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Help with a cdf question

  1. Oct 20, 2010 #1
    X1,X2,...,Xn is a sample from a distribution with cdf F. Show
    FX(i)(x)= [tex]\sum[/tex][tex]\stackrel{n}{j=i}[/tex] ([tex]\stackrel{n}{j}[/tex]) Fj(x)(1-F(x))(n-j)


    How would I start on this?
     
    Last edited: Oct 20, 2010
  2. jcsd
  3. Oct 21, 2010 #2
    Fx(n)(x) = Px(n)(X(n)<=x)
    Fx(n)(x) = Px1...xn(max{x1,...xn}<=x)
    Fx(n)(x) = Px1...xn(x1<=x,...,xn<=x)
    Fx(n)(x) = Px1(x1<=x)*...*Pxn(xn<=x)
    Fx(n)(x) = Fx1(x)*...*Fxn(x)
    Fx(n)(x) = Fx1(x)*...*Fxi(x)*...*Fxn(x)
    Fx(n)(x) = Fx1(x)*...*Fx(n-1)(x)*Fxi(x)
    Fxi(x) = Fx(n)(x) / Fx1(x)*...*Fx(n-1)(x)
    Fxi(x) = (Fx(1)(x))n / (Fx(1)(x))(n-1)
    Fxi(x) = Fx(1)(x)

    not sure where to go from here...
     
    Last edited: Oct 21, 2010
  4. Oct 21, 2010 #3

    vela

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    You might get more responses if you say what the notation X(n) means. Others might be familiar with the notation, but I have no clue what the problem is about.
     
  5. Oct 21, 2010 #4
    example of what X(n) means

    X1=0.5 X(1) = 0.1 (so it is increasing...)
    X2=0.7 X(2) = 0.2
    x3=0.96 X(3) = 0.45
    x4=0.45 X(4) = 0.5
    x5=0.2 X(5) = 0.7
    x6=0.1 X(6) = 0.96

    Fx(n)(X) = Px(n)(Xn<=X)

    so a cumulative distribution function
     
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