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Does there exists independent non-constant random variables with $X^2 + Y^2 =1$? I think not because intuitively if there is a relation between them it must mean they are dependent but I can't think of a proof.

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    $\begingroup$ There is another trivial example that you possibly meant to exclude, which is if $X$ and $Y$ are not themselves constant, but $X^2$ and $Y^2$ are. (for example, take $|X| = |Y| = \frac 12 \sqrt 2$ and flip two coins to determine the signs). So maybe you'd like to ask: must $X^2$ and $Y^2$ be almost surely constant? The standard way to then argue is: given that $X^2$ and $Y^2$ are independent, show that this implies $X^2$ is independent of itself. $\endgroup$ Commented May 11 at 19:30

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Example ... $X$ and $Y$ are independent, and both have the scaled Radermacher distribution $$ P(X=1/\sqrt2) = P(X=-1/\sqrt2) = 1/2. $$ Then $X$ and $Y$ are not constant, they are independent, but their squares are constnat $X^2 = 1/2 = Y^2$, so $X^2+Y^2 = 1$.

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  • $\begingroup$ I tried something similar but couldn't check the independence. Don't I need the explicit construction of probability space $\Omega$ to check it? $\endgroup$
    – Invincible
    Commented May 11 at 19:42
  • $\begingroup$ I guess I answered my own question. $\Omega = \{1,2,3,4 \}$, all points have weight $1/4$ and $X(1) = X(2) = 1/ \sqrt{2}$, $Y(1) = Y(3) =1/ \sqrt{2} $ $\endgroup$
    – Invincible
    Commented May 11 at 19:47
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If $X$ and $Y$ are independent and $X^2+Y^2=1$, then $X^2$ and $Y^2$ are almost surely constant. Indeed, let $U=X^2$ and $V=Y^2$. Then $U=1-V$ and the random variables $U$ and $V$ are independent because so are $X$ and $Y$. Therefore, $1-V$ is independent of $V$ hence $V$ is independent of itself.

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Consider two random variables $X$ and $Y$. Assume for the sake of contradiction that $X$ and $Y$ are independent and satisfy the equation $X^2 + Y^2 = 1$ almost surely. This implies that the pair $(X, Y)$ lies on the unit circle in $\mathbb{R}^2$ with probability one.

Recall the definition of independence for random variables $X$ and $Y$: they are independent if and only if for all measurable sets $A$ and $B$ in the Borel $\sigma$-algebra, the equation $$ \mathbb{P}(X \in A, Y \in B) = \mathbb{P}(X \in A) \mathbb{P}(Y \in B) $$ holds. This means that the joint distribution of $X$ and $Y$ can be expressed as the product of their marginal distributions.

The constraint $X^2 + Y^2 = 1$ defines a circle, and therefore, the values of $X$ and $Y$ are not independent since knowing $X$ determines $Y$ up to a sign: $$ Y = \pm \sqrt{1 - X^2}. $$ This relationship implies that if $X$ takes a particular value $x$, $Y$ is constrained to only two possible values, $\pm \sqrt{1 - x^2}$, contradicting the requirement for independence unless $Y$ is deterministic given $X$ (and vice versa), which contradicts the assumption of non-constancy.

In fact, you can consider a parametrization in terms of an angle $\Theta$, where $X = \cos(\Theta)$ and $Y = \sin(\Theta)$. For $X$ and $Y$ to remain independent under this parametrization, $\Theta$ would have to be uniformly distributed over $[0, 2\pi)$, and even then, $\cos(\Theta)$ and $\sin(\Theta)$ are not independent, as evident from their covariance in the standard uniform distribution on the circle.

Therefore, it is impossible for $X$ and $Y$ to be both independent and non-constant while simultaneously satisfying $X^2 + Y^2 = 1$. The relationship by the unit circle creates a dependency between $X$ and $Y$, demonstrating that any such $X$ and $Y$ must indeed be dependent.

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  • $\begingroup$ What if I take the probability space $\{0,1\}$ with $p(1)=1$ and $p(0)=0$. Then take the function $f(0)=0$ and $f(1)=1$ and $g(0)=1$ and $g(1)=0$. Now $g+f=1$ but $g$ and $f$ are independent and not constant. $\endgroup$
    – user1318062
    Commented May 11 at 19:33

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