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How can I compute the PDF/CDF and expected value of the following function:

$$ \frac{\alpha}{r^2} $$

where $r$ is generated as follows:

  1. draw $x$ and $y$ from a uniform distribution in the range $[-r_\max,+r_\max]$.
  2. if $r_\min \le \sqrt{x^2+y^2} \le r_\max$ then $r=\sqrt{x^2+y^2}$ else repeat step (1)

$r_\min$, $r_\max$, and $\alpha$ are given constants.

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up vote 1 down vote accepted

This is really the same as drawing the pair $(x,y)$ uniformly from inside the annulus bounded by concentric circles of radii $r_\min$ and $r_\max$. If $s$ is between $r_\min$ and $r_\max$ then the probability that $r\le s$ is the area between the circle of radius $s$ and the circle of radius $r_\min$ divided by the whole area of the annulus. I.e. $$ \Pr(r \le s) = \frac{\text{area of annulus between radii }r_\min\text{ and }s}{\text{area of annulus between radii }r_\min\text{ and }r_\max} = \frac{\pi s^2 - \pi r_\min^2}{\pi r_\max^2 - \pi r_\min^2} = \frac{s^2-r_\min^2}{r_\max^2-r_\min^2}. $$ So that's the CDF as a function of $s$. Differentiate it to get the PDF, and the numerator becomes $2s$ while the denominator stays as it was. Call the CDF (capital) $F$ and density (lower-case) $f$, and then the expected value is $$ \alpha\int_{r_\min}^{r_\max} \frac{1}{s^2} f(s)\;ds. $$

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And of course, all this assumes that $X$ and $Y$ are independent random variables, which is not stated anywhere in the problem. Also, I think the OP wanted the pdf, CDF, and expectation of $\alpha/R^2$, not of $R$. – Dilip Sarwate Mar 7 '12 at 0:18
I'd overlooked that last point. As for independence, forgetting to mention that is one of the most frequent of all oversights, even by professionals in published work. – Michael Hardy Mar 7 '12 at 0:46
....and I've edited accordingly. – Michael Hardy Mar 7 '12 at 0:48

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