# Distribution of sample standard deviation

Noting this question, it is known that the sum of independent identically distributed gaussian random variables is a Noncentral chi-squared distribution.

It is also known that the sample distribution is a normal distribution: $$\bar{X} = \frac{X_1+X_2+\cdots+X_k}{k}$$

That is great and all, but what about:

$$Z = \sqrt{\frac{(X_1-\bar{X})^2+(X_2-\bar{X})^2+\dots+(X_k-\bar{X})^2}{k}}$$

Is this also a known distribution with well known properties?

• Obscure. Who are $\overline{X}_i?$ Commented May 23 at 6:44
• @LetacGérard fair point, clarified. Commented May 23 at 6:51
• Not really..... Commented May 23 at 7:16

A "noncentral chi-squared distribution" comes from the sum of squares of $$k$$ iid $$N(\mu,1)$$ distributed random variables. If $$\mu=0$$ then you get an ordinary chi-squared distribution, with $$k$$ degrees of freedom. The square root of a noncentral chi-squared distributed random variable has a "noncentral chi distribution".
But your $$Z$$ does not have a non-central distribution even if $$\mu\not=0$$, because you still have $$\mathbb E[X_i-\bar X]=0$$. Furthermore $$\sum\limits_i (X_i-\bar X) =0$$, reducing the degrees of freedom.
If the iid $$X_i\sim N(\mu,\sigma^2)$$, then $$\frac{k}{\sigma^2}Z^2 = \frac{(X_1-\bar{X})^2+(X_2-\bar{X})^2+\dots+(X_k-\bar{X})^2}{\sigma^2}$$ has an ordinary chi-squared distribution, with $$k-1$$ degrees of freedom. Since it has an expectation of $$k-1$$, it is often divided by $$k-1$$ rather than $$k$$ to give the sample variance using Bessel's correction.
So $$\frac{\sqrt{k}}{\sigma}Z = \sqrt{\frac{(X_1-\bar{X})^2+(X_2-\bar{X})^2+\dots+(X_k-\bar{X})^2}{\sigma^2}}$$ has an ordinary chi distribution, with $$k-1$$ degrees of freedom.
You might then say $$Z$$ has a scaled chi distribution.