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Factorization Theorem for Sufficient Statistics & Indicator Function

  1. Feb 20, 2009 #1
    Problem:
    Let Y1,Y2,...,Yn denote a random sample from the uniform distribution over the interval (0,theta). Show that Y(n)=max(Y1,Y2,...,Yn) is a sufficient statistic for theta by the factorization theorem.

    Solution:
    http://www.geocities.com/asdfasdf23135/stat10.JPG
    1) While I understand that IA (x)IB (x)=IA intersect B (x), I don't understand the equality circled in red above.

    In the solutions, they say that I0,theta (y1)...I0,theta (yn)=I0,theta (y(n)). Is this really correct?
    Shouldn't the right hand side be I0,theta (y(n))I0,infinity (y(1)) ? I believe that the second factor is necessary because the largest observation is greater than zero does not guarantee that the smallest observation is greater than zero.
    Which one is correct?


    2) Also, is I0,theta (y(n)) a function of y(n), a function of theta, or a function of both y(n) and theta?
    If it is a function of both y(n) and theta, then there is something that I don't understand. Following the definition of indicator function that IA (x) is a function of x alone (it is a function of only the stuff in the parenthesis), shouldn't I0,theta (y(n)) be a function of only y(n) alone?


    Thank you for explaining! I've been confused with these ideas for at least a week.
     
  2. jcsd
  3. Feb 21, 2009 #2

    statdad

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    Homework Helper

    The left side of

    [tex]
    \prod_{i=1}^n I_{[0,\theta)} (y_i) = I_{[0,\theta)} (y_{(n)})
    [/tex]

    means that all the [tex] y_i [/tex] values are in the interval [tex] [0,\theta) [/tex].

    This is true if, and only if, the maximum of the y's is in the same interval, and that is the meaning of the right-side.

    The indicator [tex] I_{[0,\theta)} (y_{(n)}) [/tex] is a function of [tex] y_{(n)}[/tex] only, since [tex] \theta [/tex] is fixed (it's a parameter).
     
  4. Feb 21, 2009 #3
    I understand that
    0<X_1,...., X_n<theta here these are the unordered data
    is the same as (iff)
    0<X_(1)<X_(2)<.....<X_(n)<theta

    But I don't think
    0<X_(1)<X_(2)<.....<X_(n)<theta
    is EQUIVALENT to (iff)
    0<X_(n)<theta
    The => direction is true but <= is not. (the fact that the largest observation x(n) is greater than zero does not guarantee that the smallest observation x(1) is greater than zero.)

    So that's why I think we should have
    I0,theta(y1)...I0,theta(yn) = I0,theta(y(n))I0,infinity(y(1))
    instead of I0,theta(y1)...I0,theta(yn)=I0,theta(y(n)).

    Right?
     
    Last edited: Feb 21, 2009
  5. Feb 21, 2009 #4
    But we can also write it as I y(n), inf (theta), in this case theta would be in the parenthesis, so in this case, would it be a function of theta alone? (in the gerenal case, f(x) means a function of x, f(y) means a function of y, the stuff in the parenthesis)


    When we talk about functions, is it always only a function of the stuff in the parenthesis? It looks like that the restrictions/constraints are also important, so shouldn't it be a function also of the variables in the restrictions/constraints?

    e.g.)
    f(x)=x if x>y
    f(x)=x^3 if x<y
    Here not only the value of x controls f, the value of y also controls f, so is f a function of BOTH x and y here?

    Thanks!
     
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