2
$\begingroup$

The problem is the same as here.

A stick of 1m is divided into three pieces by two random points. Find the average length of the largest segment.

I tried solving it in a different way, and the logic seems fine, however I get a different result to $\frac{11}{18}$.

Here is my solution. Please let me know what I did wrong.

Let $X$ be the length of the stick from the beginning to the first cut. $Y$ be the length of the stick between the first and second cut and $1-X-Y$ the length between the second cut and the end of the stick.

We want to find the CDF of the following random variable: $Z=\max(X,Y,1-X-Y)$. (I believe that if anything is wrong, this might be it).

$$\begin{split} F_Z(z) = P(Z\leq z) & = P(\max(X,Y,1-X-Y) \leq z)\\ & = P(X\leq z, Y\leq z, 1-X-Y\leq z)\\ &= P(1-Y-z\leq X \leq z, Y\leq z) \end{split} $$

Since we have $1-Y-z\leq z$ we deduce that $Y\geq 1-2z$. Hence: $$\begin{split} F_Z(z) &= \int_{1-2z}^z\int_{1-y-z}^z 1 dx dy = \int_{1-2z}^z (z-1+y+z) dy\\ &= (2z-1)(z-1+2z) + \left. \frac{y^2}{2}\right|_{y=1-2z}^{y=z} \\ &=(2z-1)(3z-1) + \frac{1}{2}(z^2- (2z-1)^2) \\ & = (2z-1)(3z-1) +\frac{1}{2}(-3z^2 + 4z -1) \\ & = \frac{1}{2}(3z-1)^2 \end{split} $$ Now, the pdf of $Z$ is : $$f_Z(z) = \frac{d}{dz}F_Z(z) = 9z-3 $$

And now, in order to find the expected value of the largest length, we need to integrate over $(\frac{1}{3},1)$ as the largest piece needs to be greater than $\frac{1}{3}$. Hence

$$\begin{split} E[Z] = \int_{\frac{1}{3}}^{1} z f_Z(z) dz = \int_{\frac{1}{3}}^{1} z (9z-3) dz = \frac{14}{9} \end{split} $$ The result is obviously wrong as it needs to be something between $0$ and $1$, however after going over the solution multiple times, and checking the calculations with Wolfram, I cannot seem to figure out what went wrong.

$\endgroup$
  • $\begingroup$ Offhand, it looks like your setup allows $1-X-Y$ to be negative. $\endgroup$ – Barry Cipra Jun 14 '18 at 15:54
  • 2
    $\begingroup$ What are $X,Y$ supposed to represent? If they're the lengths of two of the parts, those are not uniformly distributed. I would set $X,Y$ to the coordinates of the two points, and then find the expectation of $\max(\min(X,Y), |X-Y|, 1-\max(X,Y))$. $\endgroup$ – Daniel Schepler Jun 14 '18 at 15:54
  • $\begingroup$ Note that $F_Z(1)=\frac 12\times 4=2>1$. For that matter $F_Z(0)=\frac 12$ which is already absurd. As a suggestion, walk through the calculation of $F_Z(0)$ to see where you go awry. $\endgroup$ – lulu Jun 14 '18 at 15:55
  • $\begingroup$ @DanielSchepler $X$ is the length of the segment from $0$ to the first cut, and $Y$ is the length from the first cut to the second cut. $\endgroup$ – Andrei Crisan Jun 14 '18 at 15:58
  • $\begingroup$ @lulu we must have $z\geq \frac{1}{3}$, as it is the largest part $\endgroup$ – Andrei Crisan Jun 14 '18 at 16:01
0
$\begingroup$

This is just a suggestion not an answer but could you try solving it keeping the distance of first division as x and second as y making the lengths of the segments as x, y-x ,1-y

$\endgroup$
  • $\begingroup$ This makes a lot of sense. $\endgroup$ – Andrei Crisan Jun 14 '18 at 15:57
0
$\begingroup$

The first integral is wrong, because it assumes that $X,Y$ are uniform and independent on $[0,1]^2$. They are not (for one thing, $X \le Y$).

$\endgroup$
0
$\begingroup$

Here is how I would do it.

Lets define $x$ to be the short stick, $y$ to be the medium stick and $z$ to be the long stick.

$x\le y\le z\\ z = 1-x-y\\ x\le y \le \frac {1-x}{2}\\ x\le \frac 13$

$$ \bar z = \frac {\displaystyle\int_0^\frac 13\int_x^{\frac {1-x}{2}} 1-x-y\ dy\ dx}{\displaystyle\int_0^\frac 13\int_x^{\frac {1-x}{2}} 1\ dy\ dx}$$

$\endgroup$
  • $\begingroup$ Would you care to explain the numerator and denominator of your fraction please? $\endgroup$ – Andrei Crisan Jun 15 '18 at 10:07
  • $\begingroup$ I understand the surface over which you are integrating, however I do not get the existence of the $dz$ terms, given that you only have a double integral. The bottom would simply be the CDF of the $\min(X,Y)$? $\endgroup$ – Andrei Crisan Jun 15 '18 at 10:13
  • $\begingroup$ Thanks, only integrating over 2 dimensions (and fixed above). $z = 1-x-y$ The numerator gives the average value of z per unit of surface area, and the denominator is the surface such that $x<y<z.$ $\endgroup$ – Doug M Jun 15 '18 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.