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Let $X_1$ and $X_2$ be independent random variables with c.d.f. $F_{X_i}(x_i)$, $i = 1,2$. Find the c.d.f. of $U = \min(X_1, X_2)$ and $V = \max(X_1, X_2)$.

I'm stuck at this exercise for a while and even searching for similar questions I didn't find out exactly what I'm supposed to do. Those min and max burn my brain already.

Thanks in advance.

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I'll try to give you a hint with $U$. $P(U ≤ u) = 1 − P(X_1 > u, X_2 > u) = 1 − P(X_1 > u)P(X_2 > u)$ Can you solve now? – jay-sun Nov 19 '12 at 22:15
Basically, with $min$ and $max$ you need to answer the question - when is $X_1$ and $X_2$ greater than some value (and equivalently when are they less than some value) AT THE SAME TIME. – jay-sun Nov 19 '12 at 22:20
@jay: your hint took me to $F_{X_1}(u) + F_{X_2}(u) - F_{X_1}(u)F_{X_2}(u)$. Is that really it all? Hell, I'm confused by the min/max but I think I got your point (and André's). – Giuliano Nov 19 '12 at 22:32
that is indeed the answer. – jay-sun Nov 20 '12 at 0:51
up vote 2 down vote accepted

A start: Let random variable $S$ be the maximum of $X_1$ and $X_2$. Then $S\le s$ iff both $X_1$ and $X_2$ are $\le s$. By independence, this probability is $\Pr(X_1\le s)\Pr(X_2\le s)$. Thus $F_S(s)=\dots$.

Now you can do tackle the minimum. It is somewhat harder, but goes along similar lines.

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As I just told jay, I'm trying to convince myself about the min/max argument. Is seems odd. Did I get it? – Giuliano Nov 19 '12 at 22:35
Yes. Let $W$ be the minimum. Then the event $W\le w$ is the event $(X_1\le w)\cup (X_2\le w)$. Now use the familiar formula for $\Pr(A\cup B)$, plus independence. Or else the complementary event has probability $(1-F_{X_1}(w)) (1-F_{X_2}(w))$, subtract from $1$. – André Nicolas Nov 19 '12 at 22:53
Works like a charm. Thanks :) – Giuliano Nov 19 '12 at 22:58

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