Cut a rope with unit length into two pieces randomly. Cut the longer piece of the first cut into two randomly again. Take the shorter piece from that second cut. What would be the PDF and the expected value of this piece?

I managed to drive the first longer piece's PDF as $2$ and CDF as $2x-1$ for values between $0.5$ and $1$. But I can't drive the PDFs and CDFs for the second cut since it should depend on the first cut.


Let $X$ be the length of the longer of the two pieces. As you say, $X$ is uniformly distributed on $\left[\frac12,1\right]$ with density $2$

Let $Y$ be the length of the shorter part of $X$. If $X=x$ then $Y$ is uniformly distributed on $\left[0,\frac{x}2\right]$ with density $\frac2{x}$ when $0 \le y \le \frac{x}{2}$, and $\mathbb E[Y \mid X=x] = \frac{x}{4}$

So we can say the marginal density for $Y$ on $\left[0,\frac{1}2\right]$ is $$f(y)=\int\limits_{\mathbb R}\frac2{x} \mathbb I\left[0 \le y \le \frac{x}{2}\right] 2 \mathbb I\left[\frac12 \le x \le 1\right]\,dx= \int\limits_{x=\max(2y,1/2)}^{1}\frac4{x} \,dx$$ using indicator functions, and this is $4{\log_e(2)}$ when $0 \le y \le \frac{1}{4}$ and is $-4{\log_e(2y)}$ when $\frac14 \le y \le \frac{1}{2}$

Similarly $$\mathbb E[Y] = \int\limits_{\mathbb R}\frac{x}4 2 \mathbb I\left[\frac12 \le x \le 1\right]\,dx=\int\limits_{x=1/2}^{1}\frac{x}2\,dx= \frac3{16}$$ and as a check $\int\limits_{y=0}^{1/2} y\, f(y) \, dy$ gives the same result

  • $\begingroup$ Thanks! Now I understand. Thanks for explaining clearly. $\endgroup$ – gizmos Apr 20 at 20:53

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