Randomly broken stick — probabilistic approach.

I've found interesting probabilistic excersise, it seemed easy, but somehow I'm stuck.

Let's take a rod/stick of unitary length($x \in [0; 1]$). Rod is broken into two pieces in random spot.

What is needed to be found is expected value and variance of the ratio of the shorter part to the longer one.

I tried using uniform distribution, but I failed. Looking for hints, but also for an answer.

• Notice that the ratio if broken at $x$ is the same as the ratio if broken at $1-x$. This should make your life easier – munchhausen Jun 26 '18 at 11:16

Hint.

You cut the stick according to a uniform distribution on $[0,1]$. Verify that the shortest part is uniformly distributed on $[0,\frac 1 2]$. Let $X\sim \mathcal U [0,\frac 1 2]$, then what you want to find is the expectation and the variance of the random variable $Y$ defined as: $$Y:=\frac{X} {1-X}$$ The law of unconscious statistician will help and the rest is just evaluating an integral.

• Is this $Y$ supposed to be $\frac{\min (X,1-X)}{\max (X,1-X)}$? – StubbornAtom Jun 26 '18 at 13:11
• @StubbornAtom yes, of course – Shashi Jun 26 '18 at 13:13

Hint. Let $Z=1$ if $X\ge1/2$, $Z=0$ otherwise. Let $Y$ be the desired variable (ratio of the shorter part to the longer one). Then (tower rule)

$$E[Y] = E[ E [ Y \mid Z]]=\frac12 E[Y \mid Z=0]+\frac12 E[Y \mid Z=1] \tag1$$

Because symmetry (this is not necessary, just to save some work), both terms in $(1)$ are equal, hence we can just compute $E[Y]=E[Y \mid Z=1]$

But, $X$ conditioned on $Z=1$ is uniform on $[\frac12,1]$. And $Y=\frac{1-X}{X}=\frac{1}{X}-1$

Then you just need to compute $E(1/X)$ for a uniform on $[\frac12,1]$, by a simple integral.

Computation of $E[Y^2]$, and hence variance, is similar.