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?
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.