Download The Radon-Nikodým Theorem - damtp

The Radon-Nikodým Theorem
Helge Dietert
University of Cambridge
15 March 2012
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
1 / 17
1
Radon-Nikodým Theorem
2
Standard Measure Theory
3
Functional Analysis Proof
4
Conditional Expectation
5
Martingales
6
Martingale Proof of Radon-Nikodým
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
2 / 17
Statement
(Ω, Σ) measure space
µ, ν finite measures – we will assume all measures are finite!
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
2 / 17
Statement
(Ω, Σ) measure space
µ, ν finite measures – we will assume all measures are finite!
ν is called absolutely continuous with respect to µ if ∀ > 0, ∃δ > 0
so that if ν(A) < then µ(A) < δ.
The Radon-Nikodým Theorem
states that ν has a density with
R
respect to µ (i.e. ν(A) = A f dµ) iff ν is absolutely continuous with
respect to µ.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
2 / 17
Characterisation of Absolute Continuity
Lemma
ν is absolutely continuous w.r.t µ iff every µ null set is a ν null set.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
3 / 17
Characterisation of Absolute Continuity
Lemma
ν is absolutely continuous w.r.t µ iff every µ null set is a ν null set.
Proof.
⇒ If µ(A) = 0, by absolute continuity ν(A) ≤ , ∀ > 0, so ν(A) = 0
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
3 / 17
Characterisation of Absolute Continuity
Lemma
ν is absolutely continuous w.r.t µ iff every µ null set is a ν null set.
Proof.
⇒ If µ(A) = 0, by absolute continuity ν(A) ≤ , ∀ > 0, so ν(A) = 0
⇐ Suppose not: Then there exists > 0 without appropriate δ, i.e.
∃(An ) in Σ s.t.
µ(An ) ≤ 2−n , v (An ) > Take
A = lim sup An =
∞ [
\
An
(1)
(2)
n=1 m≥n
Then µ(A) = 0, ν(A) ≥ > 0, contradiction.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
3 / 17
Easy Direction
Lemma
ν is absolutely continuous w.r.t µ iff every µ null set is a ν null set.
Corollary
If v has density f w.r.t. µ (i.e. ν(A) =
continuous with respect to µ.
Helge Dietert (University of Cambridge)
R
Af
dµ), then ν is absolutely
The Radon-Nikodým Theorem
15 March 2012
4 / 17
Difference Lemma
Lemma
Suppose σ, τ are finite measures on the sigma algebra (Ω, Σ) with
σ(Ω) < τ (Ω). Then there exits Ω0 ∈ Σ such that
σ(Ω0 ) < τ (Ω0 )
(3)
0
σ(A) ≤ τ (A) for all A ∈ Ω ∩ Σ
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
(4)
15 March 2012
5 / 17
Measure Theory Proof
Proof.
Consider set G of non-negative, measurable functions g ≥ 0 with
gµ ≤ ν
Constant function g = 0 is in G
If g , h ∈ G then sup(g , h) ∈ G
Z
∀g ∈G : g dµ ≤ ν(Ω) ⇒
Z
γ := sup
g dµ < ∞
g ∈G
R
Hence exits monotone increasing (gn ) such that
gn dµ → γ. Take
R
f := sup gn , then by monotone convergence f dµ = γ.
We claim ν = f µ. By monotone convergence f µ ≤ ν, so τ := ν − f µ is an
absolutely continuous measure.
We are done if τ (Ω) = 0.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
6 / 17
Hilbert Space
Definition (Hilbert Space)
A Hilbert space H is a complete normed
p space whose norm comes from an
inner product, i.e. for x ∈ H ||x|| = hx, xi
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
7 / 17
Hilbert Space
Definition (Hilbert Space)
A Hilbert space H is a complete normed
p space whose norm comes from an
inner product, i.e. for x ∈ H ||x|| = hx, xi
Examples
1
Take Rn and define
hx, y i =
n
X
xi yi
(5)
f (x)g (x)dx
(6)
i=1
2
For two function f , g define
Z
hf , g i =
Take L2 all functions f with ||f || < ∞ and identify functions which are
equal almost everywhere.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
7 / 17
Riesz Representation Theorem
Theorem (Riesz Representation Theorem)
Given a Hilbert space H. For every continuous linear functional f : H → R,
there exists y ∈ H s.t.
