Solution :
Let the distance of the stick from one break be X
And let us assume that [tex]$X \leq l/2$[/tex].
Here, l = length of stick
Therefore, [tex]$P(X<x) =\frac{x}{l/2}$[/tex]
We know that, [tex]$R=\frac{X}{l-X}$[/tex] , so by definition we get
[tex]$X =\frac{lR}{1+R}$[/tex]
The cumulative distribution function for R is
[tex]$P(R<r) = P (\frac{X}{l-X}<r) = P(X<\frac{lr}{1+r})=\frac{lr}{(1+r)(l/2)}=\frac{2r}{1+r}$[/tex]
When it starts at zero, then r =0. It ends at one when the r has a maximum
value of one.
The probability density function is given by
[tex]\frac{d}{dr}(\frac{2r}{1+r})= \frac{2}{(r+1)^2}[/tex]
Now integrating, we find E(R) and [tex]$E(R)^2$[/tex] gives :
[tex]$E(R) =\int\limits^1_0 \frac{2r}{(1+r)^2} \, dr = 2 \ln 2-1 $[/tex]
[tex]$E(R)^2= 3 - 4\ln 2$[/tex]
Therefore, Var(R)= [tex]$2-4(\ln \ 2)^2$[/tex]
