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I want to prove that: if $E[M_t\mid\mathcal{F}_s]=0$ where $\mathcal{F}_s$ is the filtration generated by a stochastic process X knowing that $E[M_t\prod_0^n h_i(X_{t_i})]=0$ for all $n\in N,\quad 0\leq t_0<t_1<\dots<t_n,\quad h_i$ bounded functions.

It seems obvious if $\mathcal{F}_t$ is generated by the finite dimensional rectangles $(X_{t_1}\in A_1,\dots,X_{t_n}\in A_n$) but is this true and how can it be proved? I would appreciate it if you could help. Thank you in advance.

user62138
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2 Answers2

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Hint It is not difficult to see that

$$\mathcal{F}_s := \sigma(X_u; u \leq s) = \sigma \left( \bigcup_{n \geq 1} \bigcup_{u_1<\ldots<u_n \leq s} \sigma(X_{u_1},\ldots,X_{u_n}) \right),$$

i.e. that

$$\mathcal{G} := \bigcup_{n \geq 1} \bigcup_{u_1<\ldots<u_n \leq s} \sigma(X_{u_1},\ldots,X_{u_n})$$

is $\cap$-stable generator of $\mathcal{F}_s$. Now prove that

$$\mathcal{D} := \{A \in \mathcal{F}; \mathbb{E}(M_t \cdot 1_F)=0\}$$

is a Dynkin system. Then, since $\mathcal{G} \subset \mathcal{D}$, it follows that $$\mathcal{F}_s = \delta(\mathcal{G}) = \sigma(\mathcal{G}) \subset \mathcal{D}.$$

(Here $\delta(\mathcal{G})$ denotes the Dynkin system generated by $\mathcal{G}$.)

saz
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  • How are the $u_n$ chosen? For a fixed sequence $(u_n){n\in\mathbf N}$, the RHS of the first displayed equation is $\sigma(X{u_n},n\in\mathbf N)$. – Davide Giraudo Aug 11 '14 at 21:43
  • @DavideGiraudo No, the union is over all $u_1,\ldots,u_n$ such that $u_1<\ldots<u_n \leq s$. But since this union is not countable, this leads to the same problem as in your answer, I guess. Damn. – saz Aug 12 '14 at 05:44
  • I'm not sure it's a problem. We have a generating algebra of $\mathcal F_s$, namely, the collection in your second displayed equation (we don't require a countable union in the definition, just check that the intersection of two elements belong to this), and $E[M_t\chi_A]=0$ for each element of this algebra. Then we approximate. – Davide Giraudo Aug 12 '14 at 08:13
  • @DavideGiraudo Ah, right. Seems that I shouldn't write comments when I just got up. – saz Aug 12 '14 at 09:03
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You noticed that $\mathbb E[M_t\cdot \chi_S]=0$ for any $S$ of the form $\{X_{t_1}\in A_1\}\cap\cdots\cap \{X_{t_n}\in A_n\}$ (choosing $h_i:=\chi_{A_i}$). It also holds for finite unions of such sets. These finite unions form an algebra, hence by an approximation argument we can show that the equality $\mathbb E[M_t\cdot \chi_S]=0$ holds for each $S\in\sigma(X_{t_1},\dots,X_{t_n})$. By an other approximation argument, we extend this equality to $S$ in the $\sigma$-algebra generated by countably many $X_{t_i}$, $t_i\lt s$.

To conclude, observe that elements of $\mathcal F_s$ may be written as $\{\omega,(X_{t_i}(\omega))\in B\}$, where $(t_i)_{i\in\mathbf N}$ is a sequence of real numbers smaller than $s$ and $B$ is a Borel subset of $\mathbf R^{\infty}$.

Davide Giraudo
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  • "To conclude, observe that elements of Fs may be written as..." Exercise: Find some event in Fs which cannot be written as you say. – Did Aug 09 '14 at 12:13
  • @Did I should review the definition of $\sigma$-algebra generated by a process. Nevertheless, I keep the "answer": in the way, the other reader may not make this mistake. – Davide Giraudo Aug 09 '14 at 12:49
  • @Did After having thought for a while on your exercise, I failed to exhibit an element of $\mathcal F_s$ which is not of the form mentioned in my attempt. I think I managed to show that the collection $\mathcal C:={{\omega, (X_{t_i}(\omega))\in B}, (t_i)\in (0,s]^{\mathbf N},B\in\mathcal B(\mathbf R^{\infty}) }$ is a $\sigma$-algebra which makes for each $u\leqslant s$ the random variable $X_u$ measurable. Do you have a hint please? – Davide Giraudo Aug 09 '14 at 13:49
  • I would not know how to write the event ${\exists u\leqslant s,X_u=0}$ as you say... but maybe $\mathcal F_s$ is defined such as to not contain it? Dunno. – Did Aug 09 '14 at 13:59
  • Indeed, in general, the set you mentioned does not depend on countably many coordinates. I believed that $\mathcal F_s$ is the smallest $\sigma$-algebra making all the random variables $X_u, u\leqslant s$ measurable and I'm not sure if ${\exists u\leqslant s,X_u=0}$ necessarily belong to this $\sigma$-algebra. – Davide Giraudo Aug 09 '14 at 14:12
  • Yes. In the end this all depends on how $\mathcal F_s$ is defined and if it is defined as you say, my first comment is moot. Sorry for the noise. – Did Aug 09 '14 at 14:30
  • @DavideGiraudo : But how can I prove that $\sigma(X_1,\dots,X_n)$ is generated by the rectangle? I know only that it's generated by the cilinders $\lbrace (X_1,\dots,X_n)\in A\rbrace$ where A is a borel set of $E^n$ (if the process is E-valued). – user62138 Aug 11 '14 at 19:45
  • Assuming for example that $E=\mathbf R$, we can use this: http://math.stackexchange.com/questions/7881/preimage-of-generated-sigma-algebra. – Davide Giraudo Aug 11 '14 at 21:42