conditional probabilityprobabilityprobability distributionsstatistics
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.
My approach is maybe more naive than the others posted.
Break the unit interval at $x$ and $y$ where $x < y$. Our lengths are then $x$, $y - x$, and $1 - y$. It's not hard to show that they all have probability $1/3$ of being the shortest. In any case, our joint PDF is given by $f(x,y) = 6$ (since $x$ and $y$ remain uniform random variables on $1/6$th of the square $[0,1] \times [0,1]$). Each triangle in the diagram below corresponds to the domain of the PDF for one of the three cases.
I'll take care of the case when $x$ is shortest, that is, $x \leq y - x$ and $x \leq 1 - y$. This is the leftmost triangle. Since we're assuming $x$ is least, we are looking for $$E[x] = \int_0^{1/3} \int_{2x}^{1 - x} 6x \;dy \;dx = 1/9$$
The cases when $y - x$ and $1 - y$ are shortest are similar.
Imagine your rope as one unit long. Putting it on the number line, the endpoints will be at 0 and 1. Cutting at $1/4$, the ratio is $1:3$, and any cuts left of that also give smaller ratios. By symmetry, the cuts at greater or equal to $3/4$ also give $3:1$ ratios. Thus, we have two quarter length segments that work, in a segment of unit length, giving the probability $1/2$.
Best Answer
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