Random vector $(X, Y)$ has a uniform distribution on the unit circle.

Question

Faced with the following problem, I do not understand how to solve this problem:

Random vector $(X, Y)$ has a uniform distribution on the unit circle. Will its components be independent?

It is not very clear to me how to approach such tasks, is it necessary to look for a vector distribution function here? But it's probably clear that we need to check the definition somehow. $X,Y$ independent $\Leftrightarrow$ $\mathbb P(X < x, Y < y) = \mathbb P(X < x) \cdot \mathbb P(Y<y)$ But how to do this is not very clear.

Answer 1

Regardless of whether we're talking about just the boundary or including the interior of the circle, it is very easy to see that the components are not independent by considering a region between the unit square and the unit circle, e.g. $A = [0.8, 1] \times [0.8, 1]$. Clearly $P(0.8 \leq X \leq 1) > 0$ and $P(0.8 \leq Y \leq 1) > 0$, but $P((X, Y) \in A) = 0$.

Answer 2

Let $$ A=\left[-\tfrac1{\sqrt{2}},\tfrac1{\sqrt{2}}\right]\times\left[-\tfrac1{\sqrt{2}},\tfrac1{\sqrt{2}}\right] $$ be a square of area two that is inscribed in the circle with radius one.

Simple geometric considerations show that the area of each of the four regions that is not covered by the square is $$ \frac{\pi-2}{4}\,. $$ Therefore \begin{align} \mathbb P\left\{-\tfrac1{\sqrt{2}}<X<\tfrac1{\sqrt{2}},-\tfrac1{\sqrt{2}}<X<\tfrac1{\sqrt{2}}\right\}&=\frac{2}{\pi}\,,&\text{(normalized area of square)}\\[4mm] \mathbb P\left\{-\tfrac1{\sqrt{2}}<X<\tfrac1{\sqrt{2}}\right\}&=\frac2{\pi}+2\frac{\pi-2}{4\pi}\,,&(\text{square plus two extra regions})\\[4mm] \mathbb P\left\{-\tfrac1{\sqrt{2}}<Y<\tfrac1{\sqrt{2}}\right\}&=\frac2{\pi}+2\frac{\pi-2}{4\pi}\,.&(\text{ditto})\\[4mm] \end{align} It is easy to check that the first probability is not the product of the last two. Therefore $X,Y$ are dependent.

Answer 3

They are not independent: A mathematical reasoning can be:

$f_{x,y} = \frac{1}{\pi}$ (given)

From here we can find $f_{x} = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f_{x,y} \, dy = \frac{2\sqrt{1-x^2}}{\pi}$

We know, for condition of independence to hold, $f_{x|y} = f_x$ i.e. x and y are independent random variables.

Now, $f_{x|y} = \frac{1}{\text{Area of the square formed when Y=y}}=\frac{1}{2y\sqrt{1-y^2}}$

Since, $f_{x|y} \neq f_x$, therefore x and y are not independent random variables.

If you find any issues in my answer, please do let me know.

Happy to cooperate!