A stick of length L is broken at a uniformly chosen random point. Let $X$ be the length of the smaller piece. Find the density of $X$.


Notice that the probability that the stick is broken a certain is equally likely. So, we have that smaller piece is either $[0,x]$ or $[x,L]$

$$ P(X \leq x) = \frac{ \text{length from origin to x}+\text{length from x to L}}{\text{total length} }= \frac{x + |x-L|}{L}$$

But, this leads anywhere. Instead I was thinking that the only way the $X$ is the smaller piece is if $X$ is less than $L/2$ thus $X$ is uniform on $L/2$ and s o

$$ f_X(x) = \frac{2}{L} $$

Is this correct?

  • $\begingroup$ $Y \sim U(0,L)$ Find $Y=min(X,L-X)$ $\endgroup$ – Daman deep Dec 15 '18 at 5:03
  • $\begingroup$ "But, this leads anywhere" This very much leads to the result, only one should be more careful: for $x$ in $(0,L/2)$, the event $\{X<x\}$ corresponds to the location of the break being in $[0,x]$ or in $[L-x,L]$, thus, $$P(X<x)=\frac{x+(L-(L-x))}L=\frac{2x}L$$ and you are done. $\endgroup$ – Did Dec 15 '18 at 6:37

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