Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question concerning point 3 of the following problem:

Let $(\Omega,\mathfrak{F},\mathbb{P},\mathfrak{(F_n})_{n\in\mathbb{N}})$ be a filtered probability space and $\mu$ a finite measure on $\mathfrak{F_\infty}:=\sigma{(\cup_{n=1}^{\infty}\mathfrak{F_n}})$. Assume that, for every $n\geq0$, the measure $\mu$ is absolutely continuous with respect to $\mathbb{P}$ on $\mathfrak{F_n}$, and denote by $M_n$ the Radon-Nikodym density: $M_n$ is thus non-negative, $\mathfrak{F_n}$-measurable and for every $A \in \mathfrak{F_n}$,$$\mu(A)=\int_{A}M_n \;d\mathbb{P}$$

  1. Prove that $(M_n)_{n\geq0}$ is a martingale. (OK, I showed it)
  2. Prove that $(M_n)_{n\geq0}$ converges a.s. to an integrable random variable $M_{\infty}$. (Ok, I showed it using a corollary of Doob's martingale convergence theorem)
  3. Prove that $\mu$ is absolutely continuous with respect to $\mathbb{P}$ on $\mathfrak{F_\infty}$ if and only if the martingale $(M_n)$ is closed and, that, in this case $M_{\infty}$ is the Radon-Nikodym density.

I have a problem with point 3. I do not know how to start the proof knowing that $\mu$ is absolutely continuous with respect to $\mathbb{P}$ on $\mathfrak{F_\infty}$. I would like to show that $M_n$ is uniformly integrable (bounded in $L^1$ is already good, but $\mathbb{P}$-continuous?) because then $M_n$ converges in $L^1$ and this would help a lot. But this is perhaps the wrong way to attack the problem. Perhaps somebody knows the solution.

Thanks in advance.

share|improve this question
    
I don't know the complete solution, but if $\mu \ll \Bbb P$ on $\mathfrak F_\infty$ wouldn't it help if you just take $M:= \frac{\mathrm d\mu}{\mathrm d\Bbb P}$ and show that in fact $M_n = \Bbb E[M|\mathfrak F_\infty]$ since this it a Levy martingale? –  Ilya Jan 25 '13 at 10:47
add comment

1 Answer

up vote 1 down vote accepted
  • $\Leftarrow$: Assume that $(M_n)_n$ is closed, i.e. $\mathbb{E}(Y|\mathcal{F}_n)=M_n$ for some (integrable) random variable $Y$. Hence $$\mu(F) \stackrel{\text{Def}}{=} \mathbb{E}(M_n \cdot 1_F) = \mathbb{E}(Y \cdot 1_F)$$ for all $F \in \mathcal{F}_n$, i.e. the measure $F \mapsto \mu(F)$, $F \mapsto \mathbb{E}(Y \cdot 1_F)$ coincide on $\bigcup_n \mathcal{F}_n$. This is a generator of $\mathcal{F}_{\infty}$!
  • $\Rightarrow$: By the Radon-Nikodym theorem there exists a $\mathcal{F}_{\infty}$ measurable (posiitve) random variable $Y$ such that $$\mu(F) = \mathbb{E}(Y \cdot 1_F) $$ for $F \in \mathcal{F}_{\infty}$. Hence $$\mathbb{E}(M_n \cdot 1_F)=\mu(F) = \mathbb{E}(Y \cdot 1_F)$$ for $F \in \mathcal{F}_n$, i.e. $\mathbb{E}(Y|\mathcal{F}_n)=M_n$.

Moreover, we know that $(M_n)_n$ is equi-integrable (since $M_n = \mathbb{E}(Y|\mathcal{F}_n)$), so there exists $Z \in L^1(\mathcal{F}_{\infty})$ such that $M_n \to Z$ in $L^1$, almost surely. Since $M_n \to M_{\infty}$ almost surely, we conclude $Z=M_{\infty}$. By the $L^1$-convergence we obtain

$$\mathbb{E}(M_{\infty} \cdot 1_F) = \lim_{n \to \infty} \mathbb{E}(M_n \cdot 1_F) = \mathbb{E}(Y \cdot 1_F)$$

for all $F \in \bigcup_n \mathcal{F}_n$ and therefore $M_{\infty} = Y$ almost surely (since the random variables are $\mathcal{F}_\infty$-measurable).

share|improve this answer
    
Thank you! I am just wondering why you have that $M_n$ converges to $M_\infty$ in $L^1$ –  Mathoman Jan 25 '13 at 17:33
    
@Mathoman I added some more hints concerning the $L^1$-convergence. –  saz Jan 25 '13 at 22:51
    
Thank you for your perfect answer! It helps me a lot! –  Mathoman Jan 26 '13 at 9:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.