Independent random variables with $X^2 + Y^2 =1$

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.

• 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. Commented May 11 at 19:30

3 Answers

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 $$X^2+Y^2 = 1$$.

• I tried something similar but couldn't check the independence. Don't I need the explicit construction of probability space $\Omega$ to check it? Commented May 11 at 19:42
• I guess I answered my own question. $\Omega = \{1,2,3,4 \}$, all points have weight $1/4$ and $X(1) = X(2) = 1/ \sqrt{2}$, $Y(1) = Y(3) =1/ \sqrt{2}$ Commented May 11 at 19:47

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.

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.

• 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.
– user1318062
Commented May 11 at 19:33