A random point is chosen uniformly in the unit disk {$(x, y) : x^2 + y^2 \le 1$}. Let R be its
distance from the origin.(a) Find E(R) using 2D LOTUS.
(b) Find the CDFs of $R^2$ and of R without using calculus, using the fact that for a Uniform distribution on a region, probability within that region is proportional to area. Then get the PDFs of $R^2$ and of R, and find E(R) in two more ways: using the definition of expectation, and using a 1D LOTUS by thinking of R as a function of $R^2$.
Please, help me to solve the problem.
I did the following so far:
$f(x,y)=\frac1{2r^2\pi}$, when $x^2+y^2 \le1$, then $f(y)=\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac1{2r^2\pi}$dx=$\frac{2\sqrt{1-x^2}}{\pi r^2}$;
$f(x|y)=\frac{1}{4\sqrt{1-x^2}}$
since R=$x^2+y^2$; $P(R \le r)=f(\sqrt{x^2+y^2} \le r|Y=y)f(y)=f(-\sqrt{r^2-y^2} \le x \le \sqrt{r^2-y^2}|Y=y)f(y)$
What to do next? I want to find PDF to find $E[R]$
Withut calculus CDF of R will be $\frac {\pi r^2 }{\pi}=r^2 => f(R)=2r $
$E[R]$-?
Best Answer
For part (a) with LOTUS, you do not try to find the pdf for R. Rather use the joint pdf, $f(x,y)={1\over\pi}$ for $x^2+y^2\le1$ and $0$ otherwise: $$E[R]=\int\int\sqrt{x^2+y^2}f(x,y)\,dxdy$$ To do the integral, you can now transform to $(r,\theta)$, giving, $$E[R]={1\over\pi}\int_0^1\int_0^{2\pi}r\,rdr\,d\theta=2\int_0^{1}r^2dr$$ The form is the same as for part (b), where $$E[R]=\int_0^1rf(r)dr$$ using the pdf $f(r)$ that you found.
Hopefully you do the rest of the problem now...