Is the Square Root of an Unbiased Variance Estimator Also Unbiased?

[thrillhouse86]

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:
\mu_{s^{2}} = \frac{N-1}{N}\sigma^{2}

which I am cool with. It then says that the modified variance given by:
\hat{s} = \frac{N}{N-1}s^{2}
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 \hat{s}^{2} is an unbiased estimator of the variance, Schaum's outline claims that \hat{s} 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

 

[statdad]

What happens if you calculate

<br /> E(\hat{s}^2) = E\left(\frac{N}{N-1} s^2\right)<br />

using the first result you mention.

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

<br /> s = \sqrt{s^2}<br />

it is not true that

<br /> E(s) = \sqrt{E(s^2)}<br />

which would have to be true to have s as an unbiased estimator of \sigma.

 

[mathman]

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:
\mu_{s^{2}} = \frac{N-1}{N}\sigma^{2}

which I am cool with. It then says that the modified variance given by:
\hat{s} = \frac{N}{N-1}s^{2}
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 \hat{s}^{2} is an unbiased estimator of the variance, Schaum's outline claims that \hat{s} 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

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.