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I am solving an exercise in Rogers and Williams and want to ask if my solution is correct. Let me first introduce the notation. The space $b\mathcal{E}$ is the space of processes of the form $$H(t,\omega)=\sum_{i=1}^nZ_i(\omega)(S_i(\omega),T_i(\omega)]$$ where $(S_i(\omega),T_i(\omega)]$ is the characteristic function of the set $\{t:S_i(\omega)<t\le T_i(\omega)\}$. We assume $S_1\le T_1\le S_2\le \dots$ are stopping times and $Z_i\in b\mathcal{F}_{S_i}$ (bounded r.v. measurable w.r.t $\mathcal{F}_{S_i}$. The predictable sigma algebra $\mathcal{P}$ on $(0,\infty)\times\Omega$ is generated by all LCRL adapted processes. I know that $\sigma(b\mathcal{E})=\mathcal{P}$. I want to prove that the sets of the form $$(u_\Gamma,\infty)=\{(t,\omega):t>u,\omega\in\Gamma\}$$ with $u\ge 0$ and $\Gamma\in \mathcal{F}_u$ also generate the predictable sigma algebra. Let's denote with $\mathcal{C}$ the set of all these sets. What I did:

A characteristic function of a set above is of the form $\mathbf1_{(u,\infty)}\mathbf1_{\Gamma}$, which is clearly LCRL and adapted. Therefore $\sigma(\mathcal{C})\subset \mathcal{P}$. For the reverse inequality I am not sure. A hint says that $\mathcal{P}$ is also generated from processes of the form $Z(s,\infty)$, where $Z$ is $\mathcal{F}_s$ measurable. Why is this true? Clearly the generated sigma algebra of this processes is contained in $\mathcal{P}$. But why is the reverse inequality true?

Assuming this hint, I would prove the original statement like this: First let $Z=\mathbf1_\Gamma$, where $\Gamma\in\mathcal{F}_s$. Then $Z(s,\infty)$ is $\sigma(\mathcal{C})$ measurable. By measure theoretic induction, we can find a sequence of linear combinations of functions of the form $\mathbf1_\Gamma(s,\infty)$ which converges to every $H=Z(s,\infty)$ with $Z\in b\mathcal{F}_s$. This proves the reverse inequality, assuming the claim.

Is my proof correct? And if so, why is the hint true? Moreover, they claim as Corollary of the exercise: $\mathcal{P}=\{(T,\infty):T \mbox{ stopping time }\}$. I guess this should mean that the RHS is also a generating set for $\mathcal{P}$. But again, why is this true? Thanks for your help.


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1 Answer 1

up vote 2 down vote accepted

For the hint: For $Z \in \mathcal{F}_s$, the process $Z ]s,\infty[$ is obviously left-continuous, has right-limits and is adapted. So the $\sigma$-algebra generated by these processes - let's call it $\tilde{\mathcal{P}}$ - is contained in $\mathcal{P}$. On the other hand, every LCRL, adapted process $Y$ can be written as the limit of $$Y^n_t := n \int_{t-1/n}^t {\mathbb{1}_{\{|Y_s| \leq n\}} X_s} \mathrm{d}s,$$ which define continuous, adapted processes. If $(X_t)_{t > 0}$ is a continuous, adapted process, write $$X^n_t := \sum_{k=1}^{n\cdot 2^n} {X_{(k-1)/n} \cdot \mathbb{1}_{\left]\frac{k-1}{n}, \frac{k}{n}\right]}}.$$ Since $Z ]s,t] = Z ]s,\infty[ - Z ]t,\infty[$, we see that $X^n$ is $\tilde{\mathcal{P}}$-measurable. Since it converges pointwisely to $X$, the two limit procedures show that $\mathcal{P} \subseteq \tilde{\mathcal{P}}$.

Your proof using the hint looks good.

For the Corollary: doesn't this follow from $\sigma(\mathcal{bE}) = \mathcal{P}$?

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Thanks for your answer. Shouldn't it be $\int_{t-\frac{1}{n}}^t\cdots$? Your definition is for RCLL processes if I'm not wrong. –  math Apr 10 '13 at 17:12
Sorry, you are right. I thought I had corrected that before posting. Anyway, I have done it now. –  Thomas Apr 10 '13 at 18:04

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