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Apologies if I am just having a mental block and missing something very obvious. Here is a conjecture that I think is obviously true, and yet I cannot prove it:

Let $X_1, X_2, \ldots, X_n, Y, Z$ be mutually independent, real-valued, non-constant random variables. (They need not be identically distributed.)

Let $A = \sum_{j=1}^n a_j X_j + Y, B = \sum_{j=1}^n b_j X_j + Z$ where all the coefficients $a_j, b_j$ are non-zero.

Prove or provide counter-example:** For $X_j, Y, Z, A, B$ defined as above, $A, B$ cannot be independent.

Further thoughts: If some coefficients are zero, the subset of $X_j$'s that actually "affects" $A$ can be distinct from the subset that actually "affects" $B$, and then $A,B$ can of course be independent. But my statement explicitly rules this out.

Also, if instead of summation, we have general functions e.g. $A' = f(\vec{X}) + Y, B' = g(\vec{X}) + Z$, then even if each function $f, g$ must be affected by all components, we can still define them s.t. they are independent and therefore $A', B'$ are independent. However, I am not allowing arbitrary functions, but instead summations (linear combinations). To be clear, the summation is over reals.

(I would be curious to see a counter example in finite field, but that's not my main question, and even so, you cannot have the $+$ in $A$ be in a different field than the $+$ in $B$, so to speak.)

YuiTo Cheng
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antkam
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2 Answers2

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The joint distribution of two i.i.d. normal variables $X, Y$ is radially symmetric, so actually $$X \cos \theta + Y \sin \theta, \; - X \sin \theta + Y \cos \theta$$ are independent i.i.d. normal variables, by applying a rotation matrix. It should be clear how to construct a counterexample to your conjecture from here.

hdighfan
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  • Thanks for the answer!! I used to know this, but had forgotten. :) If the key is rotational symmetry, it would seem normal variables are the only ones with this property, at least for the $n=2$ case. Do you agree? And what about larger $n$ - would that allow another counter-example based NOT on normal variables? – antkam Jun 07 '20 at 04:03
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    @antkam The $a_i, b_j$ condition is simply saying that we want to find a joint distribution which has the same projection onto two lines, and which factorises with respect to two bases that contain the given lines as one of the coordinates. It is actually possible to prove that the only such distributions that work are multivariate normal; see here for more information: https://projecteuclid.org/euclid.aoms/1177728796 – hdighfan Jun 07 '20 at 04:40
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It is well known that if $X_1,X_2$ are i.i.d $N(0,1)$ then $X_1-X_2$ and $X_1+X_2$ are independent. This can be proved easily using characteristic functions. Let $Y$ and and $Z$ be any non-constant random variables such that the foursome $\{X_1,X_2,Y,Z\}$ is independent. Then the corresponding variables $A=X_1-X_2+Y$ and $B=X_1+X_2+Z$ are independent.

Kavi Rama Murthy
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  • Thanks for the answer!! I used to know this, but had forgotten. :) If the key is rotational symmetry, it would seem normal variables are the only ones with this property, at least for the $n=2$ case. Do you agree? And what about larger $n$ - would that allow another counter-example based NOT on normal variables? – antkam Jun 07 '20 at 04:04