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Determine the probability density function on the unit circle $U:=\{(x,y)\in\mathbb{R}^2:x^2+y^2 =1\}$ with respect to the Lebesgue measure $\lambda$. Calculate the marginal distributions $f_X$ and $f_Y$ if the random positions on the unit circle in the euclidian coordinate system are interpreted as a two-dimensional random variable $(X,Y)$.

Now, if we set $(x,y) = (\cos\theta,\sin\theta)$, the probability density function would be $f(\theta) = \frac{1}{2\pi}$. But I cannot figure out how to determine the marginal distributions then, as this is only a one-dimensional object and with respect to $\lambda$, not $\lambda^2$.

Analysis801
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  • $\theta ∼ U[0,{1 \over 2 \pi}]$. $X$ and $Y$ are just functions of this variable. – N74 May 12 '18 at 11:25

1 Answers1

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It is possibly easier to do this as a cumulative distribution function first

Clearly $\mathbb P(X\le x)=0$ for $x \le -1$ and $\mathbb P(X\le x)=1$ for $x \ge 1$

You have not specified the support of $\Theta$, though $(-\pi, \pi]$ and $[0, 2\pi)$ are common choices; either will do here

For $-1 \lt x \lt 1$ we have: $$\mathbb P(X \le x) = \mathbb P(\cos(\Theta) \le x) = 2\mathbb P(\cos^{-1}(x) \le \Theta \le \pi) = 1 - \tfrac1\pi \cos^{-1}(x)$$

Taking the derivative we get $$f_X(x) = \frac{1}{\pi\sqrt{1-x^2}}$$ for $|x| \lt 1$, and $f_X(x)=0$ for $|x| \gt 1$, and the marginal density for $Y$ is the same by symmetry

Henry
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  • can you please explain, why $\mathbb{P}(\cos \theta < x) = 2 \mathbb{P}(\arccos x < \theta < \pi)$? Did you take the domain of $\theta$ here as $[-\pi, \pi]$ or $[0, 2\pi]$? – Joitandr Nov 19 '19 at 06:38
  • @Joitandr On the left hand side $\mathbb P(\cos(\Theta) \le x)$ the support for $\Theta$ can be either $(-\pi, \pi]$ or $[0, 2\pi)$. On the right hand side $2\mathbb P(\cos^{-1}(x) \le \Theta \le \pi)$ it can only be in the image of $\cos^{-1}$ i.e. $[0,\pi]$, which explains the $2$ (it also reverses direction because here $\cos$ and $\cos^{-1}$ are decreasing functions) – Henry Nov 19 '19 at 08:46
  • No, I mean - could you please explain the whole transformation: $\cos \theta < x \Rightarrow \dots \Rightarrow \arccos x < \theta < \pi$ step by step, because when i did it, i got to different bounds on $\theta$ – Joitandr Nov 19 '19 at 09:55
  • @Joitandr what bounds did you get? For example with $\mathbb P(\cos(\Theta) \le \frac12)$, which I think should end up being $\frac23$ – Henry Nov 19 '19 at 10:39
  • $\cos \theta < 1/2 \Rightarrow \arccos(1/2)< \theta < \pi + \arccos(-1/2)$ – Joitandr Nov 19 '19 at 11:06
  • @Joitandr OK so $\mathbb P(\cos \theta < 1/2) $ $= \mathbb P(\arccos(1/2)< \theta < \pi + \arccos(-1/2)) $ $= \mathbb P(\arccos(1/2)< \theta \le \pi)+ \mathbb P(\pi < \theta < \pi+ \arccos(-1/2)) $ $= 2 \mathbb P(\arccos(1/2)< \theta \le \pi)$ – Henry Nov 19 '19 at 11:13
  • and why $\pi < \theta < \pi + \arccos(-1/2) \Leftrightarrow \arccos(1/2) < \theta < \pi$? – Joitandr Nov 19 '19 at 11:33
  • Did you calculate $\mathbb P(\cos(\Theta) \le \frac12)$ your way and my way? Did you get $\frac23$ both ways? If so, can you see why? – Henry Nov 19 '19 at 12:00
  • Okay, and what about such a probability: $\mathbb{P}(\cos \theta < x, \sin \theta < y)$? – Joitandr Nov 20 '19 at 07:13