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

  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
They are not independent. The maximum is always larger than the minimum ...
 
  • #3
1,789
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Yeah, nice observation. Thanks :smile:
 
  • #4
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Suppose I wanted to show it using the factorization of the joint pdf or joint pmf, how would I do that?
 
  • #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
statdad
Homework Helper
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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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