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Let $X_1, X_2, ..., X_n$ be independent continuous random variables each with cumulative distribution function $F$. Show that the joint cdf of $X_{(1)}$ and $X_{(n)}$ is

$$ F(x, y) = F^n(y) - [F(y) - F(x)]^n, \quad x \leq y $$

My attempt: $$ F_{X_{(1)}}(x) = 1 - [1 - F(x)]^n , \quad F_{X_{(n)}}(y) = F^n(y) $$

$$ F(x, y) = F_{X_{(1)}}(x)F_{X_{(n)}}(y) = \{1 - [1 - F(x)]^n\}[F^n(y)] $$

$$ = F^n(y) - F^n(y)[1 - F(x)]^n$$

The only problem is that $$F^n(y)[1 - F(x)]^n \neq [F(y) - F(x)]^n.$$

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    By multiplying $F_{X_(1)}(x)$ and $F_{X_(n)}(y)$ you're assuming that they are independent which is clearly not true. Try to start with the probability $P\{X_{(1)}>x,X_{(n)}\le y\}$ for $x\le y$. – 2017-01-08
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    Ahh I got it now, $P(X_{(1)} \leq x, X_{(n)} \leq y) = P(X_{(n)} \leq y) - P(X_{(1)} > x, X_{(n)} \leq y) = [P(X \leq y)]^n - [P(x < X \leq y)]^n = [P(X \leq y)]^n - [P(X \leq y) - P(X \leq x)]^n = F^n(y) - [F(y) - F(x)]^n$. Thanks for the help! – 2017-01-08

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