Break a stick into two pieces, and break the shorter one again, what's the pdf of the shortest pieces.

Question

Break a stick of unit length at a uniformly chosen random point. Then take the shorter of the two pieces and break it into two again at a uniformly chosen random point. Let X denote the shortest of the final three pieces. Find the probability density function of X.

I let the shorter one on the first break be x, which should be between (0, $1/2$), so the pdf should be $$f(x)=2, 0<y<\frac{1}{2}$$ then the conditional probability of $f(x|y)$ should be (since it is another uniform distribution) $$f(x|y)=\frac{1}{y/2-0}=\frac{2}{y}$$ by applying conditional probability formular $$f_X(x)=\int_0^{1/2}f_{X|Y}(x|y)f_Y{y}dy=\int_0^{1/2}\frac{2}{y}\times2dy=4ln(y)|^{1/2}_0$$

I stopped here since it is impossible. Can someone help?

Answer 1

The last integral should be taken from $x$ to $1/2$, since $f(x \mid y)$ is zero when $x > y$.

Answer 2

$$p(y) = 8 - 32 y\quad {\rm for}\ 0\leq y \leq 1/4$$

and $0$ otherwise.

After the first break the short leg is uniformly likely between $0 \leftrightarrow 1/2$, as shown by the range of the abscissa in the graph.

After this short piece is broken, it longer remaining part is uniformly distributed in $1/4 \leftrightarrow 1/2$. The other (shorter) part obeys a triangle distribution between $0 \leftrightarrow 1/4$. So just write down a triangle distribution that goes between $0 \leftrightarrow 1/4$.