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 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.

math

share|improve this question

1 Answer 1

up vote 2 down vote accepted
+100

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 $\sigma(\mathcal{bE}) = \mathcal{P}$?

share|improve this answer
    
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

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.