Proof of the Halmos-Savage Factorization Theorem (Theorem 3 of the Stat 643 Summary).
We’ll take Theorem 1 as given, prove Theorem 3 given Theorems 1 and 2,
and then prove Theorem 2.
So first assume that Theorems 1 and 2 have been established.
Suppose that T is sufficient. Pθ ¿ λ ¿ µ implies that
dPθ dλ
dPθ
=
·
a.e. µ
dµ
dλ dµ
(and both functions on the right of this equality are non-negative). By Theorem
2, ∃ a nonnegative function gθ such that
dPθ
= gθ ◦ T a.s. λ
dλ
I claim that as measures on X ∗ , λ and µ are
Let X ∗ = {x| dλ
dµ (x) > 0}.
equivalent. Clearly, λ ¿ µ on X ∗ . So suppose that B is a B-measurable subset
of X ∗ that has λ (B) = 0. But
Z
dλ
dµ
λ (B) =
B dµ
while since B ⊂ X ∗ the integrand here is positive. So it must be that µ (B) = 0.
Thus λ and µ are equivalent on X ∗ .
So then consider
¸
·
dPθ
dλ dPθ
dλ
∆=
− (gθ ◦ T ) ·
=
− gθ ◦ T
dµ
dµ
dµ dλ
Letting
dPθ
(x) 6= gθ (T (x))}
dλ
since the second term in this intersection has λ probability 0, µ (D) = 0 and
thus
dλ
dPθ
= (gθ ◦ T ) ·
a.e. µ
dµ
dµ
D = {x| ∆ (x) 6= 0} = X ∗ ∩ {x|
and taking h =
dλ
the forward implication is proved.
dµ
Conversely, suppose that the factorization holds, that is that fθ =
dPθ
dµ = (gθ ◦ T ) · h a.e. µ.
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Now
dλ
dµ
=
∞
X
i=1
=
∞
X
i=1
= h·
for k =
ci
dPθi
dµ
ci h · (gθi ◦ T )
∞
X
i=1
ci (gθi ◦ T )
= h · (k ◦ T )
P∞
i=1 ci gθi .
Then the fact that Pθ ¿ λ ¿ µ again implies that
dPθ dλ
dPθ
=
·
a.e. µ
dµ
dλ dµ
so that
(gθ ◦ T ) · h =
Then define
dPθ
· h · (k ◦ T ) a.e. µ.
dλ

 gθ ◦ T
gθ∗ ◦ T =
k◦T
 0
On X ∗
dλ
>0
dµ
otherwise.
if
dPθ
= gθ∗ ◦ T
dλ
and this set has λ probability 1, i.e.
dPθ
= gθ∗ ◦ T a.e. λ
dλ
so that Theorem 2 then implies that T is sufficient.
So Theorem 3 has been proved (conditional on Theorems 1 and 2).
Now consider the proof of Theorem 2 (assuming the truth of Theorem
1).
Suppose that T is sufficient. Let B ∈ B and note that by sufficiency ∃ a
B0 -measurable function P (B|B0 ) such that for all θ
Eθ (IB |B0 ) = Pθ (B|B0 ) = P (B|B0 ) a.s. Pθ
Then for B0 ∈ B0
Z
B0
P (B|B0 ) dPθ = Pθ (B ∩ B0 ) for all θ ∈ Θ
2
(1)
by the definition of conditional expectation. Then since λ =
Z
P (B|B0 ) dλ = λ (B ∩ B0 )
P
ci Pθi ,
B0
and so (again by the definition of conditional expectation)
Eλ (IB |B0 ) = P (B|B0 ) a.s. λ
and thus
Eλ (IB |B0 ) = P (B|B0 ) a.s. Pθ for all θ
(2)
(that is, this object serves as the conditional expectation wrt each Pθ and wrt
λ).
Now consider Pθ0 and λ0 , respectively the restrictions of Pθ and λ to B0 and
dP 0
the R-N derivative dλθ0 on (X , B0 ). By Lehmann’s Theorem, ∃ non-negative
F-measurable functions gθ such that
dPθ0
= gθ ◦ T
dλ0
(3)
We need to show that gθ ◦ T also serves as an R-N derivative of Pθ wrt λ. So
let B ∈ B.
