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Unformatted text preview: Lecture 13 13.1 Minimal jointly sufficient statistics. When it comes to jointly sufficient statistics ( T 1 , . . . , T k ) the total number of them ( k ) is clearly very important and we would like it to be small. If we don’t care about k then we can always find some trivial examples of jointly sufficient statistics. For instance, the entire sample X 1 , . . . , X n is, obviously, always sufficient, but this choice is not interesting. Another trivial example is the order statistics Y 1 ≤ Y 2 ≤ . . . ≤ Y n which are simply the values X 1 , . . . , X n arranged in the increasing order, i.e. Y 1 = min( X 1 , . . . , X n ) ≤ . . . ≤ Y n = max( X 1 , . . . , X n ) . Y 1 , . . . , Y n are jointly sufficient by factorization criterion, since f ( X 1 , . . . , X n | θ ) = f ( X 1 | θ ) × . . . × f ( X n | θ ) = f ( Y 1 | θ ) × . . . × f ( Y n | θ ) . When we face different choices of jointly sufficient statistics, how to decide which one is better? The following definition seems natural. Definition. (Minimal jointly sufficient statistics.) ( T 1 , . . . , T k ) are minimal jointly sufficient if given any other jointly sufficient statistics ( r 1 , . . . , r m ) we have, T 1 = g 1 ( r 1 , . . . , r m ) , . . . , T k = g k ( r 1 , . . . , r m ) , i.e. T s can be expressed as functions of r s. How to decide whether ( T 1 , . . . , T k ) is minimal? One possible way to do this is through the Maximum Likelihood Estimator as follows....
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This note was uploaded on 10/11/2009 for the course STATISTICS 18.443 taught by Professor Dmitrypanchenko during the Spring '09 term at MIT.
- Spring '09