Distribution of a point farther away from a certain point Related to this question, 
Suppose that two points are independently distributed according to some CDF $F$ (and pdf $f$) over $[0,1]$. 
What would be the distribution of the point farther away from a certain point $a$?
If I follow the derivation of the distribution of the closest point, let $f(x|a)$ denote the distribution of the point which is farther than the other point from $a$.
We have $$P[\textrm{the other point is inside of the interval }[a-|a-x|,a+|a-x|]]\\=F[a+|a-x|]-F[a-|a-x|].$$
There are two possible cases: point 1 is closer or point 2 is closer, so we multiply the probability by 2. 
So, the distribution of the point farther away from $a$ is given by 
$$2(F[a+|a-x|]-F[a-|a-x|])f(x).$$
I'm wondering whether this is a right derivation of the distribution or not.. I mechanically followed the derivation which is shown in the above link, but I don't understand why I need to multiply the probability to the baseline pdf $f$. Can anyone confirm the answer or explain the behind idea of the multiplication? 
 A: It will become easier to understand when you master the common technique in deriving the joint pdf of order statistics. Let me call this a multinomial argument.
Some common results can be founded here:
https://en.wikipedia.org/wiki/Order_statistic#The_joint_distribution_of_the_order_statistics_of_an_absolutely_continuous_distribution
Now in your question, 
Let $U_1, U_2$ be the original points which is unordered, and thus i.i.d. Assume they are absolutely continuous with common pdf $f_U$ and CDF $F_U$. Let $X$ be the point which is farther than the other point from $a$ and we want to derive its pdf $f_X$. 
Consider the realization of $X$ being $x$. Next we partition the whole real line into $4$ sets accordingly:
The first set contains all the points which is closer to $a$ than $x$, that is $(a - |a - x|, a + |a - x|)$. And 
$$p_1 \triangleq \Pr\{U \in (a - |a - x|, a + |a - x|)\} = F_U(a + |a - x|) - F_U(a - |a - x|)$$
The second set is the point $x$. Since $U$ is absolutely continuous, the probability that it fall into an infinitesimal small interval $(x, x+dx)$ is $p_2 \triangleq f_U(x)dx$, and the probability that more than one point fall into the identical point is $0$.
The third set is $\{a - |a - x|\} \cup \{a + |a - x|\} \backslash \{x\}$, the point of reflection which has the identical distance to $a$ as $x$. Similarly the probability is $p_3 \triangleq f_U(x^*)dx$
The fourth set is the complement, which contains all the points which farther away:
$(-\infty, a - |a - x|) \cup (a + |a - x|, +\infty)$. And
$$ \begin{align} p_4 
&\triangleq \Pr\{U \in (-\infty, a - |a - x|) \cup (a + |a - x|, +\infty)\} \\
&= 1 - F_U(a + |a - x|) + F_U(a - |a - x|) \end{align} $$
By multinomial argument, the number of points fall into these $4$ sets jointly follows a multinomial distribution. For $X$ to be the farther point, we require exactly $(1, 1, 0, 0)$ of points in the $4$ sets respectively. The probability is 
$$ \frac {2!} {1!1!0!0!}p_1^1p_2^1p_3^0p_4^0 = 2[F_U(a + |a - x|) - F_U(a - |a - x|)]f_U(x)dx $$
Note that as $X$ is also continuous, $\Pr\{X \in (x, x+dx)\} = f_X(x)dx$ and this is equivalent to the above probability. Comparing both sides, by eliminating the $dx$ we obtain the desired result.
The good point of this argument is that it can easily extend to cases with general sample size $n$, and you can derive the joint pdf in general.
