Editing Chi-squared distribution (section)

=== Cochran's theorem ===
{{Main|Cochran's theorem}}
The following is a special case of Cochran's theorem.

'''Theorem.''' If <math>Z_1,...,Z_n</math> are [[independence (probability theory)|independent]] identically distributed (i.i.d.), [[standard normal]] random variables, then
<math>\sum_{t=1}^n(Z_t - \bar Z)^2 \sim \chi^2_{n-1}</math>
where <math>\bar Z = \frac{1}{n} \sum_{t=1}^n Z_t.</math>

{{hidden begin|style=width:100%|ta1=center|border=1px #aaa solid|title=[Proof]}}
'''Proof.''' Let <math>Z\sim\mathcal{N}(\bar 0,1\!\!1)</math> be a vector of <math>n</math> independent normally distributed random variables,
and <math>\bar Z</math> their average.
Then
<math>
  \sum_{t=1}^n(Z_t-\bar Z)^2 ~=~ \sum_{t=1}^n Z_t^2 -n\bar Z^2 ~=~ Z^\top[1\!\!1 -{\textstyle\frac1n}\bar 1\bar 1^\top]Z ~=:~ Z^\top\!M Z
</math>
where <math>1\!\!1</math> is the identity matrix and <math>\bar 1</math> the all ones vector.
<math>M</math> has one eigenvector <math>b_1:={\textstyle\frac{1}{\sqrt{n}}} \bar 1</math> with eigenvalue <math>0</math>,
and <math>n-1</math> eigenvectors <math>b_2,...,b_n</math> (all orthogonal to <math>b_1</math>) with eigenvalue <math>1</math>,
which can be chosen so that <math>Q:=(b_1,...,b_n)</math> is an orthogonal matrix.
Since also <math>X:=Q^\top\!Z\sim\mathcal{N}(\bar 0,Q^\top\!1\!\!1 Q) =\mathcal{N}(\bar 0,1\!\!1)</math>,
we have
<math>
  \sum_{t=1}^n(Z_t-\bar Z)^2 ~=~ Z^\top\!M Z ~=~ X^\top\!Q^\top\!M Q X ~=~ X_2^2+...+X_n^2 ~\sim~ \chi^2_{n-1},
</math>
which proves the claim.
{{hidden end}}