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I pick a uniformly random point on a stick of length 1 and sunder the stick there. What is the cumulative distribution function for the ratio of the shorter and longer lengths?

If I let

$S \rightarrow $ shorter length
$L \rightarrow $ longer length
$R \rightarrow $ ratio of shorter and longer lengths ($\frac{S}{L}$)

then

$P(R \le k) = P(\frac{S}{L} \le k) = P(S \le Lk) = P(S \le k(1 - S)) = P(S \le k - kS) = P(S \le \frac{k}{k + 1})$

However, I am not sure where to go from here... any help would be appreciated!

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3 Answers

up vote 2 down vote accepted

Hint: First find the cumulative distribution function or density function of $S$. The rest is as in your approach.

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HINT: $$\Pr\left\lbrace S>c\right\rbrace=\Pr\left\lbrace \lvert X-0.5\rvert<\max\left(0.5 -c,0\right)\right\rbrace,$$ where $X$ is uniformly chosen from $\left[0,1\right]$.

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As $X$ increases from $0$ to $\frac{1}{2}$ to $1$, $R = \begin{cases} \frac{X}{1-X}, &0 \leq X \leq \frac{1}{2},\\ &\\ \frac{1-X}{X}, &\frac{1}{2} < X \leq 1,\end{cases}$ increases from $0$ at $X=0$, attains a maximum value of $1$ when $X = \frac{1}{2}$, and then decreases to $0$ when $X = 1$.

Thus, for $0 \leq \alpha \leq 1$, $$ 1-F_R(\alpha) = P\{R > \alpha\} = P\left\{\frac{\alpha}{1+\alpha} < X < \frac{1}{1+\alpha}\right\} = F_X\left(\frac{1}{1+\alpha}\right) - F_X\left(\frac{\alpha}{1+\alpha}\right). $$

Can you take it from there?

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