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?

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

1 Answer 1


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.


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