# Let w i 2 y i then the w i and x i are iid and

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(Let W i = 2 Y i . Then the W i and X i are iid and Corollary 4.6 applies.)
CHAPTER 4. SUFFICIENT STATISTICS 130 Rule of thumb 4.1: A k –parameter minimal sufficient statistic for a d –dimensional parameter where d<k will not be complete. In the following example d = 1 < 2 = k. (A rule of thumb is something that is frequently true but can not be used to rigorously prove something. Hence this rule of thumb can not be used to prove that the minimal sufficient statistic is not complete.) Warning: Showing that a minimal sufficient statistic is not complete is of little applied interest since complete sufficient statistics are rarely used in applications; nevertheless, many qualifying exams in statistical inference contain such a problem. Example 4.14, Cox and Hinckley (1974, p. 31). Let X 1 ,...,X n be iid N ( μ,γ 2 o μ 2 ) random variables where γ 2 o > 0 is known and μ> 0 . Then this family has a one dimensional parameter μ , but f ( x | μ ) = 1 radicalbig 2 πγ 2 o μ 2 exp parenleftbigg - 1 2 γ 2 o parenrightbigg exp parenleftbigg - 1 2 γ 2 o μ 2 x 2 + 1 γ 2 o μ x parenrightbigg is a two parameter exponential family with Θ = (0 , ) (which contains a one dimensional rectangle), and ( n i =1 X i , n i =1 X 2 i ) is a minimal sufficient statistic. (Theorem 4.5 applies since the functions 1 and 1 2 do not satisfy a linearity constraint.) However, since E μ ( X 2 ) = γ 2 o μ 2 + μ 2 and n i =1 X i N ( nμ,nγ 2 o μ 2 ) implies that E μ [( n summationdisplay i =1 X i ) 2 ] = 2 o μ 2 + n 2 μ 2 , we find that E μ [ n + γ 2 o 1 + γ 2 o n summationdisplay i =1 X 2 i - ( n summationdisplay i =1 X i ) 2 ] = n + γ 2 o 1 + γ 2 o 2 (1 + γ 2 o ) - ( 2 o μ 2 + n 2 μ 2 ) = 0 for all μ so the minimal sufficient statistic is not complete. Notice that Ω = { ( η 1 2 ) : η 1 = - 1 2 γ 2 o η 2 2 } and a plot of η 1 versus η 2 is a quadratic function which can not contain a 2–dimensional rectangle. Notice that ( η 1 2 ) is a one to one function of μ ,
CHAPTER 4. SUFFICIENT STATISTICS 131 and thus this example illustrates that the rectangle needs to be contained in Ω rather than Θ . Example 4.15. The theory does not say that any sufficient statistic from a REF is complete. Let Y be a random variable from a normal N (0 2 ) dis- tribution with σ 2 > 0 . This family is a REF with complete minimal sufficient statistic Y 2 . The data Y is also a sufficient statistic, but Y is not a function of Y 2 . Hence Y is not minimal sufficient and (by Bahadur’s theorem) not complete. Alternatively E σ 2 ( Y ) = 0 but P σ 2 ( Y = 0) = 0 < 1 , so Y is not complete. Theorem 4.9. Let Y 1 ,...,Y n be iid. a) If Y i U ( θ 1 2 ) , then ( Y (1) ,Y ( n ) ) is a complete sufficient statistic for ( θ 1 2 ) . See David (1981, p. 123.) b) If Y i U ( θ 1 2 ) with θ 1 known, then Y ( n ) is a complete sufficient statistic for θ 2 . c) If Y i U ( θ 1 2 ) with θ 2 known, then Y (1) is a complete sufficient statistic for θ 1 .