f (x) = hx, y i ∀x ∈ H
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
(7)
15 March 2012
8 / 17
Lebesgue Decomposition Theorem
Theorem (Lebesque Decomposition Theorem)
Let µ, ν be finite measures on (Ω, Σ). Then there
exists non-negative
R
f ∈ L1 (µ) and B ∈ Σ with µ(B) and ν(A) = A f dµ + ν(A ∩ B) for all
A∈Σ
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
9 / 17
Lebesgue Decomposition Theorem
Proof.
π = µ + ν defines a measure on Σ.R
Define L : L2 (π) → R with L(g ) = g dν. Using Cauchy-Schwarz
sZ
p
p
|L(g )| ≤ ν(Ω)
|g |2 dν ≤ ν(Ω) ||g ||L2 (π)
Hence L is continuous. By Riesz representation theorem find h such that
for all g ∈ L2 (π) holds L(g ) = hg , hi.
Z
Z
Z
Z
Z
g dν = ghdµ + ghdν ⇒ g (1 − h)dν = ghdµ
Z
Z
Z
ν(A) =
dν =
hdµ +
hdν Take g indicator function
A
Helge Dietert (University of Cambridge)
A
A
The Radon-Nikodým Theorem
15 March 2012
9 / 17
Lebesgue Decomposition Theorem
Proof.
Z
Z
Z
Z
Z
g dν = ghdµ + ghdν ⇒ g (1 − h)dν = ghdµ
Z
Z
Z
ν(A) =
dν =
hdµ +
hdν Take g indicator function
A
A
A
N = (h < 0), Gn = (0 ≤ h ≤ 1 − 1/n), G = (0 ≤ h < 1), B = (h ≥ 1)
Z
Z
ν(N) =
hdµ +
hdν < 0 ⇒ µ(N) = ν(N) = 0
Z N
Z N
ν(B) =
hdµ +
hdν ≥ ν(B) + µ(B) ⇒ µ(B) = 0
B
Helge Dietert (University of Cambridge)
B
The Radon-Nikodým Theorem
15 March 2012
9 / 17
Lebesgue Decomposition Theorem
Proof.
Z
Z
Z
Z
Z
g dν = ghdµ + ghdν ⇒ g (1 − h)dν = ghdµ
Z
Z
Z
ν(A) =
dν =
hdµ +
hdν Take g indicator function
A
A
A
N = (h < 0), Gn = (0 ≤ h ≤ 1 − 1/n), G = (0 ≤ h < 1), B = (h ≥ 1)
h(x)
for x ∈ G . On Gn , f is bounded by n so
Take f (x) = 1−h(x)
R
ν(A ∩ Gn ) = A∩Gn f dµ. By monotone convergence
Z
ν(A ∩ G ) =
f dµ
A
Hence
Z
ν(A) = ν(A ∩ G ) + ν(A + B) =
f dµ + ν(A ∩ B)
A
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
9 / 17
Definition of Conditional Expectation
Definition (Conditional Expectation)
Given a probability space (Ω, Σ, P) and a random variable X (E(|X |) < ∞).
Let G be a sub-σ-algebra of Σ, then the conditional expectation
Y = E(X |G) is a random variable such that
1
Y is G measurable,
2
E(|Y |) < ∞,
3
E(1A Y ) = E(1A X ) for all A ∈ G.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
10 / 17
Definition of Conditional Expectation
Definition (Conditional Expectation)
Given a probability space (Ω, Σ, P) and a random variable X (E(|X |) < ∞).
Let G be a sub-σ-algebra of Σ, then the conditional expectation
Y = E(X |G) is a random variable such that
1
Y is G measurable,
2
E(|Y |) < ∞,
3
E(1A Y ) = E(1A X ) for all A ∈ G.
