1
$\begingroup$

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.

  • 1
    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?2012-11-19
  • 1
    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.2012-11-19
  • 0
    @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).2012-11-19
  • 0
    that is indeed the answer.2012-11-20

1 Answers 1

2

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.

  • 0
    As I just told jay, I'm trying to convince myself about the min/max argument. Is seems odd. Did I get it?2012-11-19
  • 1
    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 $1$.2012-11-19
  • 0
    Works like a charm. Thanks :)2012-11-19