martingale proof of the Radon-Nikodym theorem
We apply the martingale convergence theorem to prove the Radon-Nikodym theorem, which states that if and are -finite measures on a measurable space and is absolutely continuous with respect to then there exists a non-negative and measurable such that for all measurable sets .
As any -finite measure is equivalent to a probability measure (http://planetmath.org/AnySigmaFiniteMeasureIsEquivalentToAProbabilityMeasure), it is enough to prove the result in the case where and are probability measures. Furthermore, by the Jordan decomposition, the result generalizes to the case where is a signed measure. So, we just need to prove the following.
Let and be probability measures on the measurable space , such that is absolutely continuous with respect to . Then, there exists a non-negative random variable such that and for every .
Here, is called the Radon-Nikodym derivative of with respect to .
More generally, for any sub--algebra of we can restrict the measures and to and ask if the Radon-Nikodym derivative of with respect to exists. If it does we shall denote it by , which by definition is a non-negative -measurable random variable satisfying for all .
We note that if does exist, then it is uniquely defined (-almost everywhere). Suppose that also satisfied the required properties, then
so almost surely. Similarly, and therefore (almost surely).
First, the easy case. For a finite -algebra, the Radon-Nikodym derivative can be written out explicitly.
If is a finite sub--algebra of then the Radon-Nikodym derivative exists.
Note that whenever then , and we adopt the convention that . Clearly, is -measurable, and
Here, we have used and . By linearity, this equality remains true if both sides are replaced by any union of the , and therefore is the required Radon-Nikodym derivative. ∎
Next, martingale convergence is used to prove the existence of the Radon-Nikodym derivative in the case where the -algebra is separable. By separable, we mean that there is a countable sequence of sets generating . Note that if we let be the -algebra generated by , then is an increasing sequence of finite sub--algebras such that generates . The following result is general enough to apply in many useful cases, such as with the Boral -algebra on .
Let be a separable sub--algebra of . Then, the Radon-Nikodym derivative exists. If furthermore, is an increasing sequence of finite -algebras satisfying then as .
Let us set . If then the conditional expectation is -measurable, and for every ,
This equality just uses the definition of the conditional expectation and then the definition of as the Radon-Nikodym derivative restricted to . So, is the Radon-Nikodym derivative restricted to , and equals (almost-surely).
Therefore, is a martingale and the martingale convergence theorem implies that the limit
exists almost surely. We now show that the sequence is uniformly integrable. Choose any . As is absolutely continuous with respect to , there exists a such that whenever . Using
we see that whenever and, therefore, . So
for every , showing that is a uniformly integrable sequence with respect to . Therefore, convergence in (1) is in , and as . So, for any ,
By linearity and the monotone convergence theorem, the collection of sets satisfying (2) is a Dynkin system containing the -system so, by Dynkin’s lemma, is satisfied for every and, by definition, is the Radon-Nikodym derivative restricted to . ∎
Finally, by approximating by finite -algebras we can prove the Radon-Nikodym theorem for arbitrary inseparable -algebras .
Proof of the Radon-Nikodym theorem:
First, we use contradiction to show that for any there exists a finite -algebra satisfying for every finite -algebra with . If this were not the case, then by induction we could find an increasing sequence of finite sub--algebras of satisfying . However, letting , Lemma 2 shows that exists and
— a contradiction.
So, there exists a sequence of finite sub--algebras of such that for every finite sub--algebra of containing . Let be the (separable) -algebra generated by , and set . By Lemma 2, the Radon-Nikodym derivative restricted to , , exists, and we show that it is the required derivative of with respect to .
Choose any set and let be the (finite) -algebra generated by . Then, exists and satisfies and,
So, as required.
- 1 David Williams, Probability with martingales, Cambridge Mathematical Textbooks, Cambridge University Press, 1991.
|Title||martingale proof of the Radon-Nikodym theorem|
|Date of creation||2013-03-22 18:34:10|
|Last modified on||2013-03-22 18:34:10|
|Last modified by||gel (22282)|