I am currently struggling with the fact that the answer in the back of the text states this probability is $\frac{5}{16}$, but I got $\frac{1}{2}$ using a transformation method and using the cdf method for the order statistics $Y_1$ and $Y_n$.
CDF Method:
The joint pdf of $Y_1$ and $Y_4$ is
$g_{1,4}(y_1,y_4) = \frac{4!}{(1-1)!(4-1-1)!(4-4)!}(F_{X}(y_4)-F_{X}(y_1))^{4-1-1} = 12(y_4-y_1)^2.$
Then the probability is:
$P(Range\{X_i\} < \frac{1}{2}) = P(Y_4 – Y_1 < \frac{1}{2}) = P(Y_1 < Y_4 < Y_1+\frac{1}{2}) = \int_{0}^{1}\int_{y_1}^{y_1+\frac{1}{2}} 12(y_4-y_1)^2 dy_4dy_1 = 12\int_{0}^{1} \int_{y_1}^{y_1+\frac{1}{2}} y_4^2-2y_4y_1+y_1^2 dy_4dy_1 =12 \int_{0}^{1} \left[\frac{y_4^3}{3} – y_4^2y_1 + y_4y_1^2\right] \Bigg{|}_{y_1}^{y_1+1/2} dy_1 = 12\int_{0}^{1}\frac{1}{3}\left(y_1+\frac{1}{2}\right)^2-\left(y_1+\frac{1}{2}\right)^2y_1+\left(y_1+\frac{1}{2}\right)y_1^2-\frac{1}{3}y_1^3+y_1^3-y_1^3 dy_1 = \ldots = \frac{1}{2}$
Best Answer
It should be \begin{align*} &P(Y_1 < Y_4 < \min\Bigl(Y_1+{\small{\frac{1}{2}}},1\Bigr)\\[4pt] =& \int_{0}^{\frac{1}{2}}\int_{y_1}^{y_1+\frac{1}{2}} 12(y_4-y_1)^2 dy_4dy_1 \;+\; \int_{\frac{1}{2}}^{1}\int_{y_1}^1 12(y_4-y_1)^2 dy_4dy_1 \\[4pt] =& \frac{1}{4}+\frac{1}{16} \\[4pt] =& \frac{5}{16} \\[4pt] \end{align*}