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 closed-form expressions of $a$ and $b$?

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    $\begingroup$ In case $X,Y$ are jointly normal, then $S,T$ are independent if and only if they are orthogonal. $\endgroup$ Apr 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. $\endgroup$ Apr 16, 2021 at 22:19

1 Answer 1


The classical theorem of Darmois-Skitovich says that two linear forms (with all non-zero 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.


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