A stick of length $L$ is broken uniformly at random. Let $X$ denote the length of the smaller stick. What is the CDF $F_X(s)$ of $X$?
my solution:
Let $Y$ be a random variable denoting the spot where the stick is broken, then $Y ~ Unif[0, L]$. Furthermore, $X = g(Y) = min(Y, L-Y)$ – not sure if this helps with the CDF though.
Now I am struggling to find the CDF. Clearly, for $s > \frac L 2$, $F_X(s) = 1$, and $F_X(s) = 0$ for $s < 0$. But for $F_X(s) = P(X \leq s), 0 \leq s \leq \frac L 2$, I'm not sure how to write the probability.
$P(X \leq s) = P(min(Y, L-Y) \leq s)$ but then not sure how to proceed with the $min$ any hints would be appreciated!
Best Answer
Here we have $X = \left\{\begin{array}{lr} Y, & Y \leq \frac{L}{2}\\ L-Y, & \text{otherwise}\\ \end{array}\right\}$, where $Y \sim U(0,L)$
$P(X \leq s) = P(min(Y,L-Y)\leq s)$
$=P(Y \leq s) + P(L-Y \leq s)$, since they correspond to 2 disjoint events, namely, $Y \leq \frac{L}{2}$ and $Y > \frac{L}{2}$
$=\frac{s}{L}+1-\frac{L-s}{L}$, since $P(L-Y \leq s)=P(Y\geq L-s)=1-P(Y \leq L-s)$
$=\frac{s}{L}+1-1+\frac{s}{L}$
$=\frac{2s}{L}$
Thinking in another way, $X$ is smaller part of the stick implies $X \sim U(0,\frac{L}{2})$
Hence, $P(X\leq s)=\frac{s}{L/2}=\frac{2s}{L}$