Continuous uniform distribution over a circle with radius R

Question

I started to do this problem with the standard integration techniques, but I cant help but think that there has got to be something I am not seeing. Since it is a uniform distribution, even though x and y are not independent, it seems like there should be some shortcut. Here is the problem:

Take a random point (x, y) which is uniformly distributed over the circle with radius R

$f(x,y) = \begin{cases} \frac{1}{\pi R^2} & x^2 + y^2 \le R^2, \\ 0 & \text{otherwise} \end{cases}$

calculate $g(x)$, $h(y)$, the mean of $x$, $y$, and $xy$.

Answer 1

Let the pair $(X,Y)$ of random variables have uniform distribution on the disk with centre the origin and radius $R$. Then the joint density function $f_{X,Y}$ is $\frac{1}{\pi R^2}$ on the disk, and $0$ elsewhere.

For the density function of $x$, "integrate out" $y$. So we are integrating a constant from $-\sqrt{R^2-x^2}$ to $\sqrt{R^2-x^2}$. The result is $\frac{2\sqrt{R^2-x^2}}{\pi R^2}$ (for $-R\le x\le R$, and $0$ elsewhere).

The density function of $Y$ can be read off by symmetry.

For the mean of $X$, we don't need to calculate at all. By symmetry it is $0$.

The mean of $XY$ is also $0$. The second and fourth quadrant parts cancel the first and third quadrant parts.

Answer 2

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 :

$f(x,y)dydx$

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.