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


2 Answers 2


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 $\dfrac{1}{\pi}$.

  • $\begingroup$ 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, 2013 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. $\endgroup$ Sep 19, 2013 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, 2017 at 5:18
  • $\begingroup$ @monalisa your question doesn't make sense: the original question is about two random variables, $X$ and $Y$. $\endgroup$
    – baibo
    Jun 30, 2019 at 7:49

I will outline what I do. I first create a small piece inside the circle, evaluate the probability of observing a point within that small piece.

Then I use the fact that since all such pieces are equally likely, this probability is equal to some constant 'k'.

Then I add up the probabilities of all these small pieces, clearly this involves calculating the area of the circle as the the areas of all the pieces adds up to the area of the circle.

First let us create Cross sections perpendicular to the $ x $ -axis. Consider the circle $ x^{2}+y^{2}=R^{2} $. Its top and bottom halves can be given explicitly via $ y=\pm \sqrt{R^{2}-x^{2}} $, and $ x $ ranges from $ -R $ to $ R $. Thus, the area in question is given by the integral $ \int_{-R}^{R} \int_{-\sqrt{R^{2}-x^{2}}}^{\sqrt{K^{2}-x^{2}}} dy dx $.

Now let us assume the pdf is f(x,y), hence the probability of choosing a piece with area dydx is :


Now remember by DEFINITION the pdf must be a constant as no matter which cross section you take the probability of choosing it must be equal (think of this as chopping up the disc into these small pieces given by dydx and each of these pieces is equally likely to be chosen). This implies that the the pdf is a constant i.e. $f(x,y)$ is equal to some constant. Be careful here, as this is in fact the intuition that you are seeking here.

Now what does the integration look like :

$ \int_{-R}^{R} \int_{-\sqrt{R^{2}-x^{2}}}^{\sqrt{K^{2}-x^{2}}} f(x,y) dy dx $

Replace f(x,y) by some constant 'k' (Had each piece been not equally likely then we CANNOT make this statement and the integral would demand this specification in terms of x and y).

$ \int_{-R}^{R} \int_{-\sqrt{R^{2}-x^{2}}}^{\sqrt{K^{2}-x^{2}}} k dy dx $

Remember now that this whole integral must evaluate to 1.

$ \int_{-R}^{R} \int_{-\sqrt{R^{2}-x^{2}}}^{\sqrt{K^{2}-x^{2}}} k dy dx = 1$

The LHS is simply the area of a circle times 'k'. You can Google how to derive the area of a circle by integration.

$f(x,y)$ = k = $1/\pi R^{2}$

Hence, 'k' is now simply $1/\pi R^{2}$. Which is your pdf. You have used some conditions on the problem to get an explicit value for the constant.


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