TITLE: $X,Y$ independent then $X+Y$, $X-Y$ independent as well?
QUESTION [7 upvotes]: My question is simple: If $X$, $Y$ are independent random variables then $X+Y$, $X-Y$ independent as well?

REPLY [3 votes]: $\newcommand{\cov}{\operatorname{cov}}\newcommand{\var}{\operatorname{var}}$
$$
\cov(X+Y,X-Y) = \var X -\var Y \\[8pt] \begin{cases} =0 & \text{if }\var X=\var Y, \\  \ne 0 & \text{if }\var X\ne\var Y. \end{cases}
$$
In the cases where this $\ne0$, the random variables $X+Y$ and $X-Y$ cannot be independent.
In some of the cases where this $=0$, they are independent, but as "David"'s answer has shown, not in all.