If we pick randomly two points inside a circle centred at $O$ with radius $R$, and draw two circles centred at the two points with radius equal to the distance between them, what is the expected area of the intersection of the two cirlces that contain the origin $O$.
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3I don't see how the first circle (the one with radius R) has any relevance to the problem. – bubba Mar 09 '13 at 13:26
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2What have you tried? Check this PDF. It'll guide you to proceed a bit. – hjpotter92 Mar 09 '13 at 13:26
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1@ bubba: $R$ restrict the location of the circle. – JSCB Mar 09 '13 at 13:29
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2@bubba For instance the distance of the two points cannot be more than $2R$, and almost surely it is not $2R$. – AndreasT Mar 09 '13 at 13:29
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2I think it depends on how you measure the randomness of the points. Check for instance the Bertrand's paradox – AndreasT Mar 09 '13 at 13:32
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1The expected area is $(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})R^2$. Let $\vec{x}_1, \vec{x}_2$ be the two points, the key is $E[|\vec{x}_1-\vec{x}_2|^2] = E[|\vec{x}_1|^2+|\vec{x}_2|^2]=R^2$. – achille hui Mar 09 '13 at 13:42
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1@achille: Nice observation -- but that's a full answer; if you write it as an answer it can be accepted so the question doesn't remain open. – joriki Mar 09 '13 at 13:44
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Yes, but give us more detail on the answre – sys Mar 09 '13 at 13:51
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What should be the expected area of the intersection in the case where the intersection must contain the origin of the circle of radius $R$ – sys Mar 09 '13 at 13:55
3 Answers
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Let $\vec{x}_1$ and $\vec{x}_2$ be the two points. Let $r = |\vec{x}_1 - \vec{x}_2|$ be the distance between them. By elementary geometry, if you draw two circle of radius $r$ using these two points as center, the area of their intersection is given by $(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})r^2$. Notice the picking of two points are independent, we have: $$E\left[ \vec{x}_1 \cdot \vec{x}_2 \right] = E\left[\vec{x}_1\right] \cdot E\left[\vec{x}_2\right] = \vec{0} \cdot \vec{0} = 0$$ This implies $$E\left[|\vec{x}_1 - \vec{x}_2|^2\right] = E\left[|\vec{x}_1|^2 + |\vec{x}_2|^2\right] = 2\frac{\int_0^R r^3 dr}{\int_0^R rdr} = R^2$$
As a result, the expected area of the intersection is $(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})R^2$.
Update for those who are curious
Let $\mathscr{C}$ be the set of events such that the intersection contains the origin, then: $$\begin{align} \operatorname{Prob}\left[\,\mathscr{C} \right] &= \frac{2\pi + 3\sqrt{3}}{6\pi}\\ E\left[\,|\vec{x}_1 - \vec{x}_2|^2 : \mathscr{C}\right] &= \frac{20\pi + 21\sqrt{3}}{6(2\pi + 3\sqrt{3})} \end{align}$$ and the expected area of intersection conditional to containing the center is given by: $$\frac{(4\pi - 3\sqrt{3})(20\pi + 21\sqrt{3})}{36(2\pi + 3\sqrt{3})}$$
To evaluate $E\left[ \varphi(\vec{x}_1,\vec{x}_2) ) : \mathscr{C} \right]$ for any function $\varphi( \vec{x}_1, \vec{x}_2 )$ which is symmetric and rotational invariant w.r.t its argument, you need to compute an integral of the from:
$$\int_{\frac{\pi}{3}}^{\pi} \frac{d\theta}{\pi} \left[2\int_{0}^{R} \frac{2udu}{R^2} \left( \int_{\alpha(\theta)u}^{u} \frac{2vdv}{R^2} \phi( \vec{x}_1, \vec{x}_2 ) \right) \right] $$
where $u \ge v$ are $|\vec{x}_1|$ and $|\vec{x}_2|$ sorted in descending order. $\theta$ is the angle between $\vec{x}_1$ and $\vec{x}_2$. The mysterious $\alpha(\theta)$ is $\max(2\cos(\theta),0)$ for $\theta \in [\frac{\pi}{3},\pi]$.
The integral is a big mess and I need a computer algebra system to crank that out. I won't provide more details on this part not relevant to the main answer.
achille hui
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Thank you for the answer, do you have an idea about what should be the expected area of the intersection if we pick $K$ points inside a circle of radius $R$ and we draw all possible two circles $\binom K2$ – sys Mar 09 '13 at 18:56
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@sys No idea, once you put constraint of the circles, even the cases of 2 circles start to look ugly. – achille hui Mar 09 '13 at 19:03
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Please, can you explain more how did you get \begin{align} \operatorname{Prob}\left[,\mathscr{C} \right] &= \frac{2\pi + 3\sqrt{3}}{6\pi}\ E\left[,|\vec{x}_1 - \vec{x}_2|^2 : \mathscr{C}\right] &= \frac{20\pi + 21\sqrt{3}}{6(2\pi + 3\sqrt{3})} \end{align} Also, the expected ditance between two points inside a circle is given by:$d= {128 r\over 45\pi} $ but you said that is $r$ ?? – sys Mar 22 '13 at 21:54
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May be you can try to integrate the Circular Segment the half one and then multiply by 2 (since the R of each circles are the same). Since minimum angle when the intersection happen exactly at each center point, so the minimum angle is 120 degree, and the maximum angle is exactly 180 degree.
Based on that, we just put that into integration :
$$E[A] =2\int_{2\pi/3}^\pi \frac{R^2}2(\theta-sin\theta)d\theta$$
Where E[A] is expected area, then just integrate that formula like basic integral.
thanks. hope that can help, if not someone can fix that.
thanks
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Sharing a Geometrical Solution.
We can see that this region is simply :
- 4 Cone(1/6 of Circle) - 2Equilaterral Triangle
So you can simply calculate those Area: (T = Triangle, C = Cone)
Mr Rubix
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