Bertrand's paradox is a neat problem that shows what happens when we take the words 'at random' a bit too casually. In this post, we will be looking at some problems in the same setup and how the different methods yield answers for the 'seemingly' same question.
Bertrand's paradox asks for the probability of a 'random' chord in a unit circle being longer than the side of an inscribed triangle. We are interested in the expected length of such a chord. We now see how the expected length varies if we choose the chord according to the three methods in the paradox.
(1) Circumference method
In this method, we choose two points uniformly randomly on the circumference of the circle and connect those points to create the chord. After the first point is chosen, the second point is uniformly distributed in the circumference.
With this method, if we draw a tangent at the first point, then the angle between the chord and the tangent is uniform in the range $(0,\pi)$. If we let that angle be $\theta$, then the length of the chord is simply $2\sin\theta$. Therefore,
$\displaystyle\mathbb{E}(L)=\int_0^{\pi/2}2\sin\theta\,\frac{d\theta}{\pi/2}=\frac{4}{\pi}\approx 1.2732$
(2) Radial method
In this method, we first choose a random radius. We then choose a point uniformly on this radius and draw a chord perpendicular to this line. If we let $x$ be the length of the point from the origin, then the length of the chord is $2\sqrt{1-x^2}$.
Because $x$ is uniform in $(0,1)$, we have
$\displaystyle \mathbb{E}(L)=\int_0^12\sqrt{1-x^2}dx=\frac{\pi}{2}\approx1.5708$
But here's something interesting. If we use $x=\cos\theta$ in the above integral, the integral becomes,
$\displaystyle \mathbb{E}(L)=\int_0^{\pi/2}2\sin\theta\cdot\sin\theta\,d\theta$
From what we saw in the 'Circumference method', we know that the $2\sin\theta$ denotes the length of the chord. Hence, whatever remains after that must be the PDF of the variable $\theta$.
Therefore, the PDF, $f(\theta)$, for this case is $\displaystyle f(\theta)=\sin\theta$
(3) Midpoint method
Here, a point is chosen randomly in the circle and the unique chord which has this point as the midpoint is drawn. The expected length is slightly tricky here. The trick is to identify the parameter of interest here is the distance of the point from the origin.
If we let $r$ be that distance, then the length of the chord is $2\sqrt{1-r^2}$. But $r$ is not uniform. It is well known that the density of function of $r$ is $2r$ for a unit circle. We then have
$\displaystyle \mathbb{E}(L)=\int_0^12\sqrt{1-r^2}\cdot 2r\,dr=\frac{4}{3}\approx 1.3334$
If we use the substitution $r=\cos\theta$ in the above integral, we can see that the PDF of $\theta$ in this case is $f(\theta)=\sin2\theta$
(4) Random point and line at random angle method
The video also specifies another method of creating a chord which involves choosing a point randomly in the circle and drawing a line at a random angle through this point. To make things clear, let the $r$ be the distance between the point and other centre of the circle.
Now we draw a line perpendicular to the radial line containing this point. The angle, $t$, that the random line (that we draw to create the chord) makes with this perpendicular line is uniformly distributed between $(0,\pi)$.
It can shown then the distance of the random chord and the centre of the circle is then $r\cos t$. Therefore,
$\displaystyle \mathbb{E}(L)=\int_0^{\pi/2}\int_0^12\sqrt{1-(r\cos t)^2}\cdot 2r\,dr\frac{dt}{\pi/2}=\frac{16}{3\pi}\approx 1.69765$
The above integral was solved using Wolfram Alpha.
Here we use the substitutions $u=r\cos t$ and $v=r\sin t$. Then we know that $dudv=rdrdt$. The limits of $r$ and $t$ covers the first quadrant of the unit circle and so should the limits of $u$ and $v$. Therefore,
$\displaystyle \mathbb{E}(L)=\int_0^{1}\int_0^{\sqrt{1-u^2}}2\sqrt{1-u^2}\cdot 2\,\frac{dvdu}{\pi/2}=\int_0^{1}2\sqrt{1-u^2}\cdot 2\sqrt{1-u^2}\,\frac{du}{\pi/2}$
If we now use $u=\cos\theta$, then we see that
$\displaystyle f(\theta)=\frac{2\sin^2\theta}{\pi/2}$
(5) Random point on a radial line and a line at random angle method
Just for the sake of needlessly complicating things, we look at another method where we choose a point randomly in a randomly chosen radial line and draw a at a random angle through this point (the number of randoms in this sentence!). This is almost exactly like the previous case.
$\displaystyle \mathbb{E}(L)=\int_0^{\pi/2}\int_0^12\sqrt{1-(x\cos t)^2}\cdot \,dx\frac{dt}{\pi/2}=\frac{2}{\pi}(2G+1)\approx 1.80286$
where $G$ is the Catalan constant. Note that this integral can be expressed as the integral of the Elliptic integral of the second kind. With that and some known results, this can also be calculated manually.
Again using the subtitution $u=x\cos t$ and $v=x\sin t$ with the Jacobian $dudv=xdxdt$, we see
$\displaystyle \mathbb{E}(L)=\int_0^{1}\int_0^{\sqrt{1-u^2}}2\sqrt{1-u^2}\cdot \,\frac{dvdu}{x\cdot\pi/2}$
But $x=\sqrt{u^2+v^2}$. Using this, the fact that the integral of the secant function is the inverse hyperbolic sine of the tangent function and the substitution $u=\cos t$, we can see that the PDF in this case as
$\displaystyle f(\theta)=\frac{\sin\theta\sinh^{-1}(\tan\theta)}{\pi/2}$
That sums up the Expected values. But we can do more. There was a recent Numberphile video on similar lines. The question discussed was to find the probability that two points selected at random in a circle is lies on different sides of a 'random' chord. Here's where the PDFs we've found so far will be really useful. We'll see those in the next post.
Hope you enjoyed this.
Until then
Yours Aye
Me
