A stick of length 1 is broken at a uniformly random point, yielding two pieces. Let X and Y be the lengths of the shorter and longer pieces, respectively, and let R = X/Y be the ratio of the lengths X and Y . (a) Find the CDF and PDF of R. (b) Find the expected value of R (if it exists). (c) Find the expected value of 1/R (if it exists)
I'm stuck on finding a PMF and CDF of R - how would I set this up?
On
$z = \frac{x}{y} = \frac {x}{1-x}$ With x uniformly distributed over $[0,\frac 12]$
$E[z] = \int_0^1 zf(z) \ dz = \int_0^{\frac 12} \frac {2x}{1-x}\ dx$
$z = \frac {x}{1-x}\\ x = \frac{z}{1+z}\\ dx = \frac {1}{(1+z)^2} dz\\ f(z) = \frac {2}{(1+z)^2}$
$F(z) = \int_0^z \frac {2}{(1+z)^2} \ dz = -\frac {2}{1+z}+2 = \frac {2z}{1+z}$
$E[z] = \int_0^{\frac 12} \frac {2x}{1-x}\ dx = \int_0^1 \frac {2z}{(1+z)^2} \ dz = \ln 4 - 1$
$E[\frac 1z] = \lim_\limits{x\to 1}\int_{\frac12}^x \frac {x}{1-x} \ dx = \infty$
Hints: let $Z$ have uniform distribution on $(0,1)$, $X=\min \{Z,1-Z\}$ and $Y=\max \{Z,1-Z\}$. Also $R=\frac X Y=\frac Z {1-Z}$ if $Z <\frac 1 2$ and $R=\frac {1-Z} Z$ otherwise. Can you now do the computations?.