Z
Pθ (B) =
P (B|B0 ) dPθ
(by fact (1))
ZX
Eλ (IB |B0 ) dPθ
(by fact (2))
=
X
Z
Eλ (IB |B0 ) dPθ0
(since Eλ (IB |B0 ) is B0 -measurable)
=
X
Z
Eλ (IB |B0 ) gθ ◦ T dλ0 (by relationship (3))
=
X
Z
Eλ (IB |B0 ) gθ ◦ T dλ
=
X
Z
Eλ (IB · gθ ◦ T |B0 ) dλ (since gθ ◦ T is B0 -measurable)
=
ZX
IB · gθ ◦ T dλ (X ∈ B0 and the definition of conditional expectation)
=
ZX
gθ ◦ T dλ
=
B
That is, gθ ◦ T works as the R-N derivative of Pθ wrt λ and the first half of the
theorem is proved.
So conversely, suppose that ∃ non-negative F -measurable functions gθ
such that
dPθ
= gθ ◦ T a.s. λ
dλ
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Since gθ ◦ T is B0 -measurable,
dPθ0
= gθ ◦ T a.s. λ0
dλ0
(4)
To show that T is sufficient, it will be enough to show that for any B ∈ B
Eθ (IB |B0 ) = Eλ (IB |B0 ) a.s. Pθ for all θ
So for fixed θ and B ∈ B, define a measure ν on (X , B) by
Z
IB dPθ = Pθ (C ∩ B)
ν (C) =
C
Now ν ¿ Pθ and
dν
= IB a.s. Pθ
dPθ
For all B0 ∈ B0 ,
Z
IB dPθ
ν (B0 ) =
B0
Z
=
Eθ (IB |B0 ) dPθ
(5)
(by the defintion of conditional expectation)
B0
So for ν 0 the restriction of ν to B0
dν 0
= Eθ (IB |B0 ) a.s. Pθ0
dPθ0
(since Eθ (IB |B0 ) is B0 -measurable and has the right integrals).
ν 0 ¿ Pθ0 ¿ λ0
dν 0 dPθ0
dν 0
=
·
a.s. λ0
dPθ0 dλ0
dλ0
(6)
Then since
So since Pθ0 ¿ λ0 and facts (4) and (6) hold
dν 0
= Eθ (IB |B0 ) · gθ ◦ T a.s. Pθ0
dλ0
On the other hand, from (5), we still have
dν
dPθ
= IB a.s. Pθ , so that
dPθ
dν
= IB ·
= IB · gθ ◦ T a.s. Pθ
dλ
dλ
This then implies by the definition of conditional expectation that
dν 0
= Eλ (IB · gθ ◦ T |B0 ) a.s. Pθ0
dλ0
4
(7)
(The object on the right isR obviously B0 -measurable, and if we multiply by IB0
and integrate dλ0 , we get IB∩B0 gθ ◦ T dλ = ν (B0 ).) So
dν 0
= Eλ (IB |B0 ) · gθ ◦ T a.s. Pθ0
dλ0
(8)
So from (7) and (8)
Eθ (IB |B0 ) · gθ ◦ T = Eλ (IB |B0 ) · gθ ◦ T a.s. Pθ0
This is enough to imply that
Eθ (IB |B0 ) = Eλ (IB |B0 ) a.s. Pθ0
(9)
Why? Let A = {x| gθ (T (x)) = 0} and note that
Z
gθ ◦ T dλ = 0
Pθ (A) =
A
So X − A has Pθ probability 1 and so does {x| Eθ (IB |B0 ) · gθ ◦ T =Eλ (IB |B0 ) ·
gθ ◦ T } ∩ (X − A). And on this set, Eθ (IB |B0 ) =Eλ (IB |B0 ).
So (9) holds and Theorem 2 has been proved (conditional on Theorem 1).
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