Define $X$ as the random variable of interest. We observe a sequence of $T$ realized values for $x$, $x_{1},x_{2},\ldots, x_{T}$. The sample mean is \begin{equation*} m=\hat\mu = \frac{1}{T}\sum_{i=1}^{T}x_{i} \end{equation*} The sample variance is defined as \begin{equation*} s^{2}=\hat\sigma^{2} = \frac{1}{T-1}\sum_{i=1}^{T}(x_{i}-\hat\mu)^{2} \end{equation*} Suppose $x_{1},x_{2},\ldots, x_{T}$ have a normal distribution with mean $\mu$ and variance $\sigma^{2}$, it has been proved that \begin{equation*} (T-1)\hat\sigma^{2}/\sigma^{2} \sim \chi^{2}(T-1) \end{equation*} If the sample size $T$ is large enough, by chi-square distribution converges to a normal distribution, then: \begin{equation} \hat\sigma^{2} \sim N(\sigma^{2},\sigma^{4}\frac{2}{T-1}) \end{equation} The central limit theorem states that this distribution is only valid asymptotically. How come the result that, approximately, the sample standard deviation $\hat\sigma$ has a normal distribution with a standard error of \begin{equation} se(\hat\sigma)=\sigma \sqrt {\frac{1}{2T}} \end{equation} ? By definition of standard error, $se(\hat\sigma)=\sqrt{E[\hat \sigma^{2}-E(\hat\sigma^{2})]^{2}}=\sqrt{E[\hat\sigma^{2}]-(E[\hat\sigma])^2}=0$ I am wondering how to get the standard error $\sigma\sqrt{\frac{1}{2T}}$? Thanks!

  • $\begingroup$ How do you go from a chi square distribution to a normal distribution for the estimate. The chi square distribution depends on the normality of the data in the first place. $\endgroup$ – Michael Chernick May 15 '17 at 12:30
  • $\begingroup$ chi square distribution converges to normal distribution as the degree of freedom goes to infinity. $\endgroup$ – cmd1991 May 16 '17 at 1:26
  • $\begingroup$ The question is essentially that given the distribution of $\hat\sigma^{2}$, is there a way to get the standard error of $\hat\sigma$? $\endgroup$ – cmd1991 May 16 '17 at 3:15

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