Suppose the joint PDF of $R$ and $\Theta$ is $$ f_{R\Theta}(r,\theta) = Cg(r^2)r\quad, \ \text{where } C \text{ is some constant and } g \text{ is some function}$$

The joint distribution does not depend on $\Theta$, but why is $\Theta$ distributed Uniformly?


1 Answer 1


We have $$f_{R,\Theta}(r,\theta) = C\,g(r^2)\,r \quad\Big[\theta\in[0;2\pi), r\in S_R, S_R\subseteq (0;\infty)\Big]$$

If $G$ exists such that $g(r)=\dfrac{\mathrm d G(r)}{\mathrm d r}$ for all values of $r$ in the support of $R$.

Then $$\begin{align}f_{\Theta}(\theta) &= \int_0^\infty C~g(r^2)~r\operatorname d r\\[1ex] & = {\left. \frac {C~G(r^2)}2\right\vert}_{r\to\inf S_R}^{r\to\sup S_R}\quad\Big[\theta\in[0;2\pi)\Big] \\&= D\quad\Big[\theta\in[0;2\pi)\Big]\end{align}$$

Which will be some constant value, let us call it, $D$.   (Indeed, I assert that $D=1/2\pi$ if $f_\Theta$ is in fact a probability density function.)

Long story short.   If the marginal density of $\Theta$ is a constant value within the support $[0;2\pi)$ then $\Theta$ is uniformly distributed over that support, by definition of that distribution.

  • $\begingroup$ I was stuck integrating the marginal, but your explanation makes sense to me now, thank you!! $\endgroup$
    – Zoey A
    Nov 17, 2016 at 0:05

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