# Probability density of the distance from a centre of a unit circle

Consider a point chosen randomly (and uniformly) somewhere within the unit disk, let the point have coordinates $$x$$ and $$y$$. I am trying to find the probability density $$f(x,y)$$. How can we show that $$f(x,y) = \frac{1}{\pi}$$?

I am trying to work out the expected value of the distance of the point from the center. So I would like to compute the integral

$$\int_0^1 \int_0^1 f(x,y) \sqrt{x^2+y^2} \, {\rm d} x {\rm d} y$$

In polar co-ordinates, if I choose a point $$r, \theta$$, then the probability of landing here, i.e. $$f(r, \theta) dr d \theta$$ is $$r dr d\theta / \pi$$ since the radius of the disk is $$1$$.

Taking the expected value of $$r$$, I get

$$\frac{1}{\pi}\int_0^1 \int_0^{2 \pi} r^2 \, {\rm d} r {\rm d} \theta = \frac23$$

which I believe is the correct answer.

• You might want to just do the computation in polar coordinates then, the problem is incredibly easier that way.
– Qise
Apr 8, 2023 at 22:05
• Thanks, just posted an attempt. What do you think? Apr 8, 2023 at 22:22
• In this question, you first ask how to show that $f(x,y) = \frac{1}{\pi}.$ Then you ask about the expected value of the distance from the center. Which of these is the actual question? Apr 9, 2023 at 4:23

Being uniformly distributed over a finite area $$A$$ means that the probability density $$f$$ of $$(X,Y)$$ is $$f(x,y)=\frac1{A}$$ for $$(x,y)\in A$$ and $$f(x,y) = 0$$ for $$(x,y) \notin A$$. This has to be the case in order to ensure that $$\int\int_{(x,y)\in \Re}f(x,y)dxdy = 1$$. Can you take it from here to show that $$f(x,y)=\frac{1}{\pi}$$? Note that here, $$A$$ is just $$\pi r^2$$ with $$r=1$$ so $$A=\pi$$.
Once you have that, we get that the expected distance is $$\int_0^{1} \int_0^{1} f(x,y) \sqrt{x^2+y^2} \, {\rm d} x {\rm d} y$$
which after changing to polar coordinates, and noting that our bounds of integration tell us that we are inside $$A$$, so $$f(x,y)$$ can be substituted by above expression, the preceding double integral becomes $$\frac{1}{\pi}\int_0^{2\pi} \int_0^{1} r^2 \, {\rm d} r {\rm d} \theta = \frac{1}{3\pi}\int_0^{2\pi} \, {\rm d} r {\rm d} \theta =\frac{2\pi}{3\pi}=\frac{2}{3}$$