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Let $X_1, \dots , X_n$ be an independent sample from a normal distribution with variance $\sigma^2$, and define $$S^2 = \frac{1}{n-1}\sum_{i=1}^{n}(X-\bar X)^2$$ We know that $S^2$ is unbiased for $\sigma^2$, i.e. $\Bbb{E}[S^2] = \sigma^2$.

Show however, that $\Bbb{E}[S] = c\sigma$, finding $c$ in terms of $n$. Is $c$ ever equal to $1$?

Any hints on how to start this question off would be appreciated.

Galc127
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Eoin
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  • Presumably $S$ is the positive square root of $S^2$ – Henry Feb 08 '18 at 17:51
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    $S^2$ has dist'n related to chi-sq. Set up the integral for $E(S)$ and manipulate the constant so the integrand is that of a known density function. $c = \sqrt{\frac{2}{n-1}}\Gamma(n/2)/\Gamma(\frac{n-1}{2}).$ (That's from memory, think it's right). Never 1, but converges. – BruceET Feb 08 '18 at 18:40
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    See here – V. Vancak Feb 08 '18 at 20:25
  • https://math.stackexchange.com/questions/2777526/expectation-of-sample-standard-deviation/ – StubbornAtom Jan 14 '20 at 15:21

1 Answers1

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Knowing that $\dfrac{(n-1)S^2}{\sigma^2}\sim\chi^2_{(n-1)}$, it can be easily shown that its $r$th order raw moment about $0$ is

$\displaystyle\mathbb{E}\left(\frac{(n-1)S^2}{\sigma^2}\right)^r=2^r\frac{\Gamma\left(\frac{n-1}{2}+r\right)}{\Gamma\left(\frac{n-1}{2}\right)}$ for $r>\frac{1-n}{2}$ and $n>1$.

Choose $r=\frac{1}{2}$, so that for $n>1$,

$\displaystyle\mathbb{E}(S)=c(n)\sigma$, where $\displaystyle c(n)=\sqrt{\frac{2}{n-1}}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}$.

StubbornAtom
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