Suppose $X=(X_1,X_2,X_3)^T$ is a random vector with a Gaussian distribution having mean vector $(0,0,0)$ and variance-covariance matric
$$
\begin{pmatrix}
1 & 0 & 0 \\
0 & 2 & 1 \\
0 & 1 & 2 \\
\end{pmatrix}.
$$
By definition, the $\chi^2 $ is the sum of squares of independent standard normal distributions. In my question, I was given different options on which one has the $\chi^2 $ distributions:
a) $(X_2 + X_3)^2/6 +(X_2-X_3)^2/2$
b) $(X_1+X_2)^2/5$
c) $X_1^2 + X_2^2 + X_3^2$
d)$X_1^2 + X_2^2$
e)$X_3^2/2 + (2X_2 -X_3)^2/6$
f) $(X_2^2+X_3^2)/2.$
The correct answers are (a) and (e). This doesn't make sense to me since X2 and X3 are not independent or standard Gaussian?
Similarly, X1 and X2 have a covariance of 0, so I don't understand how B, the sum of the two squared over 5, or even D, the sum of the squares of $X_1$ and $X_2$, aren't $\chi^2 $.
Any thoughts would be appreciated.
Best Answer
(a) is $\chi^2$. Note that although $X_2$ and $X_3$ aren't standard normal, the rvs $$A=\frac{X_2+X_3}{\sqrt{6}}\quad\text{and}\quad B=\frac{X_2-X_3}{\sqrt{2}}$$ are normal (as the vector $(X_1,X_2,X_3)$ is Gaussian). You can compute the means and variances of $A$, $B$ to show that they are in fact standard normal. They are independent, as you can compute the covariance $$\operatorname{Cov}(A,B)=\frac{1}{\sqrt{12}}\left[\operatorname{Cov}(X_2,X_2)-\operatorname{Cov}(X_2,X_3)+\operatorname{Cov}(X_3,X_2)-\operatorname{Cov}(X_3,X_3)\right]=0.$$ So $A^2+B^2$ is the sum of the squares of two independent standard normals, hence $\chi^2$.
If you expand (e), you will get the same expression as (a), proving that (e) is also $\chi^2$.
(b) is not $\chi^2$. If you compute the mean, $$\mathbb E\left[(X_1+X_2)^2/5\right]=(1+0+2)/5=3/5.$$ So it doesn't even have the same mean as any $\chi^2$ rv.
(d) isn't $\chi^2$ either, as $X_2$ is not standard normal. However, $X_1^2+\frac{X_2^2}{2}$ is $\chi^2$.
For completeness: for similar reasons as (d), the rv in (c) isn't $\chi^2$. Finally (f) isn't $\chi^2$ as $X_2$, $X_3$ aren't independent.