Suppose a point has a random location in the circle of radius 1 around the origin. The coordinates $(X,Y)$ of the point have a joint density

$$f_{X,Y}(x,y) = \begin{cases}\frac{2}{\pi}(x^2+y^2)&\mathrm{\ if \ } x^2+y^2\le1\\ 0&\mathrm{\ otherwise\ }\end{cases}$$

Let $D$ be the distance from the random point to the center of the circle. How do I compute the $nth$ moment of $D$, $E(D^n)$, for $n = 1,2,...m$?

  • $\begingroup$ Are you sure you have copied this problem correctly, that is, are you sure the joint density is indeed a valid density function? Hint: use LOTUS, the Law of the Unconscious Statistician. $\endgroup$ – Dilip Sarwate Sep 28 '12 at 3:58

We have $D=\sqrt{X^2+Y^2}$, so the $n$-th moment of $D$ is the integral of $$(x^2+y^2)^{n/2}\left(\frac{2}{\pi}\right)(x^2+y^2)$$ over the unit disk. Thus we want to integrate $\displaystyle\frac{2}{\pi}\displaystyle(x^2+y^2)^{1+\frac{n}{2}}$ over the unit disk.

Change to polar coordinates.

  • $\begingroup$ Where did you get that first term? $(x^2+y^2)^{n/2}$? Or more precisely, why is it take to power $n/2$? $\endgroup$ – Denys S. May 15 '14 at 7:46
  • $\begingroup$ In general, if $W$ is a random variable, the $n$-th moment of $W$ is $E(W^n)$. So the $n$-th moment of $(X^2+Y^2)^{1/2}$ is the integral of $((x^2+y^2)^{1/2})^n f_{X,Y}(x,y)\,dx\,dy$ over the region where the joint density "lives.' $\endgroup$ – André Nicolas May 15 '14 at 7:53
  • $\begingroup$ Oh indeed, sorry I missed the sqrt in the D. Thank you. $\endgroup$ – Denys S. May 15 '14 at 8:01
  • $\begingroup$ No problem, you are welcome. It gave me the chance to notice and fix a "typo," I had $x^2+y^2$ where I meant $X^2+Y^2$. $\endgroup$ – André Nicolas May 15 '14 at 12:34

For every $0\leqslant x\leqslant1$, $\mathrm P(D\leqslant x)=\int\limits_0^{2\pi}\int\limits_0^x(2/\pi)r^2\cdot r\mathrm dr\mathrm d\theta=\int_0^x4r^3\mathrm dr$ hence the density $f_D$ of $D$ is such that $f_D(x)=4x^3\cdot \mathbf 1_{0\leqslant x\leqslant1}$.

In particular, $\mathrm E(D^n)=\int\limits_0^1x^n\cdot f_D(x)\mathrm dx=\int\limits_0^14x^{n+3}\mathrm dx=4/(n+4)$.

One can note that $D^4$ is uniform on $(0,1)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.