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I have some question about the Monotone Class Theorem and its application. First I state the Theorem:

Let $\mathcal{M}:=\{f_\alpha; \alpha \in J\}$ be a set of bounded functions, such that $f_\alpha:N\to \mathbb{R}$ where $N$ is a set. Further we suppose that $\mathcal{M}$ is closed under multiplication and define $\mathcal{C}:=\sigma(\mathcal{M})$. Let $\mathcal{H}$ be a real vector space of bounded real-valued functions on $N$ and assume:

  1. $\mathcal{H}$ contains $\mathcal{M}$
  2. $\mathcal{H}$ contains the constant function $1$.
  3. If $0\le f_{\alpha_1}\le f_{\alpha_2}\le \dots$ is a sequence in $\mathcal{H}$ and $f=\lim_n f_{\alpha_n}$ is bounded, then $f\in \mathcal{H}$

Then $\mathcal{H}$ contains all bounded $\mathcal{C}$ -measurable functions.

My first question: I know the Dynkin lemma which deals with $\sigma$-Algebras and $\pi$-Systems. Which is the stronger one, i.e. does Monotone Class imply Dynkin or the other way around?(or are they equal?) A reference for a proof would also be great!

My second question is about an application. Let $X=(X_t)$ be a right continuous stochastic process with $X_0=0$ a.s. and denote by $F=(F_t)$ a filtration, where $F_t:=\sigma(X_s;s\le t)$. I want to show:

If for all $0\le t_1<\dots<t_n<\infty$ the increments $X_{t_{i}}-X_{t_{i-1}}$ are independent then $X_t-X_s$ is independent of $F_s$ for $t>s$.

The hint in the book is to use Monotone Class Theorem. So $$\mathcal{H}:=\{Y:\Omega\to \mathbb{R} \mbox{ bounded };E[h(X_t-X_s)Y]=E[h(X_t-X_s)]E[Y] \forall h:\mathbb{R}\to\mathbb{R} \mbox{ bounded and Borel-measurable}\}$$

This choice is clear. Now they say $$\mathcal{M}:=\{\prod_{i=1}^n f_i(X_{s_i});0\le s_1\le \dots\le s_n\le s,n\in \mathbb{N},f_i\colon\mathbb{R}\to\mathbb{R} \mbox{ bounded and Borel-measurable}\}$$

Two question about this choice:

  1. Why is $\sigma(\mathcal{M})=\mathcal{F}_s$?
  2. Why do they define $\mathcal{M}$ like this? (As family of products?)

Thanks in advance!

hulik

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$F$ is a filtration and $\sigma(\mathcal M)$ is a $\sigma$-algebra/family of measurable maps, right? What do you mean then by $F = \sigma(\mathcal M)$? –  Ilya Mar 13 '12 at 9:24
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@ Hulik : Regarding your first question I think you should compare Dynkin Lemma with the 'set ' version of the Monotone class Theorem. You can find definitions, axioms and theorems at planetmath.org. But from the proof of Functional MCT that you can find there, Dynkin's lemma is used in the proof of FMCT (theorem 1). Best regards –  TheBridge Mar 13 '12 at 15:16
    
@ The Bridge: Thank you for your answer! I will check this right now :) –  user20869 Mar 14 '12 at 8:25
    
@ Hulik : I think there is a typo in FMCT item 3, the limit should be in $\mathcal{H}$ not in $\mathcal{M}$. Regards –  TheBridge Mar 14 '12 at 14:38
    
@ hulik : I nmy opinion $\mathcal{M}$ is not the easiest set of function, you should try with $\mathcal{M}$ built rather with products of functions of increments $X_{t_i}-X_{t_{i-1}}$ not $X_{t_i}$'s in my opinion. Best regards –  TheBridge Mar 14 '12 at 14:45
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