Assume you choose randomly two points $A,B$ in the unit disk. Let $R_1$ be the distance of $A$ from $(0,0)$. Let $R_2$ be the distance of $B$ from $(0,0)$.
how would you calculate $P(R_1>2R_2)$?
thanks.
$\begingroup$
$\endgroup$
12
-
$\begingroup$ It's $\|frac{1}{2}$, by symmetry. $\endgroup$– gnometoruleJan 26, 2013 at 18:25
-
$\begingroup$ @gnometorule No it's not. I think you missed the factor of two. $\endgroup$– Jonathan ChristensenJan 26, 2013 at 18:27
-
$\begingroup$ what do you mean by frac? I dont get this word.. $\endgroup$– adamcoJan 26, 2013 at 18:27
-
$\begingroup$ @adamco it's a LaTeX command for fraction. But he's wrong anyway... $\endgroup$– Jonathan ChristensenJan 26, 2013 at 18:28
-
1$\begingroup$ @JonathanChristensen You're absolutely right, I totally agree with you now. $\endgroup$– RustynJan 26, 2013 at 18:38
|
Show 7 more comments
1 Answer
$\begingroup$
$\endgroup$
2
Here's how you solve the problem:
Find $P(R_1 > 2R_2 | R_1)$ for a fixed value of $R_1$. This will be a function of $R_1$, call it $g(R_1)$. Remember that since the points are distributed uniformly on the disc, probability is proportionate to area. This is pretty easy.
Find the density function of $R_1$, call it $f(R_1)$.
Finally, integrate the function you found in part (1) over the distribution you found in part (2) to find the marginal probability $P(R_1 > 2R_2)$: $$P(R_1 > 2R_2) = \int_0^1 g(R_1) f(R_1) \, dR_1.$$
answered Jan 26, 2013 at 18:38
-
-
1$\begingroup$ @adamco Easiest via the CDF: There's a nice closed form for $P(R_1 \leq r)$. Then differentiate to find the pdf. $\endgroup$ Jan 26, 2013 at 19:47