Let $X,Y$ be independent multivariant random variables on $\mathbb{R}^d$. Let $\alpha,\beta$ be two positive real values such that $\alpha^2+\beta^2=1$. Then, $S=\alpha X+\beta Y$ is a new random variable. The question is: do there exist two real values $a,b$, such that $T=aX+bY$ is independent with $S$? If so, are there closedform expressions of $a$ and $b$?

1$\begingroup$ In case $X,Y$ are jointly normal, then $S,T$ are independent if and only if they are orthogonal. $\endgroup$– Gerald EdgarApr 16, 2021 at 20:22

$\begingroup$ If $S,T$ are independent then each component of $S$ and $T$ are independent, hence uncorrelated. For the case $d = 1$ you get for the covariance $C(S,T) = \alpha a V(X) + \beta b V(Y)$ and this can be easily solved for $a$ and $b$. If the variances of the components of $X$ and $Y$ are not identical, these solutions for $a$ and $b$ will differ. Thus for the case $d > 1$ there seem to exist no solutions except in trivial cases. $\endgroup$– Dieter KadelkaApr 16, 2021 at 22:19
1 Answer
The classical theorem of DarmoisSkitovich says that two linear forms (with all nonzero coefficients, and of at least two variables) of independent random variables are independent only if the random variables are normal. Combined with the comment of @Gerald Edgar, this completely settles the question.
Ref. A. Kagan, Yu. Linnik, C. Rao, Characterization problems of mathematical statistics, Wiley, NY, 1973, Chapter 3.