Question

A piece of lumber of fixed length ?? is broken into two sub pieces at position ?? ~ ??(0,??). Let ?? denote the length of the shorter sub piece.
a. Find the cumulative distribution function of ??. Hint: First express ?? as a piecewise function in terms of ??.
b. Find the probability distribution function of ??. Identify the name of the distribution that ?? follows. Also write down the parameter value(s) of this distribution.

Answer 1

Since $X\sim U\left ( 0,l \right )$ , then is defined as $Y=\left\{\begin{matrix} X, & 0<X\leqslant \frac{l}{2}\\ l-X, & \frac{l}{2}<X<l \end{matrix}\right.$

a) Now

For $0<X\leqslant \frac{l}{2}$

$P\left ( Y<y \right )=P\left ( X<y \right )\\ P\left ( Y<y \right )=\frac{y}{l},0<y<\frac{l}{2}$

For $\frac{l}{2}<X<l$

$P\left ( Y<y \right )=P\left ( l-X<y \right )\\ P\left ( Y<y \right )=P\left ( X>l-y \right )\\ P\left ( Y<y \right )=1-P\left ( X\leqslant l-y \right )\\ P\left ( Y<y \right )=1-\frac{l-y}{l}\\ P\left ( Y<y \right )=\frac{y}{l},0<y<\frac{l}{2}$

Thus, the CDF of is ${\color{Blue} F_Y\left ( y \right )=\frac{2y}{l},0<y<\frac{l}{2}}$

b) The pdf is

$f_Y\left ( y \right )=\frac{\mathrm{d} }{\mathrm{d} y} F_Y\left ( y \right )\\ {\color{Blue} f_Y\left ( y \right )=\frac{2}{l},0<y<\frac{l}{2}}$

The distribution of is uniform in $\left ( 0,\frac{l}{2} \right )$ . That is ${\color{Blue} Y\sim \left ( 0,\frac{l}{2} \right )}$