Let samples be taken from a population with central moments . The sample variance is then given by
(1)
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where is the sample mean.
The expected value of for a sample size is then given by
(2)
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Similarly, the expected variance of the sample variance is given by
(3)
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(4)
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(Kenney and Keeping 1951, p. 164; Rose and Smith 2002, p. 264).
The algebra of deriving equation (4) by hand is rather tedious, but can be performed as follows. Begin by noting that
(5)
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so
(6)
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The value of is already known from equation (◇), so it remains only to find . The algebra is simplified considerably by immediately transforming variables to and performing computations with respect to these central variables. Since the variance does not depend on the mean of the underlying distribution, the result obtained using the transformed variables will give an identical result while immediately eliminating expectation values of sums of terms containing odd powers of (which equal 0). To determine , expand equation (6) to obtain
(7)
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(8)
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(9)
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(10)
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Working on the first term of (10),
(11)
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(12)
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(13)
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(14)
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The second term of (◇) is given by
(15)
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(16)
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and the third term by
(17)
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(18)
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(19)
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Plugging (◇)-(19) into (◇) then gives
(20)
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(21)
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(22)
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(23)
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(Kenney and Keeping 1951, p. 164). Plugging (◇) and (23) into (◇) then gives
(24)
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(25)
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as before.
For a normal distribution, and , so
(26)
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(27)
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The third ane fourth moments of are given by
(28)
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(29)
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giving the skewness and kurtosis excess of the distribution of the as
(30)
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(31)
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as computed by Student. Student also conjectured that the underlying distribution is Pearson type III distribution
(32)
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where is the gamma function--a conjecture that was subsequently proven by R. A. Fisher. Curves are illustrated above for and varying from to 10.