# How do we derive that $S^2$ is chi-squared distributed (with $n-1$ df)?

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 $n$.

• 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 $x$. Nov 12 '17 at 16:18
• @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 Nov 12 '17 at 16:20
• It is the sum of squares of independent normal distributions. The dependency is taken into account by the loss of 1 degree of freedom. Nov 12 '17 at 18:48
• @michael, sure, that is plausible. But how is this derived? Nov 13 '17 at 4:39
• Wasn't this question just answered at stats.stackexchange.com/questions/312337/…?
– whuber
Nov 13 '17 at 21:26

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