Given $X_1, X_2,... ,X_n$ be independent variables, each having a uniform distribution over (0,1). Let M = maximum $(X_1,X_2,...,X_n)$. Show that the distribution function of M is given by $$F_M (x) = x^n, 0 \leq x \leq 1$$ What is the probability density function of M?
Solution: Having $X_1, X_2,... ,X_n$ be independent variables, each having a uniform distribution over (0,1) implies $P(X_i < x) = x$ Why this implication?
$$P(X_1 < x, X_2 < x, X_3 < x,...) = P(X_1 < x ) P(X_2 < x)P(X_3 < x)...$$ Let M = maximum $(X_1,X_2,...,X_n)$. To find the distributions function of M calculate P(M < x)
$$F_M (x) = P(M < x )$$ $$ = P(maximum(X_1,X_2,...,X_n) < x )$$ $$ = P(X_1 < x, X_2 < x,...,X_n < x )$$ $$ = P(X_1 < x)P( X_2 < x)...P(X_n < x )$$ $$ = x.x...x (n times)$$ $$ = x^n$$
Since, $X_i$ lies between 0 and 1, and maximum of $X_i'$s lies between 0 and 1. How do you know that the maximum also lies between 0 and 1? So, $$F_M (x) = x^n, 0 \leq x \leq 1$$
The probability density function of M is $f_M(x)$. $$f_M(x) = \dfrac{d}{dx}(F_M(x))$$ $$= \dfrac{d}{dx}(x^n)$$ $$= nx^n-1$$
Why do we derivate $F_M(x)$ to get the probability density function of M?
For question 1, by definition, the maximum of a discrete finite set must be one of the members of the set. Since all members are between 0 and 1, so must the maximum lie in that interval.
For question 2, as you noted $F_M(x)$ was relatively easy to derive. Deriving $f_M(x)$ directly is much more difficult.