# Sample variance of a normal distribution

If my sample variance of a random sample from a $$\mathop{\mathcal N}\left(\mu,\sigma^2\right)$$ distribution is

$$S_2^2=\frac{1}{n}\sum_{i=1}^n{\left(X_i-\bar{X}\right)}^2,$$

then

$$Q \mathrel {:=}\frac{\sum_{i=1}^n{\left(X_i-\bar{X}\right)^2}}{\sigma^2}\frac{n}{n} = \dfrac{nS_2^2}{\sigma^2}.$$

Does this mean that $$Q$$ has a $$\chi^2$$-distribution with $$n$$ degrees of freedom?

• No. It is a classical result that $nS_1^2/\sigma^2$ has $\chi^2$ distribution with $\color{red}{n - 1}$ degrees of freedom. Commented Feb 20, 2023 at 14:52
• @Zhanxiong but if I change it to $nS_2^2/\sigma^2$ does that make it have a $\chi^2$ -distribution with n degrees of freedom?
– user379637
Commented Feb 21, 2023 at 0:21
• what is your definition of $S_2^2$? Commented Feb 21, 2023 at 0:32
• $S_2^2=\dfrac{1}{n}\sum{(X_i-\bar{X})}^2$ which makes $Q = \dfrac{nS_2^2}{\sigma^2}$. Since I have read in a book that $S_1^2$ results to $\chi^2$-distribution with $n-1$ degrees of freedom. Or it does not make a difference?
– user379637
Commented Feb 21, 2023 at 0:49
• Aren't your $S_2^2$ in the comment and $S_1^2$ in the post exactly the same? Commented Feb 21, 2023 at 0:50

With $$\mathbf X \mathrel{:=} \left(X_1, \ldots, X_n\right)^\top$$ we have $$\mathbf X \sim \mathop{\mathcal N}\left(\mu \mathbf 1_n, \sigma^2 I_n\right)$$ and $$\sum_{i=1}^n \left(X_i - \bar X\right)^2 = \left(C_n \mathbf X\right)^\top \left(C_n \mathbf X\right) = \mathbf X^\top C_n \mathbf X,$$ where $$C_n$$ denotes the centering matrix of order $$n$$.
Idempotency and symmetry of $$C_n$$ allow for the decomposition $$C_n = AA^\top$$ with $$A^\top A = I_{\mathop{\mathrm{rk}}\left(C_n\right)}$$ and $$A \in \mathbb R^{n \times \mathop{\mathrm{rk}}\left(C_n\right)}$$.
With $$\mathop{\mathrm{rk}}\left(C_n\right) = \mathop{\mathrm{tr}}\left(C_n\right) = n-1$$ this gives $$\frac{1}{\sigma}A^\top \mathbf X \sim \mathop{\mathcal N}\left(\frac{\mu}{\sigma} A^\top \mathbf 1_n, I_{n-1}\right)$$ and hence $$Q = \sum_{i=1}^n \frac{\left(X_i - \bar X\right)^2}{\sigma^2} = \frac{1}{{\sigma^2}} \mathbf X^\top C_n \mathbf X = \left\|\frac{1}{\sigma}A^\top \mathbf X\right\|_2^2 \sim \mathop{\chi_{n-1}^2}\left(\lambda\right),$$ where $$\lambda = \left\|\frac{\mu}{\sigma} A^\top \mathbf 1_n\right\|_2^2$$ denotes the non-centrality parameter of the non-central chi-squared distribution with $$n-1$$ degrees of freedom.
Finally, $$\left\|\frac{\mu}{\sigma} A^\top \mathbf 1_n\right\|_2^2 = \frac{\mu^2}{\sigma^2}\mathbf 1_n^\top C_n \mathbf 1_n = 0$$ yields $$Q \sim \mathop{\chi_{n-1}^2}$$, i.e., $$Q$$ follows a (central) chi-squared distribution with $$n-1$$ degrees of freedom.