Not clear if such a random variable exists
traditionally done using Radon-Nikodým
can be done directly using orthogonal projection in L2
Unique in almost sure sense
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
10 / 17
Adapted Process
Definition (Filtration)
A filtration (Fn ) is an increasing family of sub-σ-algebras F0 ⊂ F1 . . .
Definition (Process)
A process X = (Xn : n ≥ 0) is a sequence of random variables.
The process is adapted if Xn is Fn -measurable.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
11 / 17
Definition of Martingales
Definition (Martingale)
A process X is a martingale if
1
X is adapted,
2
E(|X |n ) < ∞, ∀n,
3
E(Xn |Fn−1 ) = Xn−1 , a.s.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
12 / 17
Convergence of Martingales
Theorem (Doob’s Convergence Theorem)
Let X be martingale bounded in L1 (supn E(|Xn |) < ∞). Then almost
surely X∞ = lim Xn exists and is finite.
For definiteness can define X∞ (ω) = lim sup Xn (ω)
Corollary
If X is a non-negative martingale, then X∞ = lim Xn exists almost surely.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
13 / 17
Uniform Integrability
Gives a criterion when convergence in probability and in L1 are equivalent.
Definition
Uniform Integrability (UI) A family X of random variables is UI if for every
> 0 there exits K such that
sup (E(|X | 1|X |≥K )) < X ∈X
Theorem
Let X be a rondom variable, (Xn ) be a sequence of random variables. Then
the two following statement are equivalent
Xn ∈ L1 for all n, X ∈ L1 and Xn → X in L1
{Xn : n ∈ N} is UI and Xn → X in probability.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
14 / 17
Countable Sigma Algebra
Proof Radon-Nikodým-Theorem under additional assumption that Σ is
separable Σ = σ(Fn : n ∈ N). By scaling make µ to the probability.
Proof.
Define Fn := σ(F1 , F2 , . . . Fn ).
Each set in Fn consists of a union of ‘atoms’ An , 1, . . . An,r (n) .
Define Xn : Ω → [0, ∞) for ω ∈ An,k by
(
0
if µ(An,k ) = 0
Xn (ω) =
ν(An,k )/µ(An,k ) if µ(An,k ) > 0
(8)
Xn ∈ L1 (Ω, Fn , µ) and
Z
Xn dµ ∀A ∈ Fn
ν(A) =
A
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
15 / 17
Countable Sigma Algebra
Proof Radon-Nikodým-Theorem under additional assumption that Σ is
separable Σ = σ(Fn : n ∈ N). By scaling make µ to the probability.
Proof.
Z
Xn dµ ∀A ∈ Fn
ν(A) =
A
X = (Xn ) is a non-negative martingale, so X∞ = lim Xn exists almost
surely.
S
Find X is UI, so Xn → X in L1 and hence for all A ∈ Fn
Z
ν(A) =
X∞ dµ
A
This is a generating π-system so it holds on Σ.
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
15 / 17
Extending Result
Let Sep be the class of countable sub-σ-algebras G. Let XG be the
corresponding density.
We claim that XG → X in L1 in the sense that given > 0 there exists
K ∈ Sep s.t. if K ⊂ G ∈ Sep, then ||XG − X || < .
Suppose not, thenit is not Cauchy, so
there exists 0 and K(0) ⊂ K(1), . . .
in Sep such that XK(n) − XK(n+1) > 0 . However, (XK(n) ) is u UI
martingale, contradiction.
For A ∈ Σ and > 0 find appropriate K. Then σ(K, A) ∈ Sep, so
Z
Z
ν(A) −
=
X
dµ
≤
+
ν(A)
−
X
dµ
σ(A,K
)
A
Helge Dietert (University of Cambridge)
A
The Radon-Nikodým Theorem
15 March 2012
16 / 17
Acknowledgement
I would like to thank the lectures Perla Sousi (Advanced Probability) and
Ben Garling (Functional Analysis) for showing their proofs.
Details are from
Bauer, Heinz: “Wahrscheinlichkeitstheorie”
Williams, David: “Probability with Martingales”
Helge Dietert (University of Cambridge)
The Radon-Nikodým Theorem
15 March 2012
17 / 17