Statistics 550 Notes 6 Addendum
Reading: Sections 1.5, 1.6.1
I. I will e-mail the take home midterm by tomorrow night.
It will be due by the beginning of class on October 20th.
II. Notes on Minimal Sufficiency
(1) Clarified statement of Theorem 1
Theorem 1 (Lehmann and Scheffe, 1950): Suppose
S ( X ) is a sufficient statistic for  . Also suppose that the
following condition holds: if for two sample points x and
y , the ratio f ( x |  ) / f ( y |  ) is constant as a function of
 (where 0/0 is defined as 1), then S ( x )  S ( y ) . Then
S ( X ) is a minimal sufficient statistic for  .
(2) Stronger version of Lehmann and Scheffe Theorem.
Suppose S ( X ) is a sufficient statistic for  . Also suppose
that the following condition holds: for two sample points
x and y , the ratio f ( x |  ) / f ( y |  ) is constant as a
function of  (where 0/0 is defined as 1) if and only if
S ( x )  S ( y ) . Then S ( X ) is a minimal sufficient statistic
for  .
If the condition in the second sentence of Theorem 1 holds,
then S ( X ) is a sufficient statistic for  .
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Proof: Let S  {s : s  S ( x) for some x  X } (where X is the
sample space) be the image of X under S ( x ) . Define the
partition sets induced by S ( x ) as As  { x : S ( x)  s} . For
each As , choose and fix one element xs  As . For any
x  X , x S ( x ) is the fixed element that is in the same set,
As , as x . Since x and x S ( x ) are in the same set As ,
S ( x )  S ( xS ( x ) ) and hence by the assumptions of the
theorem, f ( x |  ) / f ( xS ( x ) |  ) is constant as a function of
 . Thus, we can define a function on X by
h( x )  f ( x |  ) / f ( xS ( x ) |  ) and h does not depend on  .
Define a function on S by g ( s |  )  g ( xs |  ) . Then it can
be seen that
f ( xS ( x ) |  ) f ( x |  )
f (x | ) 
 g (T ( x ) |  )h( x )
f ( xS ( x ) |  )
and by the factorization theorem, S ( x ) is a sufficient
statistic for  .
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