Sum of a sum of regular normal distribution Can someone please confirm if the solution below is correct? If not, please give me some hints about what is wrong.
Let $X$ and $Y$ be two real valued stochastic variables who's joint distribution is the regular normal distribution on $(\mathbb{R}^2,\mathbb{B}_2)$ with mean $0$ and variance matrix
$$
\Sigma =
 \begin{pmatrix}
 \sigma_1^2 & \rho \\
 \rho & \sigma_2^2
 \end{pmatrix}
$$
where we assume that $\sigma_1^2,\sigma_2^2>0$ and $-\sigma_1\sigma_2<\rho<\sigma_1\sigma_2$. It can be shown that $\Sigma$ is positive definite, and can then be a variance matrix for a regular normal distribution.

  
*
  
*Find the joint distribution of $X+Y$ and $X-Y$
  

I use that if $M=(X,Y)^T$ follows a regular normal distribution, then 
$$
\alpha+BX\sim\mathcal{N}(\alpha+B\xi,B\Sigma B^T)
$$
Let 
$$
 B=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad M= \begin{pmatrix} X \\ Y\end{pmatrix} ,\qquad \xi = \begin{pmatrix}
 0 \\ 0\end{pmatrix}
$$
 Then 
 $$
BM =
 \begin{pmatrix}
  X + Y \\
  X - Y
  \end{pmatrix}
  \sim\mathcal{N}(B \xi,B\Sigma B^T)=\mathcal{N}(0_{2\times1},
  \begin{pmatrix}
   2\sigma_1^2+2\rho & 0 \\
   0 & 2\sigma_1^2-2\rho
   \end{pmatrix} )
 $$


  
*Determine all values of $\sigma_1^2,\sigma_2^2$ and $\rho$ for which $X+Y\perp \! \! \! \perp X-Y$
  

As the $Cov(X+Y,X-Y)=0$ we see that $X+Y$ and $X-Y$ are always independent. 


  
*Assume that $\sigma_1^2=\sigma_2^2=0.6$ and that $\rho=0.14$. Find the distribution of  $(X+Y)^2+\frac{1}{0.44}(X-Y)^2$
  

We know that if $X\sim \mathcal{N}(0,1)$, then
$$
 \chi^2=\sum_{i=1}^n X^2_i
$$
is $\chi^2$ distributed with $n$-degrees of freedom. So let's standarize $X+Y$ and $X-Y$.
\begin{align}
     \frac{X+Y}{\sigma_1^2+2\rho}&=\frac{X+Y}{2\times0.6^2+2\times 0.14}=X+Y \\
     \frac{1}{0.44}\frac{X-Y}{\sigma_1^2-2\rho}&=\frac{X-Y}{2\times0.6^2-2\times 0.14}=\frac{X-Y}{0.44}0.44=X-Y
 \end{align}
So we conclude that $(X+Y)^2+\frac{1}{0.44}(X-Y)^2\sim\chi^2(2)$
 A: $\newcommand{\cov}{\operatorname{cov}}$Your conclusion in $(2)$ is incorrect.
$X$ and $Y$ are independent only if $\rho=0$.  So your statement that begins by saying "If $X$ and $Y$ are independent" is not applicable in the context where you apply it.
\begin{align}
\cov(X+Y,X-Y) & =\cov(X,X-Y)+\cov(Y,X-Y) \\[6pt]
& = \underbrace{\cov(X,X)-\cov(X,Y)} + \underbrace{\cov(X,Y)-\cov(Y,Y)} \\[6pt]
& = \sigma_1^2 -\sigma_2^2.
\end{align}
Only if this covariance is $0$ are $X+Y$ and $X-Y$ independent.
In your third point, you're giving a function of two real variables as the density of a single real-valued random variable.  That isn't right.  Also, you're writing down something that looks like a normal density for a random variable that you say has a chi-square distribution.  A chi-square-distributed random variable is always positive; a normally distributed random variable is not, so that is also incorrect.
Note that $\dfrac{X^2}{\sigma_1^2}$ has a chi-square distribution with one degree of freedom.
You've said $X+Y\sim N(0,\sigma_1^2+\sigma_2^2 + 2\rho)$.  It follows that $\dfrac{(X+Y)^2}{\sigma_1^2+\sigma_2^2+2\rho}$ has a chi-square distribution with one degree of freedom.  Similarly
$$
\frac{\left( \frac{X-Y}{0.44} \right)^2}{\left(\frac{\sigma_1^2+\sigma_2^2-2\rho}{0.44}\right)^2} \sim \chi^2_1.
$$
In $\#3$ Since $\sigma_1=\sigma_2$, and since we concluded earlier that that is just what is needed for $X+Y$ and $X-Y$ to be independent, we conclude that in this case they are independent.  Since $\dfrac{(X+Y)^2}{\operatorname{var}(X+Y)}\sim\chi^2_1$ and $\dfrac{(X-Y)^2}{\operatorname{var}(X-Y)}\sim\chi^2_1$ are independent, their sum is distributed as $\chi_2^2$.  But now we have a problem: which constants do we have to multiply these squares of normally distributed random variables by to conclude that?  They're multiplied here by $1$ and $1/0.44$, and those don't appear to fit.
