Sum of a sum of regular normal distribution

Question

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.

1. 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 $$ BM = \begin{pmatrix} X + Y \\ X - Y \end{pmatrix} \sim\mathcal{N}(B \xi,B\Sigma B^T)=\mathcal{N}(0_{2\times1}, \begin{pmatrix} 2\sigma_1^2+2\rho & 0 \\ 0 & 2\sigma_1^2-2\rho \end{pmatrix} ) $$

2. 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.

3. 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 $(X+Y)^2+\frac{1}{0.44}(X-Y)^2\sim\chi^2(2)$

Answer 1

$\newcommand{\cov}{\operatorname{cov}}$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.