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Order Statistics Probabilities

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Let Xi ~ iid f(x) = (2x)I[0,1](x), i = 1,....,n.

    Find the distribution of X(1). What is the probability that the smallest one exceeds .2?
  2. jcsd
  3. Apr 24, 2012 #2

    Ray Vickson

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    What work have you done on this so far? You need to show an attempt to solve this yourself before we can help.

  4. Apr 25, 2012 #3
    FX(1)(x) = P(X(1) <= x)
    = P(X1,....,Xn <= X)
    =1-P(X1,...,Xn > X)
    =1-P(X1>X)^n since the xi's are iid.
    =1-[1-P(X1 <= X)]^n
    =1-[1-∫ from 0 to x (2tdt)]^n

    For the probability, P(X1>.2) = 1-Fx(1) (0.2) = (1-(0.2)^2)^n = (0.96)^n
  5. Apr 25, 2012 #4

    Ray Vickson

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    I hope you were just being sloppy and don't really believe that
    [tex] P(X_1, X_2, \ldots, X_n \leq x) = 1 - P(X_1, X_2, \ldots, X_n > x) \, \leftarrow \text{FALSE!}[/tex] because that is not valid. For two events A and B we have P(A) = 1 - P(B) if the events A and B are complementary (that is, have no points in common and together cover the entire sample sample space). Do the events [itex] \{X_1, \ldots,X_n \leq x \}[/itex] and [itex] \{X_1,\dots,X_n > x\}[/itex] look complementary to you? (Try drawing these for n = 2.) Of course, what is true is that [itex] P(X_i > x) = 1-P(X_i \leq x)[/itex] for each [itex] X_i[/itex] separately. You need to decide whether or not we have
    [tex] P(X_1,\ldots,X_n > x) = (1-x^2)^n \text{ or } P(X_1,\ldots,X_n > x) = 1-(1-x^2)^n \,.[/tex]

    Last edited: Apr 25, 2012
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