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If we want to chose a point $(x,y)$ uniformly at random from a unit circle in a plane, why is the joint probability density of the random variable $f(x,y) = \frac{1}{\pi}$ for $x^2+y^2\leq1$? The question linked simply states this but doesn't prove it. Can someone give an explanation please? Continuous uniform distribution over a circle with radius R

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The area of the disk is $\pi$. We have a uniform distribution on the disk, so the probability of landing in a part of the disk with area $A$ is proportional to $A$, say $kA$. Since the probability of landing in the unit disk is $1$, we have $k\pi=1$ and therefore $k=\dfrac{1}{\pi}$. Thus the probability of landing in a part of the disk with area $A$ is $\dfrac{A}{\pi}$.

Now what function $f(x,y)$ is it that integrated over any part of the disk with area $A$ gives result $\dfrac{A}{\pi}$? The constant function $\dfrac{1}{\pi}$.

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  • $\begingroup$ Why is it proportional - I intuitively know this and can come with some heuristic arguments about it although I cannot find a formal proof to this. $\endgroup$ – baibo Sep 19 '13 at 22:13
  • $\begingroup$ That is one definition of uniform distribution over a region. (The other standard definition is to say it is the distribution with density function $\dfrac{1}{C}$, where $C$ is the "content" (maybe length, maybe area, maybe volume) of the region we are uniform over, and $0$ elsewhere. That would make the density function $\frac{1}{\pi}$ in the disk by definition. $\endgroup$ – André Nicolas Sep 19 '13 at 22:20
  • $\begingroup$ @AndréNicolas ... For the same question if we have to find $P(|X|<12) $ then how to approach this problem? $\endgroup$ – monalisa Mar 9 '17 at 5:18

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