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I'm reading George Lowther's blog and have a question about the proof of lemma 2. We want to verify that the predictable sigma algebra is also generated by the continuous and adapted processes. One direction is clear, for the other the idea is to write a left continuous process $X$ as the limit of a ccontinuous process. For this purpose he defines

$$Y_t^n:=n\int_{t-\frac{1}{n}}^t\mathbf1_{|X_{s\vee 0}|\le n}X_{s\vee 0} ds$$

I've proved the following statement (see this question). For $f:\mathbb{R}\to\mathbb{R}$ Borel, bounded the map $t\mapsto \int_{t-h}^tf(s)ds$ is lipschitz continuous. Assuming additionally left continuity of $f$ we have

$$\lim_{h\downarrow 0}\frac{1}{h}\int_{t-h}^tf(s)ds=f(t)\tag{1}$$

$(1)$ implies the continuity of $Y^n_t$ and the convergence of $Y_t^n\to X_t$. This leads to the following two question:

  1. George Lowther writes $Y^n\to X$, what does this mean?
  2. I wonder why left-continuity is important comparing with right continuity. We can write $$\lim_{h\downarrow 0}\frac{1}{h}\int_{t}^{t+h}f(s)ds=f(t)\tag{1'}$$ if we assume $f$ to be right continuous. Hence we would conclude that the optional sigma algebra, which is generated by the right continuous and adapted processes is also generated by the continuous and adapted processes implying that the optional and predictable sigma algebra coincide. Therefore, I made a mistake in the right continuous case, but I can't see where. It would be appreciated if someone could explain what goes wrong in the right continuous case.
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  1. The predictable $\sigma$-algebra $\mathcal{P}$ is defined on the product space $\tilde{\Omega} := [0,\infty) \times \Omega$. In order to prove that a process $X$ is measurable with respect to $\mathcal{P}$, it suffices to show that there exists a sequence $(X^n)_n$ of $\mathcal{P}$-measurable random variables such that $X^n \to X$ pointwise as $n \to \infty$. So, in this case, $X^n \to X$ means $$X^n(t,\omega) \to X(t,\omega) \qquad \text{for all} \, t \geq 0, \omega \in \Omega.$$
  2. If we set $$Y_t^n := n \int_t^{t+1/n} 1_{|X_{s}| \leq n} X_{s} \, ds,$$ then $Y^n \to X$, but as $Y^n$ is not adapted it is not $\mathcal{P}$-measurable.
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  • $\begingroup$ thanks for pointing out that in the right continuous case, $Y^n$ is not adapted! $\endgroup$ – math Mar 1 '14 at 9:28

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