Results 1 to 4 of 4

Math Help - Factorization Theorem for Sufficient Statistics

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    Factorization Theorem for Sufficient Statistics

    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:

    1) While I understand that I_A (x)I_B (x)=I_A intersect B (x), I don't understand the equality circled in red above.

    In the solutions, they say that I_0,theta (y1)...I_ 0,theta(yn)=I_0,theta (y(n)). Is this really correct?
    Shouldn't the right hand side be I_0,theta (y(n))I_0,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 I_0,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 I_A (x) is a function of x alone (it is a function of only the stuff in the parenthesis), shouldn't I_0,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.
    Last edited by kingwinner; February 20th 2009 at 02:54 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    I'm lecturing on suff stats on monday.
    The point is that the indicator function is 0 or 1.
    It's 1 as long as each X_i is between 0 and theta.
    NOW look at the order stats
    YOU need to convince yourself that
    0<X_1,...., X_n<theta here these are the unordered data
    is the same as
    0<X_(1)<X_(2)<.....<X_(n)<theta
    which is the same as
    0<X_(n)<theta
    All we need is the largest to be less than theta
    Now if theta was a lower value for our rvs instead of an upper bound,
    then the smallest order stat would be suff for theta
    And in the case of U(a,b)
    the smallest and largest order stats are suff for the TWO
    parameters, a and b.
    ARE you using Wackerly too?
    I knew Dennis at Florida and Scheaffer was my chair there.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Quote Originally Posted by matheagle View Post
    I'm lecturing on suff stats on monday.
    The point is that the indicator function is 0 or 1.
    It's 1 as long as each X_i is between 0 and theta.
    NOW look at the order stats
    YOU need to convince yourself that
    0<X_1,...., X_n<theta here these are the unordered data
    is the same as
    0<X_(1)<X_(2)<.....<X_(n)<theta
    which is the same as
    0<X_(n)<theta
    All we need is the largest to be less than theta
    Now if theta was a lower value for our rvs instead of an upper bound,
    then the smallest order stat would be suff for theta
    And in the case of U(a,b)
    the smallest and largest order stats are suff for the TWO
    parameters, a and b.
    ARE you using Wackerly too?
    I knew Dennis at Florida and Scheaffer was my chair there.
    Yes, I am using Wackerly.

    But I don't think
    0<X_(1)<X_(2)<.....<X_(n)<theta
    is EQUIVALENT to (iff)
    0<X_(n)<theta
    => is true but <= is not.

    So that's why I think we should have I_0,theta (y1)...I_ 0,theta(yn) = I_0,theta (y(n))I_0,infinity (y(1)) instead of I_0,theta (y1)...I_ 0,theta(yn)=I_0,theta (y(n)).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    I DO think
    0<X_(1)<X_(2)<.....<X_(n)<theta
    is EQUIVALENT to 0<X_(n)<theta
    Or better yet
    it's equvivalent to
    I(0<X(1)) times I(X(n)<theta)
    and who needs the rest
    as long as the smallest order stat is greater than zero
    and the largest order stat is less than theta
    thats the same
    as all the data (ordered or not) are between
    0 and theta
    You can toss the I(0<X(1)) in the factorization theorem into the other term, h(.), but you cannot separate theta and our largest order stat
    HENCE the largest order stat is suff.

    OR get the conditional density of your vector given the largest order stat.
    That's a lot more work.
    I always have a student ask me to do that.
    And it's harder, but a lot of fun.

    -------------------------------------------------------------------------

    Likewise as I said earlier
    Look at problem 9.51, here we need the variables to be greater than theta in our underlying distribution
    In that case the smallest order stat is suff for theta.
    In 9.52 theta is larger than our y's, so the largest order stat is suff for theta...
    I may just assign all of these on Monday.

    ---------------------------------------------------------------------------

    I(0<X_1<theta) times I(0<X_2<theta) ... I(0<X_n<theta)
    is 1 iff all the rvs are between 0 and theta.
    And
    I(0<X_(1)<X_(2)<.....<X_(n)<theta)
    is equal to 1 iff all the rvs are between 0 and theta.
    Likewise
    I(0<X_(1)) times I(X_(n)<theta)
    is 1 iff all the rvs are between 0 and theta.
    So throw I(0<X_(1)) into h(X_1,..., X_n).
    BUT that shows that the largest order stat is suff for our parameter.
    Last edited by mr fantastic; February 22nd 2009 at 12:56 AM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find min sufficient statistics for a continuous random sample
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: September 10th 2011, 09:06 PM
  2. sufficient statistics
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: August 26th 2011, 08:09 AM
  3. factorization criterion and sufficient statistics
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 30th 2011, 06:01 AM
  4. Bayesian Statistics - Factorization Theorem
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: February 24th 2011, 05:22 PM
  5. sufficient statistics and MVUE's
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 23rd 2010, 02:30 PM

Search Tags


/mathhelpforum @mathhelpforum