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Let

  • $T>0$
  • $I:=(0,T]$
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\in\overline I}$ be a filtration of $\mathcal A$
  • $\tau_n:\Omega\to\overline I$ be an $\mathcal F$-stopping time with $$\tau_n\le\tau_{n+1}\tag1$$ for all $n\in\mathbb N_0$ and $$\tau_n\xrightarrow{n\to\infty}T\tag2$$
  • $X_n:\Omega\to\mathbb R$ be $\mathcal F_{\tau_{n-1}}$-measurable for all $n\in\mathbb N$

Now, let $$\Phi_t:=\sum_{n\in\mathbb N}1_{\left\{\:\tau_{n-1}\:<\:t\:\right\}}X_n\;\;\;\text{for }t\in\overline I\;.\tag3$$ How can we show that $$\Phi_t=\sum_{n\in\mathbb N}1_{(\tau_{n-1},\tau_n]}(t)Y_n\;\;\;\text{for all }t\in\overline I\tag4$$ for some $\mathcal F_{\tau_{n-1}}$-measurable $Y_n$?

I guess we can simply rewrite $(3)$ in order to obtain $(4)$. However, I didn't find the right way to do that.

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Since

$$1_{\{\tau_{n-1}<t\}} = 1_{\{\tau_{n-1} <t \leq \tau_n\}} + 1_{\{\tau_n<t\}} = \dots = \sum_{k \geq n} 1_{\{\tau_{k-1} <t \leq \tau_k\}} $$

we have

$$\begin{align*} \Phi_t = \sum_{n=1}^{\infty} 1_{\{t>\tau_{n-1}\}} X_n = \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} 1_{\{\tau_{k-1}<t \leq \tau_k\}} X_n &= \sum_{k=1}^{\infty} \sum_{n=1}^k 1_{\{\tau_{k-1}<t \leq \tau_k\}} X_n \\ &= \sum_{k=1}^{\infty} 1_{\{\tau_{k-1}<t \leq \tau_k\}} \underbrace{\sum_{n=1}^k X_n}_{=: Y_k}. \end{align*}$$

$Y_k$ is clearly $\mathcal{F}_{\tau_k}$-measurable.

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  • $\begingroup$ Thank you. I was sure that it would be easy, but I the only ideas I had were way too complicated. Please note that $Y_k$ is even $\mathcal F_{\tau_{k-1}}$-measurable. $\endgroup$ – 0xbadf00d May 17 '17 at 11:14
  • $\begingroup$ @0xbafd00d You are welcome. $\endgroup$ – saz May 17 '17 at 16:43

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