expected value of longest length of stick broken into 3 pieces

Question

I know this question has already been asked and answered here:Average length of the longest segment But I am trying to determine the answer in the following way. Let $X$ and $Y$ be two independent uniform variables on $[0,1]$ (the length of the stick is assumed to be equal to 1). If we suppose $Y>X$, then the three lengths of the stick are $X$, $Y-X$ and $1-Y$, so we just have to determine the cdf of $M=\max(X,Y-X,1-Y)$. Let $t\in[0,1]$, then: $$\mathbb{P}(M\leq t)=\mathbb{P}(X\leq t,Y-X\leq t,1-Y\leq t).$$ However, when I try to determine this probability graphically in the unit square, I find $0.5t^2+t-0.5$. I know this is a slightly different approach than the one used in the link above, as I am not going through the laws of $\min(X,Y)$ and $\max(X,Y)$, but rather considering the two symmetric cases where $X<Y$ and $X>Y$. Does anyone understand where I am making a mistake? Thanks.

Answer 1

By your method, you should be calculating :

$$\begin{align}\mathsf P(M\leq t, Y>X) ~=~& \mathsf P(X\leq t, Y-X\leq t, 1-Y\leq t, \color{blue}{Y>X})\\[1ex]~=~& \left(\int_0^1\int_0^1 \mathbf 1_{y-t\leq x\leq t\,,\, x<y\,,\, 1-t\leq y}\;\mathrm d\, x\;\mathrm d\,y\right)\mathbf 1_{0< t< 1}+\tfrac 1 2\mathbf 1_{1\leq t} \\[1ex] ~=~& \left(\int_{1-t}^1\int_{\max\{0,y-t\}}^{\min\{y,t\}}\;\mathrm d\, x\;\mathrm d\,y\right)\mathbf 1_{0< t< 1} +\tfrac 1 2\mathbf 1_{1\leq t}\\[1ex] =~& \left(\int_{\min\{t,1-t\}}^t\int_{0}^{y}\;\mathrm d\, x\;\mathrm d\,y +\int_{\max\{t,1-t\}}^1\int_{y-t}^t\;\mathrm d\, x\;\mathrm d\,y \right)\mathbf 1_{0< t< 1}+\tfrac 1 2\mathbf 1_{1\leq t}\\[1ex] =~& \left(\int_{\min\{t,1-t\}}^ty\;\mathrm d\,y +\int_{\max\{t,1-t\}}^1 2t-y\;\mathrm d\,y\right)\mathbf 1_{0< t< 1}+\tfrac 1 2\mathbf 1_{1\leq t} \\[1ex] =~& \left(\int_{1-t}^1 2t-y\;\mathrm d\,y \right)\mathbf 1_{0< t< 1/2}+ \left(\int_{1-t}^ty\;\mathrm d\,y +\int_{t}^1 2t-y\;\mathrm d\,y\right)\mathbf 1_{1/2\leq t< 1}+\tfrac 1 2\mathbf 1_{1\leq t} \\[1ex] =~& (\tfrac 5 2 t^2-t)\,\mathbf 1_{0 < t < 1/2}+ (-\tfrac 32 t^2+3t-1)\,\mathbf 1_{1/2\leq t < 1}+\tfrac 1 2\mathbf 1_{1\leq t} \\[4ex] \mathsf P(M\leq t) ~=~& (5t^2-2t)\,\mathbf 1_{0 < t < 1/2}+ (-3 t^2+6t-2)\,\mathbf 1_{1/2\leq t < 1}+\mathbf 1_{1\leq t}\end{align}$$

Answer 2

Note Graham Kemp's answer is wrong. For t=0.2, we get $P(M\lt t)$ negative... However, his method is correct. Below I give the answers if his method is applied properly.

First, by symmetry:

$(\leq) = 2(\leq, \gt )$

Then, we compute

$(\leq,\gt) = (\leq,−\leq,1−\leq,\gt)$ using integrals method which yields:

• $(\leq)= 0$ if $t\leq1/3$

• $(\leq) = 9t^2 -6t + 1$ if $1/3 \leq t\leq1/2$

• $(\leq) = -3t^2 +6t - 2$ if $1/2 \leq t\leq1$

Finally, integrate $tP'(M<=t)$ for $t \in [0,1]$, which yields 11/18