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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.

Michael Hardy
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jeff123
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  • What is your reasoning for B? – Thomas Jan 06 '21 at 23:10
  • Because: Let $A$=$\frac{Z_1-2Z_2+Z_3}{\sqrt6}$ and $C$=$\frac{Z_1-Z_3}{\sqrt2}$. Then, E(A)=E(C)=0, Var(A)=Var(C)=1 and Cov(A,C)=0, showing that A and C are both independent standard normal variables. Therefore, $A^2 + C^2$ follows the $\chi_2^2$ distribution. Open to any constructive criticism – jeff123 Jan 06 '21 at 23:34
  • I don't think that your proof of $A$ not being $\chi_2^2$ holds water. Recall that $\chi_2^2$ means of the form $X^2+Y^2$, where $X$ and $Y$ are iid $N(0,1)$. Your current form of $A$ has it as the sum of four squares, not two, so analysing the variance of any of the four terms doesn't really tell you much – jlammy Jan 07 '21 at 02:38
  • However, I don't think $A$ is $\chi_2^2$. If you expand out, you can write $$A=\frac{1}{3}\left[(Z_1-Z_2)^2+(Z_2-Z_3)^2+(Z_3-Z_1)^2\right].$$ This means that (a) $A$ is completely symmetric in $Z_1$, $Z_2$, $Z_3$ and (b) $A$ is $0$ iff $Z_1=Z_2=Z_3$. I think that should be enough to imply that $A$ can't be the sum of two squares – jlammy Jan 07 '21 at 02:42
  • Thank you for your comment! However, I am not trying to prove that A is not $\chi^2$, I am rather trying to figure out how it is $\chi^2$ instead, and was wondering where to start. – jeff123 Jan 07 '21 at 02:47
  • Hi, how are you so sure it is $\chi_2^2$? Did you find it in a book? Or from lectures? – jlammy Jan 07 '21 at 02:48
  • Through lectures- our professor said that A, from my question above, is definitely $\chi^2$. However, I am not sure whether it is or isn't a $\chi^2$ with 2 degrees of freedom. I only think that it is 2 degrees of freedom because I calculated E(A)=2 and Var(A)=4, and since I do know it is $\chi^2$ as per my professor, I am deducing that it is $\chi_2^2$. – jeff123 Jan 07 '21 at 03:05

2 Answers2

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$$ \left[ \begin{array}{c} \frac{Z_1-2Z_2+Z_3}{\sqrt{12}} \\[4pt] \frac{Z_1-Z_3} 2 \\[6pt] \frac{Z_1-Z_2} 2 \\[6pt] \frac{Z_1+Z_2-2Z_3}{\sqrt{12}} \end{array} \right] = \left[ \begin{array}{rrr} 1/\sqrt{12} & -2/\sqrt{12} & 1/\sqrt{12} \\ 1/2 & 0 & -1/2 \\ 1/2 & -1/2 & 0 \\ 1/\sqrt{12} & 1/\sqrt{12} & -2/\sqrt{12} \end{array} \right] \left[ \begin{array}{c} Z_1 \\ Z_2 \\ Z_3 \end{array} \right] \tag1 $$ 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.

Michael Hardy
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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 $\operatorname{rank}(P)=\operatorname{tr}(P)=3-1$.

StubbornAtom
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