In question 9, how is the pdf of U calculated?-why are we considering 1-P(x>u) and in case of V, P(x<=v)?
In the case of U = min {X1, ....Xn}, then F(u) = P(U<=u) is the probability that the smallest value is less than or equal to u. We can't work out this directly because the smallest one might be less than u but what values do the others take? However, we can say that F(u) = P(U<=u)=1-P(U>u). If the minimum value is greater than u, then all the values must be greater than u, so P(U>u) is [P(Xi>u)]^n. The same argument then applies to V.
