variance of standard sample deviation in a normal distribution

Question

I need help. I Know that using Fisher Lemma, we know that in a normal distribution $ Var(\hat{\sigma^2})= \frac{2\sigma^4(n-1)}{n^2}$, since $\frac{n\hat{\sigma^2}}{\sigma^2}\approx \chi^2_{n-1}$. I need now to calculate $Var(\hat{\sigma})$, but i don´t know the relation between $var(X^2) $ and $Var(X)$ in a normal distribution. Thank you everyone.

Answer 1

I need now to calculate $Var(\hat{\sigma})$

Let's state $\hat{\sigma}^2=S^2$

you can express

$$\mathbb{V}[S]=\frac{\sigma^2}{n}\mathbb{V}\Bigg[\sqrt{\frac{nS^2}{\sigma^2}\Bigg]}$$

the calculation is not immediate but not difficult because now you can calculate the simple moments using the distribution of $\frac{nS^2}{\sigma^2}$

In particular, set $\frac{nS^2}{\sigma^2}=Y$ you have

$$\mathbb{V}[S]=\frac{\sigma^2}{n}\Bigg\{\mathbb{E}[Y]-\mathbb{E}^2[\sqrt{Y}]\Bigg\}=\frac{\sigma^2}{n}\Bigg\{n-1-\Bigg[\int_0^{\infty}\sqrt{y}f_Y(y)dy\Bigg]^2\Bigg\}$$

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Now the only hurdle to jump over is to calculate

$$\int_0^{\infty}\sqrt{y}\frac{\Big(\frac{1}{2}\Big)^{\frac{n-1}{2}}}{\Gamma\Big(\frac{n-1}{2}\Big)}y^{\frac{n-1}{2}-1}e^{-\frac{y}{2}}dy=\frac{\Gamma\Big(\frac{n}{2}\Big)\sqrt{2}}{\Gamma\Big(\frac{n-1}{2}\Big)}\underbrace{\int_0^{\infty}\frac{\Big(\frac{1}{2}\Big)^{\frac{n}{2}}}{\Gamma\Big(\frac{n}{2}\Big)}y^{\frac{n}{2}-1}e^{-\frac{y}{2}}dy}_{=1}=\frac{\Gamma\Big(\frac{n}{2}\Big)\sqrt{2}}{\Gamma\Big(\frac{n-1}{2}\Big)}$$

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Usually $\frac{(n-1)S^2}{\sigma^2}\sim \chi_{(n-1)}^2$ is taken into consideration because it is used to consider the unbiased estimator of $\sigma^2$. It is not forbidden to use the statistic you showed; the procedure remains the same, just modify it accordingly

Answer 2

Hint: If $Y\sim N(0,1)$ the $X \stackrel {d} {=}\mu+\sigma Y$ where $\mu$ is the mean of $X$ and $\sigma^{2}$ is the variance. Hence, $EX^{4}=E(\mu+\sigma Y)^{4}$ and $EX^{2}=E(\mu+\sigma Y)^{2}$ You can compute these by expanding the powers using Binomial Theorem.