1
$\begingroup$

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

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

OK, how about this:

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.

$\endgroup$
3
  • 2
    $\begingroup$ You might want to just do the computation in polar coordinates then, the problem is incredibly easier that way. $\endgroup$
    – Qise
    Apr 8, 2023 at 22:05
  • $\begingroup$ Thanks, just posted an attempt. What do you think? $\endgroup$
    – JohnRoper
    Apr 8, 2023 at 22:22
  • $\begingroup$ 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? $\endgroup$
    – David K
    Apr 9, 2023 at 4:23

1 Answer 1

2
$\begingroup$

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 $$ \frac{1}{\pi}\int_0^{2\pi} \int_0^{1} r^2 \, {\rm d} r {\rm d} \theta = \frac{1}{3\pi}\int_0^{2\pi} \, {\rm d} r {\rm d} \theta =\frac{2\pi}{3\pi}=\frac{2}{3} $$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .