# Prove that $\frac{\sum_{i=1}^n (X_i - \bar{X})^2}{\sigma^2}\,\sim \,\chi_{(n-1)}^2$

If $X_1, \ldots, X_n$ are iid $\sim N(\mu,\sigma^2)$,

$\frac 1 {\sigma^2} \Big((X_1-\bar X)^2 + \cdots + (X_n - \bar X)^2 \Big) \sim \chi^2_{n-1}$.

This has been shown with linear algebra projections, and is the starting point to derive the density of the $t$-Student distribution.

I would really like to get a proof without the advanced math in linear algebra projections (if possible), or with a detailed explanation of the steps followed in the derivation (if the use of projections is essential). Thanks!

Observe the following:

Since $X_i \sim N(\mu, \sigma^2) \Rightarrow X_i - \mu \sim N(0, \sigma^2)$ and that $\frac{X_i - \mu}{\sigma} \sim N(0,1)$.

Now, by definition, a $\chi^2_1$ random variable is defined as $Z^2$ where $Z \sim N(0,1)$.

So that means $(\frac{X_i - \mu}{\sigma})^2 \sim N(0,1)^2 = \chi^2_1$.

Finally, observe that $\sum_{i=1}^n \chi^2_1 = \chi^2_n$. This can be seen from the proof here:

https://onlinecourses.science.psu.edu/stat414/node/171

So now we have shown that $\sum_{i=1}^n (\frac{X_i - \mu}{\sigma})^2 \sim \chi^2_n$.

We're almost there. In your case you replace the $\mu$ with a $\bar{X}$. That's standard practice in statistics when you don't know the true value of the parameter and in that case, we reduce the degrees of freedom by one. For that same reason, we have that $\sum_{i=1}^n (\frac{X_i - \bar{X}}{\sigma})^2 \sim \chi^2_{n-1}$.

If you want a rigorous proof of why the minus one, then you will need to get into the linear algebra. A quick discussion can be found here: https://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29

Hope that helps!

• @user357015 Excellent, by the way! – Antoni Parellada Aug 25 '15 at 22:04
• @AntoniParellada : I entirely disagree with the evaluation of this answer as "excellent". It stops short of actually answering the question. Note that $(1)$ each of the terms $\dfrac{(X_i - \overline X)^2}{\sigma^2}$ fails to be distributed as $N(0,1),$ and $(2)$ there are not $n-1$ terms, but $n,$ and $(3)$ they are not independent. They are negatively correlated because of the way in which they depend on $\overline X,$ which depends on all of them. $\qquad$ – Michael Hardy Nov 14 '17 at 20:41
• @MichaelHardy I see your points... Old question... Would you consider writing up an answer? – Antoni Parellada Nov 14 '17 at 21:42
• @AntoniParellada : Do you know matrix algebra and such things as orthogonal projections? – Michael Hardy Nov 14 '17 at 21:43
• @MichaelHardy Yes, I am self-taught, and when I posted the question I had just started off learning math and stats. And I just noticed that the question linked to a proof you already wrote on this site... – Antoni Parellada Nov 14 '17 at 21:46