TITLE: Independence of sum and difference for random vectors
QUESTION [7 upvotes]: Suppose I have two independent random vectors $X,Y\in \mathbb R^n$ such that each entry $x_i\sim\mathcal{N}(0,1)$ i.i.d., $y_i\sim\mathcal{N}(0,1)$ i.i.d.  Define
$$ S = X+Y, \qquad D = X-Y .$$
I seem to remember learning that $S$ and $D$ are independent RVs but can't find a reference.  Is this even true, and how could I go about showing this?
Does the same apply for any other distributions for $X$ and $Y$ (in particular with non-independent entries)?

REPLY [3 votes]: $(X,Y)$ is a vector of dimension $2N$ whose entries are 
independent $N(0,1)$ random variables and so its covariance matrix
is the $(2N)\times (2N)$ identity matrix $I_{2N}$.  The $2N$ variables enjoy
a multivariate normal distribution, that is, they are jointly Gaussian
random variables, and $(X,Y)$ is called a Gaussian vector.
Any linear transformation
of a Gaussian random vector results in a Gaussian vector (this
is sometimes taken as the definition of a Gaussian vector), and so 
$$(X+Y,X-Y) = (X,Y)\left[\begin{matrix}1&1\\1&-1\end{matrix}\right]
= (X_1,\ldots,X_n,Y_1,\ldots Y_n)\left[\begin{matrix}I_N&I_N\\I_N&-I_N\end{matrix}\right]$$ 
is also a Gaussian vector with covariance matrix
$$\left[\begin{matrix}I_N&I_N\\I_N&-I_N\end{matrix}\right]^TI_{2N}\left[\begin{matrix}I_N&I_N\\I_N&-I_N\end{matrix}\right]
= \left[\begin{matrix}2I_N&0\\0&2I_N\end{matrix}\right]$$
which shows independence of $X+Y$ and $X-Y$ and also reveals that the
$X_i\pm Y_i$ are $N(0,2)$ random variables.
Now try and see if all the above will work when the
zero-mean Gaussian vectors $X$ and $Y$
are independent of each other but $(X_1,X_2,\ldots, X_n)$ has the
same covariance matrix $\Sigma$ as $(Y_1,Y_2,\ldots, Y_n)$ with $\Sigma$
not necessarily restricted to being the identity matrix or even a
diagonal matrix, that is, the $X_i$ (and similarly the $Y_i$) need
not be independent of each other.