If $X_1,X_2\sim U(0,1)$ and $Y$ is one of them who is closest to an end point, find distribution of $Y$. Let $X_1$ and $X_2$ be independent, $U (0, 1)$-distributed random variables, and
let $Y$ denote the point that is closest to an endpoint. Determine the distribution of $Y$.
It's a question in chapter of "order statistic",I'm thinking about solving it without using order statistics, cuz I have no idea how to solve it in order statistics way at all. For the normal way, $$P(Y<y)=1-P(Y>y)=1-P(y<X_1<1-y,y<X_2<1-y)=1-P(y<X_1<1-y)P(y<X_2<1-y)=1-(1-2y)^2$$ so $f(y)=4-8y$, when $y<1/2$, and the way when $y>1/2$ is similar. 
But the correct answer is $f(y) = 2 − 4y$ for $0 < y < 1$ , $4y − 2$ for $1 < y < 1$. 
Dose anybody can tell me how to solve this question by both order statistics way and the normal way? Thanks a lot.

We derive the density function directly, in a way analogous to the way one finds the distribution of order statistics. The only interesting value of $y$ are between $0$ and $1$. We find $f_Y(y)$ for $0\lt y\lt \frac{1}{2}$. 
The probability that $y$ is between $y$ and $y+dy$, for "small" $dy$, is approximately $f_Y(y)\,dy$. Or, if one feels like being more precise, the probability that $Y$ lies between $y$ and $y+h$ is $f_Y(y)h +o(h)$. 
In order for $Y$ to lie in this interval, we want either (i) $X_1$ lies in this interval, and $X_2$ lies roughly in the interval $(y,1-y)$, or (ii) $X_2$ lies in this interval, and $X_1$ lies roughly in the interval $(y,1-y)$. The events (i) and (ii) are equiprobable, and disjoint. We find the probability of (i). 
The probability that $X_1$ lies in the interval from $y$ to $y+dy$ is $dy$. Given this, the probability that $X_2$ lies in the interval $(y,1-y)$ is $1-2y$. Thus the probability of (i) is $\approx (1-2y)\,dy$. Taking (ii) into account, we find that the probability $Y$ lies in the interval is $\approx (2-4y)\,dy$. This is mildly inexact, there is a "second order infinitesimal" involved. 
It follows that
$$f_Y(y)\,dy\approx (2-4y)\,dy,$$
and the result follows.
Remark: The assertion that (for $y\lt \frac{1}{2}$) we have $\Pr(Y\gt y)$ is $\Pr(y\lt X_1\lt 1-y)\Pr(y\lt X_2\lt 1-y)$ was not correct. Roughly speaking, this is because we can have $Y\le y$ in many ways. Let $t\le y$. If $t$ is equal to $y$, or close, and $X_1=t$, we do indeed want $y\lt X_2\lt 1-y$. But if $t$ is smaller, say $y/2$, and $X_1=y/2$, our condition on $X_2$ changes, it can be allowed to roam up to $1-y/2$.   
