How would I show that if a distribution is equally likely to take values of $1$ or $4$, then the statistic $s_{n-1}$ forms a biased estimator of $\sigma$?
My thoughts: Do I find the expression$s_{n-1}$ in terms of the standard deviation formula:
$$s_{n-1}=\sqrt{\frac{\Sigma (x_i-2.5)^2}{n-1}}$$
Then show that: $$E\bigg(\sqrt{\frac{\Sigma (x_i-2.5)^2}{n-1}}\bigg)\ne \sigma$$
Note that $s_{n-1}$ is an unbiased estimator of $\sigma$ iff $|s_{n-1}|^2$ is an unbiased estimator of $\sigma^2$. Hence $$s_{n-1}^2={1\over n-1}\sum (x_i-2.5)^2$$It is not hard to see that $$\bar x=2.5$$therefore $$s_{n-1}^2={1\over n-1}\sum_{i=1}^{n} (x_i-\bar x)^2$$and we can write$$\Bbb E\{s_{n-1}^2\}={1\over n-1}\sum_{i=1}^{n} \Bbb E\{(x_i-\bar x)^2\}={1\over n-1}\sum_{i=1}^{n} \sigma^2={n\over n-1}\sigma^2\ne \sigma^2$$which implies a biased estimation$\blacksquare$