Let the random point $(X,Y)$ be uniformly distributed on the unit disc $D=\{(x,y):x^{2}+y^{2}<1\}$. Show that the polar coordinates $R\in [0,1)$ and $\theta \in [0,2\pi)$ of the point are independent.

Can you help me with this exercise please?

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    $\begingroup$ A first hint: $R$ and $\theta$ are independent iff their joint density function is the product of their marginal density functions. A good place to start with this kind of problem is usually to try to find these functions. $\endgroup$
    – MånsT
    Apr 30, 2012 at 15:20
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    $\begingroup$ Please make edits to your question to show what you have tried and where you are finding difficulty. It is hard for us to provide much assistance in the learning process, otherwise. $\endgroup$
    – cardinal
    Apr 30, 2012 at 16:47
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    $\begingroup$ For any $r > 0$ and any $\theta \in [0,2\pi)$, find the probability that $\{R \leq r, \Theta \leq \theta\}$. If you draw a crude sketch of the region and describe on it the region of interest, you will see that you can calculate this probability without needing to do any explicit integration (simple mensuration formulas suffice), and that this probability equals $P\{R \leq r\}P\{\Theta \leq \theta\}$ both of which can also be worked out without explicit integration. This allows you to conclude that $R$ and $\Theta$ are independent. No need for densities or Jacobians etc. $\endgroup$ Apr 30, 2012 at 18:14
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    $\begingroup$ Follow-up exercise: Prove that the answer you get for the density function for the radius is completely invariant to the choice of norm. That is, if you define the disc with respect to any other norm, you will get exactly the same answer when considering the radius with respect to the same norm. $\endgroup$
    – cardinal
    Apr 30, 2012 at 18:37
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    $\begingroup$ @Dilip: What you state is true for the Euclidean norm but will not be true for (any) other norms. The point the exercise is trying to draw out is that the answer you get for the marginal distribution of the radius is not a consequence of the particular choice of norm here. $\endgroup$
    – cardinal
    Apr 30, 2012 at 19:00

1 Answer 1


The solution may be a bit quirky, with a lot of variables, but it works fine for me.

We know that $X$,$Y$ - random variables in $\mathbb{R}^n$ and $\mathbb{R}^m$ are independent iff $$\mathbb{E}(\varphi (X) \psi (Y) )= \mathbb{E}(\varphi(X)) \cdot \mathbb{E}(\psi(Y)) $$ $ \forall \varphi \in C^{\infty}_{0} $, $ \forall \psi \in C^{\infty}_{0} $, where $C^{\infty}_{0} $ are continuos on compact smooth functions.

If $(X,Y)$ are uniformly distributed on the unit disc $D=\{(x,y):x^{2}+y^{2}<1\}$. Then we have that $$\mathbb{P}_{(X,Y)}(dxdy)=\frac{1}{\pi}\mathbb{1}_D (x,y) dxdy$$

Analogically we define two distance and angle functions: $$\begin{cases}r:\mathbb{R^2}\rightarrow \mathbb{R}\\ \vartheta:\mathbb{R^2}\rightarrow \mathbb{R}\end{cases}$$ Now we have $R=r(X,Y)$ and $\Theta=\vartheta(X,Y)$ and by passing to polar coordinates with $\begin{cases}\rho=r(x,y)\\ \gamma = \vartheta(x,y) \end{cases}$ we can show that $$\mathbb{E}(\varphi(R))=2 \int\limits^1_0 \varphi(\rho)\rho d\rho$$ $$\mathbb{E}(\psi(\Theta))=\frac{1}{2\pi}\int\limits^{2\pi}_{0} \psi (\gamma) d\gamma$$

Now we see that $$\mathbb{E}(\varphi(R)\psi(\Theta))=\frac{1}{\pi} \int \limits_{D} \varphi(r(x,y))\cdot \psi(\vartheta(x,y))dxdy=\\ \frac{1}{\pi}\int\limits^1_0[\int\limits^{2\pi}_{0} \varphi(\rho) \psi(\gamma)d\gamma]\rho d\rho=\frac{1}{\pi} \int\limits^1_0 \varphi(\rho) \rho d\rho \cdot \int\limits^{2\pi}_{0} \psi(\gamma)d\gamma= \\ \mathbb{E}(\varphi(R)) \cdot \mathbb{E}(\psi(\Theta)) $$

Using Fubini's theorem we prove that $R$ and $\Theta$ are independent.


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