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Select uniformly a point from the annulus $\{(x,y):1\leq x^2+y^2\leq 4\}$. Let R be the distance to $(0,0)$. Give the cdf and pdf of R.


I know that the sample space of choosing a point would be the area of the annulus ($3\pi$) and I know the method of getting the pdf from the cdf or vice versa. I'm just not sure how to start; How would I represent the distance R from the point (x,y) and would this be the cdf or pdf?

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  • $\begingroup$ Why delete part of the question when the answers address that part specifically? $\endgroup$ – StubbornAtom Mar 26 at 7:53
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How would I represent the distance R from the point (X,Y) and would this be the cdf or pdf?

By definition of Euclidean Distance, $R=\sqrt{X^2+Y^2}$.

So immediately we know the support is $\{r: 1\leqslant r^2\leqslant 4\}$

Now, since the points are uniformly distributed: the pdf for $R$, at any point $r$ within that support, will be equal to the ratio of the circumference of a circle with radius $r$ to the area of the annulus. Call this $f_R(r)$.

And the CDF will be the integral $\displaystyle F_R(r)=\int_1^r f_R(s)\mathrm d s$ .

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  • $\begingroup$ Thanks. Why would it be the circumference of the circle, would it not be the area? $\endgroup$ – PLC Mar 26 at 4:51
  • $\begingroup$ @Graham Kemp doesn't this give for $r=2$ which lies within that support, $\Bbb{P}(R=2)=\frac{4}{3}>1$? I interpretted the pdf for $R$ at any point within the support as $\frac{2\pi r}{3 \pi} = \frac{2r}{3}$ $\endgroup$ – Tikak Mar 26 at 5:04
  • $\begingroup$ A Probability Density Function may have values exceeding one. It is not a probability mass. Also, yes , $f_R(r)=\dfrac{2r}{3}\mathbf 1_{1\leqslant r\leqslant 2}$. $\endgroup$ – Graham Kemp Mar 26 at 12:07
  • $\begingroup$ It is the circumference because that is the derivative of the area of a circle with respect to the radius. (Which is a hint to find the CDF without integration.) $\endgroup$ – Graham Kemp Mar 26 at 12:10
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It is easier to find the CDF first (and then it is easy to find the PDF from there). Try to compute $P(R \le r)$ for any real number $r$. (This will be a ratio of areas.)

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