# Unbiased estimator of variance

1. Apr 14, 2010

### 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

Last edited: Apr 14, 2010
2. Apr 14, 2010

What happens if you calculate

$$E(\hat{s}^2) = E\left(\frac{N}{N-1} s^2\right)$$

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

$$s = \sqrt{s^2}$$

it is not true that

$$E(s) = \sqrt{E(s^2)}$$

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

3. Apr 14, 2010

### mathman

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.