Given a stick $L$ broken into 2 with UNI$(0,L)$.
I am trying to find the cdf and pdf of $Y$ which denotes the area of rectangle created from the two lengths.
( one side = $L-p$ other is $p$)
Now, I am confused about how to apply it to the area , as I think the cdf for just the stick length would be close to this:
let $$X: \min(p,L-p)\implies(X \leq y)$$
If $$y \geq \frac{L}{2} \implies P(X \leq y)=1$$
$$y\leq 0 \implies P(X \leq y) = 0$$
We conclude:
$$0 \leq y \leq \frac{L}{2} \implies P(X \leq y)= \frac{2y}{L}$$
$$ (X \leq y)= (p \leq y) \cup {p \in [L-y, L]}$$
Am I on the right path or should I form a rectangle and try work from scratch from there?
Best Answer
What you need is the probability distribution of a function of a random variable.
The process is to determine this function and its inverse. E.g. since $f$ is uniform on $[0,1]$, the cdf is $F(X)=X$; $0\leq x\leq 1$. Then $Y = h(x) = (1-x)(x)$ the area. Then determine the inverse of $h(x)$, call it $v(y)$ (simple algebra). Then $$F(y) = P(Y < y) = P(x-x^2 < y) = P(x < v(y)) = F(v(y)).$$
See Functions of one random variable