_2\mu+\mu^3$ (27)

and simplifying then gives
\begin{displaymath}
\left\langle{m_3}\right\rangle{} = {(N-1)(N-2)\over N^2} \mu_3
\end{displaymath} (28)

(Kenney and Keeping 1951, p. 189).


The fourth Moment is


$\displaystyle m_4$ $\textstyle =$ $\displaystyle \left\langle{(x-\mu)^4}\right\rangle{} = \left\langle{x^4-4x^3\mu+6x^2\mu^2-4x\mu^3+\mu^4}\right\rangle{}$  
  $\textstyle =$ $\displaystyle \left\langle{x^4}\right\rangle{}-4\mu\left\langle{x^3}\right\rangle{}+6\mu^2\left\langle{x^2}\right\rangle{}-3\mu^4$  
  $\textstyle =$ $\displaystyle {1\over N} \sum {x_i}^4-{4\over N^2} \left({\sum x_i}\right)\left...
... x_i}\right)^2\left({\sum {x_j}^2}\right)-{3\over N^4}\left({\sum x_i}\right)^4$ (29)

Now use the identities

$\left({\sum x_i}\right)\left({\sum{x_j}^3}\right)=\sum{x_i}^4+\sum{x_i}^3x_j$ (30)
$\left({\sum x_i}\right)^2\left({\sum{x_j}^2}\right)=\sum{x_i}^4+2\sum{x_i}^3x_j+2\sum{x_i}^2{x_j}^2+2\sum{x_i}^2x_jx_k$ (31)
$\left({\sum x_i}\right)^4=\sum{x_i}^4+4\sum{x_i}^3x_j+6\sum{x_i}^2{x_j}^2+12\sum{x_i}^2x_jx_k+24\sum x_ix_jx_kx_l.$ (32)
Plugging in,

$\displaystyle m_4$ $\textstyle =$ $\displaystyle \left({{1\over N}-{4\over N^2}+{6\over N^3}-{3\over N^4}}\right)\sum{x_i}^4$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({-{4\over N^2}+2\cdot{6\over N^3}-4\cdot{3\over N^4}}\right)\sum{x_i}^3 x_j$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({2\cdot{6\over N^3}-6\cdot{3\over N^4}}\right)\sum{x_i}^2{x_j}^2$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({2\cdot{6\over N^3}-12\cdot{3\over N^4}}\right)\sum{x_i}^2x_jx_k$  
  $\textstyle \phantom{=}$ $\displaystyle -24\cdot{3\over N^4}\sum x_ix_jx_kx_l.$ (33)

The expectation value is then given by
$\displaystyle \left\langle{m_4}\right\rangle{}$ $\textstyle =$ $\displaystyle \left({{1\over N}-{4\over N^2}+{6\over N^3}-{3\over N^4}}\right)N\mu'_4$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({-{4\over N^2}+{12\over N^3}-{12\over N^4}}\right)N(N-1)\mu'_3\mu$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({{12\over N^3}-{18\over N^4}}\right){\textstyle{1\over 2}}N(N-1){\mu'_2}^2$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({{18\over N^3}-{36\over N^4}}\right){\textstyle{1\over 2}}N(N-1)(N-2)\mu_2'\mu^2$  
  $\textstyle \phantom{=}$ $\displaystyle -{72\over N^4} {\textstyle{1\over 24}} N(N-1)(N-2)(N-3)\mu^4,$ (34)

where $\mu_i'$ are Moments about 0. Using the identities
$\displaystyle \mu_2'$ $\textstyle =$ $\displaystyle \mu_2+\mu^2$ (35)
$\displaystyle \mu_3'$ $\textstyle =$ $\displaystyle \mu_3+3\mu_2\mu+\mu^3$ (36)
$\displaystyle \mu_4'$ $\textstyle =$ $\displaystyle \mu_4+4\mu_3\mu+6\mu_2\mu^2+\mu^4$ (37)

and simplifying gives
\begin{displaymath}
\left\langle{m_4}\right\rangle{}={(N-1)[(N^2-3N+3)\mu_4+3(2N-3){\mu_2}^2]\over N^3}
\end{displaymath} (38)

(Kenney and Keeping 1951, p. 189).


The square of the second moment is


$\displaystyle {m_2}^2$ $\textstyle =$ $\displaystyle (\left\langle{x^2}\right\rangle{}-\mu^2)^2=\left\langle{x^2}\right\rangle{}^2-2\mu^2\left\langle{x^2}\right\rangle{}+\mu^4$  
  $\textstyle =$ $\displaystyle \left({{1\over N}\sum{x_i}^2}\right)^2-2\left({{1\over N}\sum x_i}\right)^2\left({{1\over N}\sum{x_i}^2}\right)+\left({{1\over N}\sum x_i}\right)^4$  
  $\textstyle =$ $\displaystyle {1\over N^2}\left({\sum{x_i}^2}\right)^2-{2\over N^3}\left({\sum x_i}\right)^2\left({\sum{x_j}^2}\right)+{1\over N^4}\left({\sum x_i}\right)^4.$ (39)

Now use the identities

$\left({\sum{x_i}^2}\right)^2=\sum{x_i}^4+2\sum{x_i}^2{x_j}^2$ (40)
$\left({\sum x_i}\right)^2\left({\sum{x_j}^2}\right)=\sum{x_i}^4+2\sum{x_i}^2{x_j}^2+2\sum{x_i}^3x_j+2\sum{x_i}^2x_jx_k$ (41)
$\left({\sum x_i}\right)^4=\sum{x_i}^4+6\sum{x_i}^2{x_j}^2+4\sum{x_i}^3x_j+12\sum{x_i}^2x_jx_k+24\sum x_ix_jx_kx_l.$ (42)
Plugging in,

