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Are max and min of n iid r.v.s. independent?

  1. Mar 31, 2008 #1
    Hi

    Suppose [itex]X_{1}, \ldots, X_{n}[/itex] is a sequence of i.i.d. random variables. We define

    [tex]X_{(n)} = max(X_{1}, \ldots, X_{n})[/tex]
    [tex]X_{(1)} = min(X_{1}, \ldots, X_{n})[/tex]

    Are [itex]X_{(n)}[/itex] and [itex]X_{(1)}[/itex] independent?

    Whats the best/easiest way to verify this?

    Thanks
    Vivek
     
  2. jcsd
  3. Mar 31, 2008 #2
    They are not independent. The maximum is always larger than the minimum ...
     
  4. Mar 31, 2008 #3
    Yeah, nice observation. Thanks :smile:
     
  5. Mar 31, 2008 #4
    Suppose I wanted to show it using the factorization of the joint pdf or joint pmf, how would I do that?
     
  6. Mar 31, 2008 #5
    You just have to find one example such that the cdf does not factorize.

    Let m be the minimum, M the maximum, x some real number

    What about P(m<x && M<x)

    This is equal to only M being less than x ( because then m is automatically also less than x.

    so P(m<x && M<x) = P(M<x)

    For this to be equal to the factorized probability P(m<x)P(M<x) you need to have P(m<x)=1 for all real x ...which is not true:smile:
     
  7. Apr 1, 2008 #6
    Thanks Pere :smile:
     
  8. Jul 25, 2008 #7

    statdad

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    Homework Helper

    Look for a statistics text that discusses the distributions of order statistics and sets of order statistics. You will be able to find a general formula for the p.d.f of the [tex] \min \text{ and } \max [/tex] in terms of the marginal pdfs and joint pdf of the sample. Once you see that form, you will see that they need not be independent.
     
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