Question related to the sample standard deviation bias

Question

Let $ X_{1},...,X_{n} $ be a random sample of a population with finite variance. Prove that if $ \sigma^{2} > 0 $, then $ E(S_{n}) < \sigma $.

Based on Zen's answer

$0 \leq \mathrm{Var}(S_n) = \mathrm{E}(S_n^2) - \mathrm{E}^2(S_n) \;\;\Leftrightarrow\;\; \mathrm{E}^2(S_n) \leq \mathrm{E}(S_n^2) \;\;\Leftrightarrow\;\; \mathrm{E}(S_n) \leq \sqrt{\mathrm{E}(S_n^2)} =\sigma.$

From here we have that $ E(S_{n}) \leq \sigma$.

Using the condition $ \sigma^{2} >0 $, I can not see how to arrive at the strict inequality $ E(S_{n}) < \sigma $.

Isn't $\sigma$$^2$ = Var($S_n$)? If so the the equality on the far right is wrong. — Michael R. Chernick, Apr 27 '18 at 03:04
@MichaelChernick $ \sigma^{2} $ is the variance of the population and Var($ S_{n}) $ is the variance of the statistic $ S_{n} $. — Jacob Schwartz, Apr 27 '18 at 03:24

score 1 · Accepted Answer · answered Apr 27 '18 at 02:37

1

If $\sigma > 0$ then $\mathbb{Var}(S_n) > 0$ so you can use the exact same reasoning but with strict inequality.

answered Apr 27 '18 at 02:37

Ben

124,856

All variances are >=0. Since $\sigma$.>0 and Var($S_n$} approaches $\sigma^2$, Var($S_n$)>0 for large enough n. – Michael R. Chernick Apr 27 '18 at 04:18

Question related to the sample standard deviation bias

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