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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: $$E\bigg(\sqrt{\frac{\Sigma (x_i-2.5)^2}{n-1}}\bigg)\ne \sigma$$

Aurora Borealis
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1 Answers1

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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$\blacksquare$

Mostafa Ayaz
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  • Thank you for your input. My question wants to show that it is a biased estimator though. Also why is the summation up to $n-1$ terms? – Aurora Borealis Mar 09 '20 at 09:00
  • Is your summation taken from $i=1$ to $i=n$? – Mostafa Ayaz Mar 09 '20 at 09:41
  • Well yes by definition of standard deviation, except its not divided by $n$ but $n-1$. But whatever the case is you are showing that it is an unbiased estimator, but the question asks otherwise which is probably from the different use of the summation index. – Aurora Borealis Mar 09 '20 at 13:45