Jump to Problem 2,3

Problem 3

$ \color{blue}\text{Solution 1:} $

$ \color{blue}\text{Solution 2:} $

(a)
Since $ \mathbf{Y}_{n} = min \,\{{ \mathbf{X}_1, \mathbf{X}_2, \dots \mathbf{X}_n} \}, $
We can have the CDF of $ \mathbf{Y}_{n} $
$ P(\mathbf{Y}_{n} \leq y) = P(\,\{{ \mathbf{X}_1, \mathbf{X}_2, \dots \mathbf{X}_n} \} \leq y) $
$ = 1-P(\,\{{ \mathbf{X}_1, \mathbf{X}_2, \dots \mathbf{X}_n} \} > y) $
$ = 1-P(\mathbf{X}_1 > y)P(\mathbf{X}_2 > y) \dots P(\mathbf{X}_n > y) $
$ = 1-(1-F_{X_1}(y))(1-F_{X_2}(y)) \dots (1-F_{X_n}(y))=1-(1-F_{X_1}(y))^n $

Since $ \mathbf{Y}_{n} $ is also uniform distributed on the interval [0,1]
When $ y <0 $, $ P(\{\mathbf{Y}_{n}<0 \}) =0 $ since $ F_{X_1}(y) =0 $