Expectation problem from a question paper.

Question

Suppose that a point is chosen at random on a stick of unit length and that the stick is broken into two pieces at that point. Find the expected value of the length of the longer piece.

Answer 1

Let $X$ denote the coordinate of the dividing point in the interval $(0,1)$. Then, $X \sim\mathsf{Uniform}(0,1)$.

Define $Y = \max\{X,1 − X\}$ so that $$Y =\begin{cases} 1 − X, & 0 < X < \frac12\\ X, & \frac12 \le X < 1. \end{cases} $$ It then follows that possible values for $Y$ range between $1/2$ and 1. Computing the cdf for $Y$ we then have that, for $\frac12< y < 1$, $$ \begin{align} F_Y(y)&=\Bbb P(Y\le y)\\ &=\Bbb P\left(1 − X \le y,X < \frac12 \right)+\Bbb P\left(X \le y,X > \frac12 \right)\\ &=\Bbb P\left(1 -y \le X < \frac12 \right)+\Bbb P\left(\frac12<X\le y \right)\\ &=\frac12-(1-y)+y-\frac12 \end{align} $$ since $X \sim\mathsf{Uniform}(0,1)$ so that $$ F_Y(y)=\begin{cases}2y-1&\frac12 <y< 1\\ 0& \text{elsewhere} \end{cases} $$ Differentiating with respect to $y$ we then get that $$ f_Y(y)=\begin{cases}2&\frac12 <y< 1\\ 0& \text{elsewhere} \end{cases} $$ that is $Y \sim\mathsf{Uniform}(1/2,1)$. It then follows that $$ \Bbb E(Y)=\frac{1/2+1}{2}=\frac34 $$