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Unbiased estimator of variance

  1. Apr 14, 2010 #1
    Hey all,

    In Schaum's outline it claims that the sample variance of s^2 is a biased estimate of the population variance because its mean is given by:
    [tex] \mu_{s^{2}} = \frac{N-1}{N}\sigma^{2} [/tex]

    which I am cool with. It then says that the modified variance given by:
    [tex] \hat{s} = \frac{N}{N-1}s^{2} [/tex]
    is an unbiased estimator. I know this should be really easy, but I don't know how to show it.

    Also even if I accept that [tex] \hat{s}^{2} [/tex] is an unbiased estimator of the variance, Schaum's outline claims that [tex] \hat{s}[/tex] is a biased estimator of the population standard deviation. I don't see how this could be possible. if the variance is unbiased, and we take the square root of that unbiased estimator, won't the result also be unbiased ?

    Thanks,
    Thrillhouse
     
    Last edited: Apr 14, 2010
  2. jcsd
  3. Apr 14, 2010 #2

    statdad

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    Homework Helper

    What happens if you calculate

    [tex]
    E(\hat{s}^2) = E\left(\frac{N}{N-1} s^2\right)
    [/tex]

    using the first result you mention.

    On the second point: you could work out the distribution of [tex] s [/tex] and then find the expectation and see that [itex] E(s) \ne \sigma [/itex], or simply take as explanation the fact that even thought

    [tex]
    s = \sqrt{s^2}
    [/tex]

    it is not true that

    [tex]
    E(s) = \sqrt{E(s^2)}
    [/tex]

    which would have to be true to have [itex] s [/itex] as an unbiased estimator of [itex] \sigma [/itex].
     
  4. Apr 14, 2010 #3

    mathman

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    Science Advisor
    Gold Member

    It sounds like you are overcomplicating the problem. Your first equation shows a bias factor of (N-1)/N, so simply multiplying by N/(N-1) removes the bias.
     
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