The claim is that $$(n-1)S^2/\sigma^2$$ is chi squared distributed with degrees of freedom $n-1$.

$(n-1)S^2/\sigma^2$ can be written as $$\sum_i^n \left(\frac {x_i-\mu}{\sigma}\right)^2-\left(\frac {\bar x-\mu}{\sigma/\sqrt n}\right)^2$$

I am almost there with understanding why this is $\chi^2_{n-1}$ distributed. I understand that each of these individual elements is $N(0,1)$ distributed, and that a sum of $n$ $N(0,1)$ distributed variables is $\chi^2_{n}$ distributed.

But my problem is, that the distribution of $\bar x$ is not independent of the $x_i$. How do we take this fact into account to derive the desired conclusion?

Note that the existing answers I've found did not specifically address that question.

EDIT: Note that I am not asking for an explanation of why we write $n-1$ rather than $n$. I am asking specifically how we can rigorously derive that it has the distribution that it has.

EDIT 2: Those who have marked this question as a duplicate of this one may be misunderstanding my question. I am not asking for an explanation of why the degrees of freedom are $n-1$ rather than $n$. I am asking for a derivation that it is chi squared in the first place, and that it has $n-1$ degrees of freedom. My problem is clear from the question: How do we take the dependency with $\bar x$ into account? I'm not asking for an intuitive explanation of why it has $n-1$ df rather than $n$.

  • 1
    $\begingroup$ It is also covered in numerous X validated questions, like this one. And follows from the quadratic transform $$x^\text{T}(\mathbf{I}-\frac{1}{n}\mathbf{J})x$$ of the original Normal (standardised) vector $x$. $\endgroup$
    – Xi'an
    Nov 12, 2017 at 16:18
  • $\begingroup$ @Xi'an, I specifically asked this question because the answer to the question you refer to, does not address the specific point that I don't understand namely how to take into account the dependence between $\bar x$ and the $x_i$'s $\endgroup$
    – user56834
    Nov 12, 2017 at 16:20
  • $\begingroup$ It is the sum of squares of independent normal distributions. The dependency is taken into account by the loss of 1 degree of freedom. $\endgroup$ Nov 12, 2017 at 18:48
  • $\begingroup$ @michael, sure, that is plausible. But how is this derived? $\endgroup$
    – user56834
    Nov 13, 2017 at 4:39
  • 1
    $\begingroup$ Wasn't this question just answered at stats.stackexchange.com/questions/312337/…? $\endgroup$
    – whuber
    Nov 13, 2017 at 21:26

1 Answer 1


I am mostly reproducing the argument of this excellent wiki post here.

Let $\xi_i \sim \mathcal{N}(\mu, \sigma^2)$ be $n$ independent identically distributed normally distributed random variables.

Denote the sample mean $\bar{\xi} = \frac{\sum \limits_{i=1}^{n} \xi_i}{n}$.

Also denote sample variance $S^2 = \frac{1}{n-1} \sum \limits_{i=1}^{n} (\xi_i - \bar{\xi})^2$.

Suppose that we knew the exact expectation $\mu$. Then let us construct the sum of squares of our samples:

$\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2} \sim \chi^2_n$ (sum of squares of i.i.d. standardized $\xi$ is chi-squared-distributed with n degrees of freedom)

Let us add and subtract the sample mean to the sum of squares:

$\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2} = \sum \limits_{i=1}^n \frac{(\xi_i - \bar{\xi} + \bar{\xi} - \mu)^2}{\sigma^2} = \sum \limits_{i=1}^n (\frac{(\xi_i-\hat{\xi})^2}{\sigma^2} + \underbrace{2 \frac{(\xi_i - \hat{\xi})(\hat{\xi} - \mu)}{\sigma^2}}_{0 \text{ due to }\sum \limits_{i=1}^n (\xi_i - \hat{\xi}) = 0} + \frac{(\hat{\xi} - \mu)^2}{\sigma^2}) = $ $ = (n-1)\frac{S^2}{\sigma^2} + n\frac{(\hat{\xi} - \mu)^2}{\sigma^2}$

$\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2} = (n-1)\frac{S^2}{\sigma^2} + n\frac{(\hat{\xi} - \mu)^2}{\sigma^2}$, where $\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2} \sim \chi^2_n$, $n\frac{(\hat{\xi} - \mu)^2}{\sigma^2} \sim \chi^2_1$.

By Cochran's theorem first term of the sum (random variable that is a funcion of sample variance $S^2$) is independent of the second term (function of sample mean $\hat{\xi}$), thus, probability density function of $\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2}$ is a convolution of probability density functions of $(n-1)\frac{S^2}{\sigma^2}$ and $n\frac{(\hat{\xi} - \mu)^2}{\sigma^2}$.

Now, we can directly use the convolution formula or apply one of spectral analysis tools to it to derive the distribution of $S^2$, moment-generating functions/cumulants or characteristic functions/Fourier transform.

Fourier transform of a convolution of functions is a multiple of their Fourier transforms. Thus, $\phi_{\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2}}(t) = \phi_{(n-1)\frac{S^2}{\sigma^2}}(t) \cdot \phi_{n\frac{(\hat{\xi} - \mu)^2}{\sigma^2}}(t)$.

Characteristic function of a chi-squared distribution is $\phi_{\chi^2_n}(t) = (1-2it)^{-\frac{n}{2}}$.

Thus, characteristic function $\phi_{(n-1)\frac{S^2}{\sigma^2}}(t) = (1-2it)^{\frac{-n}{2}} \cdot (1-2it)^{\frac{1}{2}} = (1-2it)^{-\frac{n-1}{2}}$. But this is the characteristic function of $\chi^2_{n-1}$ (characteristic functions are mostly reversible, so that correspondence of characteristic functions implies correspondence of distributions).

Hence, $(n-1)\frac{S^2}{\sigma^2} \sim \chi^2_{n-1}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.