Let $\displaystyle Y_1, ... , Y_n$ be the order statistics of a random sample of size $\displaystyle n$ from a uniform distribution on the interval (0,2).

What is $\displaystyle P[Y_1 < \frac{1}{2} < Y_n]$

If $\displaystyle U_1, ... , U_n$ are the uniform RVs from the sample then

$\displaystyle Y_1 = min(U_1, ... , U_n)$

$\displaystyle P(Y_1 \leq y) = 1 - P(Y > y) = 1 - [P(U_1 > y) * ... * P(U_n > y)] = 1 - [1 - \frac{y}{2}]^n$

Thus $\displaystyle P(Y_1 \leq \frac{1}{2}) = 1 - \frac{3^n}{4^n}$

$\displaystyle Y_n = max(U_1,...,U_n) $

$\displaystyle P(Y_n \leq y) = P(U_1 \leq y)*...*P(U_n \leq y) = (\frac{y}{2} )^n$

$\displaystyle P(Y_n > y) = 1 - (\frac{y}{2} ) ^ n$

Thus $\displaystyle P(Y_n > \frac{1}{2}) = 1 - \frac{1}{4^n}$

However, I am not sure how to use these probabilites to find the interval I'm interested in $\displaystyle [Y_1 < \frac{1}{2} < Y_n]$

(The end points are not real numbers, but RVs. I tried subtraction, but I don't think thats right. Is there a different way to think of this interval?)

Thank you for your assistance.