# Problem Relating to Finding the Max of a Set of Ordered Random Variable

Consider the random variables $Y_1,Y_2,...,Y_n$ which are independent random variables such that $Y_i \sim U[0,\theta]$ if $i$ is even and $Y_i \sim U[1,2\theta]$ is $i$ is odd. I've been trying to figure out the probability distribution function for $$Y_{(n)} = \min\{Y_1,Y_2,...,Y_n\}$$ Like standard order statistics, I start out by calculating the CDF as $$f_{Y_{(n)}} = P(Y_{(n)} \leq y) = 1 - P(Y_{(n)} > y) = 1-P(Y_1 > y \wedge Y_2 > y \wedge ... \wedge Y_n > y)$$, which due to independence thankfully allows for a product to form: $$1-\prod_{i=1}^n P(Y_i > y).$$ However, I am having difficulty calculating this product given the different distributions among the random variables that I need to consider. Any suggestions on this computation?

Not sure if you consider the following floor notation as "computation" - you just merely need to count the number of odd and even numbers in the first $n$ natural numbers. The last result can be expressed as $$F_{Y_{(1)}}(y) = \begin{cases} 0 & y \in (-\infty, 0) \\ \displaystyle 1 - \left(1 - \frac {y} {\theta}\right)^{\lfloor \frac {n} {2} \rfloor} & y \in (0, 1) \\ \displaystyle 1 - \left(1 - \frac {y} {\theta}\right)^{\lfloor \frac {n} {2} \rfloor} \left(1 - \frac {y-1} {2\theta - 1}\right)^{\lfloor \frac {n+1} {2} \rfloor} & y \in (1, \theta) \\ 1 & y \in (\theta, +\infty) \end{cases}$$