Biased Estimator for standard deviation

From: Fred Ma (fma_at_doe.carleton.ca)
Date: 07/18/04


Date: 18 Jul 2004 04:10:07 GMT

No question, just a comment. It is bizzare that there is an unbiased
estimator for a population's variance, but not for its standard
deviation. Especially since population's standard deviation is simply
the square root of the variance, by definition. The proof is clear to
see: http://mathforum.org/epigone/ap-stat/zayplunplald.

Or perhaps an unbiased estimator for standard deviation exists, but it
certainly is not the commonly defined sample standard deviation (which
is just square root of the sample variance). This sort of makes sense
if one remembers that the sample variance has a PDF. The square-root
conversion of a sample variance to sample standard deviation is just a
continuous one-to-one mapping, yielding an altered PDF for the sample
standard deviation. For the sample standard deviation to be unbiased,
the mean of this distribution should match the standard deviation of
the population. This in turn means that it should match the
square-root of the population variance. Since the population variance
equals the mean of the sample variance, the sample standard deviation
is unbiased if the mean of its PDF is the square-root of the mean of
the PDF for sample variance.

In short, the sample standard deviation is simply the square-root of
the sample variance. For it to be unbiased, its mean must be the
square-root of the mean for sample variance. In general, this is not
true i.e. if a random sample X undergoes mapping f(X), the mean of
f(X) is generally not f(mean(X)).

Anyway, I always wondered about this. Actually, I always wondered
about the why sample variance used N-1 for normalization, and then ran
into the above website, which also presented this caveat regarding
sample standard deviation.



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