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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

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

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