Bivariate normal distribution from independent random variables Let $X_1$ and $X_2$ random variables such that $X_1+X_2$ and $X_1-X_2$ have independent standard normal distributions. Show that $x=(X_1, X_2)$ has a bivariate normal distribution.
My work: 
Since $X_1+X_2$ and $X_1-X_2$ have independent standard normal distributions, so 
$X_1+ X_2 \sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2 )$
$X_1- X_2 \sim N(\mu_1-\mu_2,\sigma_1^2+\sigma_2^2 )$
Also $Cov(X_1,X_2)=0$.
I don't know how to relate these to prove my result. I was wondering if you could help me! Thank you so much for your time.
 A: Let $Z = X_1 + X_2$ and let $W = X_1 - X_2$.  From the question, both $Z$ and $W$ are standard normals. It is straight forward to show that
$$X_1 = \dfrac{Z+W}{2}$$
$$X_2 = \dfrac{Z-W}{2}$$
The mean of $X_1$ and $X_2$ is 0 (why?) 
Because $Z$ and $W$ are standard normal and are assumed independent, then
$$\operatorname{Var}(X_1) = 0.25(\operatorname{Var}(Z) + \operatorname{Var}(W) + \operatorname{Cov}(Z,W)) = 0.25(1 + 1 + 0)= 0.5$$ 
A similar argument can be made for $X_2$.
We know $X_1$ and $X_2$ must be normal through the properties you've listed (that is, because $Z$ and $W$ are normal and $X_1$ and $X_2$ can be obtained from $Z$ and $W$ through addition/subtraction).
So we know three things:


*

*$X_1$ and $X_2$ are normal

*The mean of $X_1$ and $X_2$ is 0

*The variance $X_1$ and $X_2$ is 0.5


Can you finish from here?

Alternatively, consider $Z$ and $W$ as defined above and that they are standard normal.  By definition of multivariate standard normals, the joint distribution of $Z$ and $W$ is multivariate standard normal. One can show that $x = Ay$ where $x$ is defined as above, $y = (Z,W)$ and 
$$ A = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} $$
By definition, $x$ is multivariable normal with mean $(0,0)$, and with covariance matrix $\Sigma = AA^T$.  If I remember correctly, $A$ is called a Cholesky Factor of $\Sigma$.
