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.
Let the random point (X,Y) be uniformly distributed on the unit disc D=\{(x,y):x^{2}+y^{2}<1\}
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probability-theory
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0To prove independency you need to show that $ P\left[R=r\wedge\Theta=\theta\right]=P\left[R=r\right] P\left[\Theta=\theta\right] $ – 2012-05-04
1 Answers
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The probability density function for $(X,Y)$, given its uniform distribution is $ f_{X,Y}(x,y) = \frac{1}{\pi} I(x^2+y^2 <1) $ where $I$ stands for Iverson bracket. Now consider a bounded function $\varphi(R) \phi(\Theta)$ and consider its expectation: $ \mathbb{E}\left(\varphi(R) \phi(\Theta)\right) = \int_D \varphi(\sqrt{x^2+y^2}) \phi(\arctan(x,y)) f_{X,Y}(x,y) \mathrm{d} x \mathrm{d}y $ Now perform the change of variables to the polar coordinates. If the expectation factors into the product of expectations $\mathbb{E}(\varphi(R)) \mathbb{E}(\phi(\Theta))$, this proves the independence.
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0@copper.hat Thanks, fixed it. – 2012-05-04