Chi Squared Clarification Suppose you have $Z_1, Z_2, Z_3$ which are all independent standard Gaussian variables.
Suppose you have
$$A=\frac{(Z_1-2Z_2+Z_3)^2}{12}+\frac{(Z_1-Z_3)^2}{4}+\frac{(Z_1-Z_2)^2}{4}+\frac{(Z_1+Z_2-2Z_3)^2}{12}.$$
$A$ is $\chi_2^2$, but I don't see how this is the case.
Particularly since $\operatorname{Var}\left(\frac{Z_1-2Z_2+Z_3}{\sqrt{12}}\right)$${}=\frac{1}{2}$, which is the same issue with the other terms, i.e since none of the terms, by themselves, are standard normal before you square them.
However, I do see how:
$$B=\frac{(Z_1-2Z_2+Z_3)^2}{6}+\frac{(Z_1-Z_3)^2}{2}.$$ follows the $\chi_2^2$ distribution. I can also see that the first two terms of $A$ multiplied by $2$ is equal to $B.$
Any feedback would be appreciated.
 A: $$
\left[ \begin{array}{c} \frac{Z_1-2Z_2+Z_3}{\sqrt{12}} \\[4pt] \frac{Z_1-Z_3} 2 \\[6pt] \frac{Z_1-Z_2} 2 \\[6pt]
\frac{Z_1+Z_2-2Z_3}{\sqrt{12}}
\end{array} \right]
= \left[ \begin{array}{rrr}
1/\sqrt{12} & -2/\sqrt{12} & 1/\sqrt{12} \\
1/2 & 0 & -1/2 \\
1/2 & -1/2 & 0 \\
1/\sqrt{12} & 1/\sqrt{12} & -2/\sqrt{12}
\end{array} \right]
\left[
\begin{array}{c}
Z_1 \\ Z_2 \\ Z_3
\end{array}
\right]
\tag1
$$
Observe that the rank of this $4\times3$ matrix is only $2,$ i.e. you have only two linearly independent columns, since the second column is $-1$ times the sum of the first and third columns.
Let us find two unit vectors orthogonal to each other that span the column space of that matrix:
$$
\left[ \begin{array}{c} -\sqrt3/\sqrt8 \\ +1/\sqrt8 \\ -1/\sqrt8 \\ +\sqrt3/\sqrt8 \end{array} \right], \qquad \left[ \begin{array}{c} +1/\sqrt8 \\ +\sqrt3/\sqrt8 \\ +\sqrt3/\sqrt8 \\ +1/\sqrt8 \end{array} \right]
$$
Write the column vector on the left side of line $(1)$ above as a linear combination of these two vectors, each of the coefficients $U_1,U_2$ being a linear combination of $Z_1,Z_2,Z_3.$
Then show that $\operatorname{var}(U_1)= \operatorname{var}(U_2)=1$ and $\operatorname{cov}(U_1,U_2)=0$ and that $U_1,U_2\sim\operatorname N(0,1),$ and that $U_1^2 + U_2^2$ is equal to the sum of squares that you say should have a $\chi^2_2$ distribution.
A: If you expand the quadratic form, you end up with
\begin{align}
A&=\frac23(Z_1^2+Z_2^2+Z_3^2-Z_1Z_2-Z_2Z_3-Z_1Z_3)
\\&=Z_1^2+Z_2^2+Z_3^2-\frac23(Z_1Z_2+Z_2Z_3+Z_1Z_3)-\frac13(Z_1^2+Z_2^2+Z_3^2)
\\&=Z_1^2+Z_2^2+Z_3^2-\frac13(Z_1+Z_2+Z_3)^2
\end{align}
In other words, if $\overline Z=\frac13\sum\limits_{i=1}^3 Z_i$, then  $$A=\sum_{i=1}^3Z_i^2-3\overline Z^2 =\sum_{i=1}^3 (Z_i-\overline Z)^2$$
So $A$ is the corrected sum of squares for a sample $Z_1,Z_2,Z_3$ drawn from a standard normal population. This is known to have a $\chi^2_{3-1}$ distribution.
One can also note that $A=Z^TPZ$ where $Z=(Z_1,Z_2,Z_3)^T$ and $P=I_3-\frac13\mathbf1\mathbf1^T$, with $\mathbf1$ being a $3\times 1$ vector of all ones. As $P$ is symmetric and idempotent, $A$ has a $\chi^2$ distribution with degrees of freedom $\operatorname{rank}(P)=\operatorname{tr}(P)=3-1$.
