Question about standard deviation and central limit theorem I have a quick question about the central limit theorem. Lets say I measure some value that comes from an arbitrary distribution N times and I repeat this M times. I understand that if I calculcate the mean from the N values I will have a set of M values that follows a normal distribution. But what if I measure the sample standard deviation from N, will my resulting distribution also be normal? Following the derivation of the CLT I do not see this to be the case, but intuitively I think this is true, at least for some distributions. Any light on the issue would be greatly appreciated.
First I will quote the CLT from wiki: 

the central limit theorem (CLT) states that, given certain conditions, the arithmetic mean of a sufficiently large number of iterates of independent random variables, each with a well-defined expected value and well-defined variance, be approximately normally distributed.

My question then is a variant on the quote from the wiki page:  will the the central limit theorem (CLT) state that, given certain conditions, the STANDARD DEVIATION of a sufficiently large number of iterates of independent random variables, each with a well-defined expected value and well-defined variance, be approximately normally distributed? Here is a series of plots I made from random numbers that follow a beta distribution. I have generated 1000 sets of 5000 points each. The first plot is a histogram of the first set. The second is a histogram of the 1000 calculated means, and the 3rd is a histogram of the 1000 calculated std.



 A: Yes, the sample standard deviation is asymptotically normal. Let the sample standard deviation be $\hat{\sigma} = \sqrt{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2}$, and let $\sigma$ be the population standard deviation. Let's use the central limit theorem to show that
$$ \sqrt{n}(\hat{\sigma} - \sigma) \xrightarrow{d} N(0, V). $$
First write things as
$$ \sqrt{n}(\hat{\sigma} - \sigma) = \sqrt{n}\left(\sqrt{ \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2} - \sqrt{\sigma^2} \right)$$
The central limit theorem tells is about how sample moments minus population moments behave. If we didn't have square roots above, we'd just have something like sample moments minus population ones, and we could use the central limit theorem. To get rid of the square roots, let's take a Taylor expansion of the first square root around $\sigma^2$.
$$ \sqrt{ \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2} = \sqrt{\sigma^2} + \frac{1}{2\sigma} \left( \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 - \sigma^2\right) + O\left(\left( \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 - \sigma^2\right)^2\right)
$$
Plugging this into the above, we have
$$\sqrt{n}(\hat{\sigma} - \sigma) = \frac{\sqrt{n}}{2 \sigma} \left( \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 - \sigma^2\right) + O\left(\frac{1}{n^{3/2}} \left(\sum_{i=1}^n (x_i - \bar{x})^2 - \sigma^2\right)^2\right)$$
Rearranging the first term on the right gives
$$\frac{\sqrt{n}}{2 \sigma} \left( \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 - \sigma^2\right) = \frac{1}{2\sigma \sqrt{n}} \sum_{i=1}^n ((x_i - E[X])^2 - \sigma^2) - \frac{\sqrt{n}}{2\sigma}(\bar{x} - E[x])^2  $$
The central limit theorem tells us that under some conditions $\frac{1}{\sqrt{n}} \sum_{i=1}^n (y_i - E[y]) \xrightarrow{d} N(0, \sigma_y^2)$. With $y_i = (x_i - E[x])^2$ and $E[y] = \sigma^2$, then the first term on the right converges in distribution to a normal.
The second term we can write as $\sqrt{n}(\bar{x} - E[x]) (\bar{x} - E[x])$, and since $\sqrt{n}(\bar{x} - E[x]) \xrightarrow{d} N$ and $(\bar{x} - E[x]) \xrightarrow{p} 0$, by Slutsky's lemma, the product converges in probability to 0.
Similarly, we could show that, $\frac{1}{n^{3/2}} \left(\sum_{i=1}^n (x_i - \bar{x})^2 - \sigma^2\right)^2 \xrightarrow{p} 0$, so the remainder from the Taylor expansion vanishes.
This Taylor expansion trick comes up often, so it has a name. It's called the delta method.
A: In the comments on Paul's answer Whuber commented that the case of a Bernoulli variable with $p=1/2$ contradicts with his argument. In this question we look further into the Delta method with a graphic. This will provide some intuition and explanation about the (multivariate) Delta method and it explains why the Bernoulli variable is an exception (the only exception along with a degenerate variable).
Graphical description and intuition

*

*The standard deviation of a sample equals $\hat\sigma = \sqrt{\hat\mu_2-{\hat\mu_1}^2}$ with $\hat\mu_1 = \frac{1}{n} \sum {x_i}$ and $\hat\mu_2 = \frac{1}{n} \sum {x_i}^2$. We can consider the sample joint distribution of $\mu_1,\mu_2$. The deviation of this sample distribution (scaled by $\sqrt{n}$) should approach a multivariate normal distribution with covariance equal to the covariance of the population distribution of a single point.


*On top of the distribution of $\hat\mu_1$ and $\hat\mu_2$ we can consider isolines for $\hat\sigma = \sqrt{\hat\mu_2-{\hat\mu_1}^2} = constant$.


*The $\hat\sigma$ is a non-linear function $\hat\mu_1$ and $\hat\mu_2$ but when we consider a small region then the non-linear function can be approximated with a linear function (the delta method does not only require that that a distribution approximates a normal distribution, but also that the variance becomes small).
If the mean $E(\hat\mu_1) = 0$ (which we can choose without loss of generality by translation of the variable) then $\hat\sigma^2 = \hat\mu_2-{\hat\mu_1}^2$ becomes approximately a normal distributed variable with mean equal to the second moment of the variable and variance equal to the difference of the fourth and second moment $$\sqrt{n}( \hat\sigma^2 - E(\hat\sigma^2)) \xrightarrow[]{P} N(0, E(X^4)-E(X^2)^2)$$
The distribution of $\hat\sigma = \sqrt{\hat\sigma^2}$ can be related to this by the series expansion around $\hat\sigma^2 = E(\hat\sigma^2)$ which is $$\sqrt{\hat\sigma^2} = \sqrt{E(\hat\sigma^2)} + \frac{\hat\sigma^2 - E(\hat\sigma^2)}{2\sqrt{E(\hat\sigma^2)}} + O\left((\hat\sigma^2 - E(\hat\sigma^2))^2\right)$$ and for the limit distribution
$$\sqrt{n}( \hat\sigma - \sqrt{E(\hat\sigma^2)}) \xrightarrow[]{P} N\left(0,\frac{ E(X^4)-E(X^2)^2}{4 E(\hat\sigma^2)}\right)$$
This is the limiting distribution. It might not be optimal for approximations of intermediate steps. For instance the mean $\sqrt{E(\hat\sigma^2)}$ is biased and not the same as $E(\hat\sigma)$ (but the bias shrinks to 0 for increasing sample size and that's why it still works as a limit distribution). A similar situation is when you take the logarithm of an approximately normal distributed variable, and you can do better than the Delta method.

The issue with the Bernoulli variable is that the limiting sample distribution of $\hat\mu_1,\hat\mu_2$ is a fully correlated multivariate normal distribution (the population distribution of the $x_i,{x_i}^2$ are just two points and the distribution of the sample mean will be on a straight line between those two points).
In the case of $p=1/2$ (and $x_i = \pm 1$ such that the mean of $x_i$ is zero) the orientation is horizontal; the variance of $\hat\mu_2$ (which is in the vertical direction) is zero, and the first order Delta method does not work (similar question/situation is here). Instead one needs to use the second order Delta method and the distribution becomes related to a chi squared distribution instead of a normal distribution.