$\displaystyle {m_2}^2$ $\textstyle =$ $\displaystyle \left({{1\over N^2}-{2\over N^3}+{1\over N^4}}\right)\sum{x_i}^4$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({2\cdot{1\over N^2}-2\cdot{2\over N^3}+6\cdot{1\over N^4}}\right)\sum{x_i}^2{x_j}^2$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({-2\cdot{2\over N^3}+4\cdot{1\over N^4}}\right)\sum{x_i}^3x_j$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({-2\cdot{2\over N^3}+12\cdot{1\over N^4}}\right)\sum{x_i}^2x_jx_k$  
  $\textstyle \phantom{=}$ $\displaystyle +{24\over N^4}\sum x_ix_jx_kx_l$ (43)

The expectation value is then given by
$\displaystyle \left\langle{{m_2}^2}\right\rangle{}$ $\textstyle =$ $\displaystyle \left({{1\over N^2}-{2\over N^3}+{1\over N^4}}\right)N\mu_4'$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({{2\over N^2}-{4\over N^3}+{6\over N^4}}\right){\textstyle{1\over 2}}N(N-1)\mu_2'^2$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({-{4\over N^3}+{4\over N^4}}\right)N(N-1)\mu_3'\mu$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({-{4\over N^3}+{12\over N^4}}\right){\textstyle{1\over 2}}N(N-1)(N-2)\mu_2'\mu^2$  
  $\textstyle \phantom{=}$ $\displaystyle +{24\over N^4}{\textstyle{1\over 24}}N(N-1)(N-2)(N-3)\mu^4$ (44)

where $\mu_i'$ are Moments about 0. Using the identities
$\displaystyle \mu_2'$ $\textstyle =$ $\displaystyle \mu_2+\mu^2$ (45)
$\displaystyle \mu_3'$ $\textstyle =$ $\displaystyle \mu_3+3\mu_2\mu+\mu^3$ (46)
$\displaystyle \mu_4'$ $\textstyle =$ $\displaystyle \mu_4+4\mu_3\mu+6\mu_2\mu^2+\mu^4$ (47)

and simplifying gives
\begin{displaymath}
\left\langle{{m_2}^2}\right\rangle{}={(N-1)[(N-1)\mu_4+(N^2-2N+3){\mu_2}^2]\over N^3}
\end{displaymath} (48)

(Kenney and Keeping 1951, p. 189).


The Variance of $k_2$ is given by

\begin{displaymath}
{\rm var}(k_2)={\kappa_4\over N}+{2\over(N-1){\kappa_2}^2},
\end{displaymath} (49)

so an unbiased estimator of ${\rm var}(k_2)$ is given by
\begin{displaymath}
\hat{\rm var}(k_2)={2{k_2}^2N+(N-1)k_4\over N(N+1)}
\end{displaymath} (50)

(Kenney and Keeping 1951, p. 189). The Variance of $k_3$ can be expressed in terms of Cumulants by
\begin{displaymath}
{\rm var}(k_3)={\kappa_6\over N}+{9\kappa_2\kappa_4\over N-1}+{9{\kappa_3}^2\over N-1}+{6{\kappa_2}^3\over N(N-1)(N-2)}.
\end{displaymath} (51)

An Unbiased Estimator for ${\rm var}(k_3)$ is
\begin{displaymath}
\hat{\rm var}(k_3)={6{k_2}^2N(N-1)\over (N-2)(N+1)(N+3)}
\end{displaymath} (52)

(Kenney and Keeping 1951, p. 190).


Now consider a finite population. Let a sample of $N$ be taken from a population of $M$. Then Unbiased Estimators $M_2$ for the population Mean $\mu$, $M_2$ for the population Variance $\mu_2$, $G_1$ for the population Skewness $\gamma_1$, and $G_2$ for the population Kurtosis $\gamma_2$ are


$\displaystyle M_1$ $\textstyle =$ $\displaystyle \mu$ (53)
$\displaystyle M_2$ $\textstyle =$ $\displaystyle {M-N\over N(M-1)} \mu_2$ (54)
$\displaystyle G_1$ $\textstyle =$ $\displaystyle {M-2N\over M-2} \sqrt{M-1\over N(M-N)}\, \gamma_1$ (55)
$\displaystyle G_2$ $\textstyle =$ $\displaystyle {(M-1)(M^2-6MN+M+6N^2)\gamma_2\over N(M-2)(M-3)(M-N)}-{6M(MN+M-N^2-1)\over N(M-2)(M-3)(M-N)}$ (56)

(Church 1926, p. 357; Carver 1930; Irwin and Kendall 1944; Kenney and Keeping 1951, p. 143), where $\gamma_1$ is the sample Skewness and $\gamma_2$ is the sample Kurtosis.

See also Gaussian Distribution, Kurtosis, Mean, Moment, Skewness, Variance


References

Carver, H. C. (Ed.). ``Fundamentals of the Theory of Sampling.'' Ann. Math. Stat. 1, 101-121, 1930.

Church, A. E. R. ``On the Means and Squared Standard-Deviations of Small Samples from Any Population.'' Biometrika 18, 321-394, 1926.

Irwin, J. O. and Kendall, M. G. ``Sampling Moments of Moments for a Finite Population.'' Ann. Eugenics 12, 138-142, 1944.

Kenney, J. F. and Keeping, E. S. Mathematics of Statistics, Pt. 2, 2nd ed. Princeton, NJ: Van Nostrand, 1951.

Kenney, J. F. and Keeping, E. S. ``The $k$-Statistics.'' §7.9 in Mathematics of Statistics, Pt. 1, 3rd ed. Princeton, NJ: Van Nostrand, pp. 99-100, 1962.



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© 1996-9 Eric W. Weisstein
1999-05-26