Lengths of randomly broken rods A continuous random variable $X$ has probability density function $f(x)$. The probability that $X \leq x$ is given by the function $F(x)$.
i) Explain why $F'(x) = f(x)$.
ii) A rod of length $2a$ is broken into two parts at a point whose position is random. State the form of the probability distribution of the length of the smaller part.
iii) Find the mean value of this length.
iv) Two equal rods, each of length $2a$, are broken into two parts at points whose positions are random. $X$ is the length of the shortest of the four parts thus obtained. Find the probability, $F(x)$, that $X \leq x$, where $0 < x \leq a$.
v) Hence or otherwise, show that the probability density function of $X$ is given by
$$f(x) = \begin{cases} (2(a-x))/a^2 & 0 < x \leq a, \\ 0 & x \leq 0 \text{ or } x > a. \end{cases}$$
vi) Show that the mean value of $X$ is $a/3$.
vii) Find the mean value of the sum of the two smaller parts and hence show the mean values of the four parts are in the proportions $1:2:4:5$.
 A: Let $U = U(0,2a)$ be a uniform distribution supported on the interval $(0,2a)$.  
For (ii), let $Y$ be the length of the shorter part of the broken rod.  Then $Y = \min\{U, 2a - U\}$.  So, for $0 < x \leq a$,
\begin{align}
\mathbb{P}(Y \leq x) &= \mathbb{P}(\min\{U, 2a - U\} \leq x) = \mathbb{P}(U \leq x \text{ or } 2a - U \leq x)\\
&= \mathbb{P}(U \leq x) + \mathbb{P}(2a - U \leq x) - \mathbb{P}(2a - x \leq U \leq x) \\
&= \frac{x}{2a} + \frac{x}{2a} - 0 = \frac{x}{a}.
\end{align}
So $Y \sim U(0,a)$ is uniformly distributed on $(0,a)$.  
Part (iii) should be standard given part (ii), since if $f_Y(x)$ is the pdf for $Y$, then $$\mathbb{E}Y = \int_0^a x f_Y(x) \, dx,$$ and I'll leave the details to you.  
For (iv), let $Y_1$ be the shorter part of the first broken rod and let $Y_2$ be the shorter part of the second broken rod.  Then $Y_1$ and $Y_2$ are iid with distribution $Y$, and $X = \min\{Y_1, Y_2\}$.  So,
\begin{align*}
\mathbb{P}(X \leq x) &= \mathbb{P}(Y_1 \leq x \text{ or } Y_2 \leq x) \\
&= \mathbb{P}(Y_1 \leq x) + \mathbb{P}(Y_2 \leq x) - \mathbb{P}(Y_1 \leq x) \mathbb{P}(Y_2 \leq x) \\
&= \frac{x}{a} + \frac{x}{a} - (\frac{x}{a})^2 = \frac{2ax-x^2}{a^2}.
\end{align*}
Using part (i), you should be able to derive the stated equation.  I'll again leave part (v) to you.  
Finally, for (vi), use the fact that expectation is additive, so that $\mathbb{E}(Y_1 + Y_2) = \mathbb{E}Y_1 + \mathbb{E}Y_2$, and proceed.  
