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Suppose I have a process $Y_t$, which is predictable. We can assume that the filtration satisfies the usual condition. Then, let $A\in \mathcal{F}_t$. I was wondering about the following. Is $\mathbf1_AY_t$ still a predictable process? This seems intuitively clear, but how can we prove this rigorously?

  • So $A\in\mathcal{F}_t$ for some fixed $t>0$? Then $Z_t=1_A Y_t$ is not necessarily adapted, is it? – Stefan Hansen Jul 18 '13 at 09:09
  • The reason why I asked this question is the following: in mathematical finance we want to prove that a set is upward directed. For example we have the set $K_x:={\theta: \int \theta_udS_u\ge -x}$, where $\theta$ is predictable and $S$-integrable. Note $S$ is here a general semimartingale (RCLL). Now we define something like this: $\theta:=\mathbf1_A\theta_1+\mathbf1_{A^c}\theta_2$ for $\theta_i\in K_x$ and claim that $\theta\in K_x$. All properties are clear, but just the predictability is not. Or can we say, well $\int\theta dS=\mathbf1_A\int\theta_1dS+\mathbf1_{A^c}\int\theta_2dS$ and... –  Jul 18 '13 at 09:18
  • everything is fine? –  Jul 18 '13 at 09:18
  • @StefanHansen sorry, forgot to ping! –  Jul 18 '13 at 11:31
  • I don't know the answer to that. Maybe you should pose it as a different question. – Stefan Hansen Jul 19 '13 at 09:24
  • Usually one has to assume some structure on the set $K_x$. This called bifurcation, which ensures that the $\theta$ of this form is still a element of $K_x$. I think in general this is not the case, as @StefanHansen pointed out. – math Jul 19 '13 at 12:12
  • @n.c. Thanks for your comment. I will check this condition! –  Jul 19 '13 at 12:13

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