Challenging advanced probability question

Let X1, X2,....Xn be a sequence of independent and identically distributed uniform random variables, with Xi ~ U[a,b].

A) What is the distribution function of M1 = min{Xi, 1 ≤ i ≤ n} ?

B) What is the probability density function of M1?

C) What is E(M1)?

A) We want P(M1 < c)
= P(min(X1, ..., Xn) < c)
= 1 - P(X1 > c, X2 > c, ..., Xn > c), via complements
= 1 - P(X1 > c) * P(X2 > c) * ... * P(Xn > c), since {Xi} are independent
= 1 - [P(X1 > c)]^n, since the Rv's are identically distributed
= 1 - [1 - P(X1 < c)]^n
= 1 - [1 - F(x)]^n, where F(x) is the distribution function of X1

In this case, U[a, b] has distribution function f(x) = (x-a)/(b-a).

Hence, the distribution function for M1 is
g(x) = 1 - [1 - (x-a)/(b-a)]^n
......= 1 - [(b-x)/(b-a)]^n for x > a (0 if x < a and 1 if x > b).
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B) The pdf is the derivative of the (c)df:
(d/dx) (1 - [(b-x)/(b-a)]^n) = n(b-x)^(n-1) / (b-a)^n for x in [a, b], and 0 otherwise.
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C) E(M1)
= ∫(x = a to b) x * (n(b-x)^(n-1) / (b-a)^n) dx
= (n/(b-a)^n) * ∫(x = a to b) x (b-x)^(n-1) dx
= (n/(b-a)^n) * ∫(u = b-a to 0) (b - u) u^(n-1) * -du, letting u = b - x
= (n/(b-a)^n) * ∫(u = 0 to b-a) (bu^(n-1) - u^n) du
= (n/(b-a)^n) * (bu^n/n - u^(n+1)/(n+1)) {for u = 0 to b-a}
= (n/(b-a)^n) * u^n (b/n - u/(n+1)) {for u = 0 to b-a}
= n (b/n - (b-a)/(n+1))
= a + (b-a)/(n+1).
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I hope this helps!