Show that $\sum_{n\in\mathbb N}X_n\left(W_t-W^{\tau_{n-1}}_t\right)$ is adapted whenever $W$ is adapted

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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$
  • $\tau_n:\Omega\to[0,\infty]$ be an $\mathcal F$-stopping time with $$\tau_n\le\tau_{n+1}\tag1$$ for all $n\in\mathbb N_0$ and $$\tau_n\xrightarrow{n\to\infty}\infty\tag2$$
  • $X_n:\Omega\to\mathbb R$ be $\mathcal F_{\tau_{n-1}}$-measurable for all $n\in\mathbb N$

Note that $$N_t:=\inf\left\{n\in\mathbb N_0:\tau_n\ge t\right\}\;\;\;\text{for }t\ge0$$ takes values in $[0,\infty)$ and hence $$\Phi_t:=\sum_{n\in\mathbb N}1_{\left\{\:\tau_{n-1}\:<\:t\:\right\}}X_n=\sum_{n=1}^{N_t}X_n\;\;\;\text{for all }t\ge0\tag3$$ is well-defined. Let $W:\Omega\times[0,\infty)\to\mathbb R$ be a stochastic process on $(\Omega,\mathcal A,\operatorname P)$ and$^1$ $$(\Phi\cdot W)_t:=\sum_{n\in\mathbb N}X_n\left(W_t-W^{\tau_{n-1}}_t\right)=\sum_{n=1}^{N_t}X_n\left(W_t-W_{\tau_{n-1}}\right)\;\;\;\text{for }t\ge0\;.\tag4$$

I'm curious which continuity and measurability properties $\Phi\cdot W$ inherits from $W$.

From the right-hand side of $(4)$ it should be clear that $\Phi\cdot W$ inherits any continuity property of $W$; or is there something that we need to worry about?

Now, I want to know if $\Phi\cdot W$ is $\mathcal F$-adapted whenever $W$ is $\mathcal F$-adapted.

It's clear that this is the case, if $W$ is left-continuous: Let $n\in\mathbb N$ and $t\ge0$. Then, $$Z:=X_n\left(W_t-W^{\tau_{n-1}}_t\right)=\begin{cases}X_n\left(W_t-W_{\tau_{n-1}}\right)&\text{on }A:=\left\{\tau_{n-1}\le t\right\}\\0&\text{on }\Omega\setminus A\end{cases}\;.\tag5$$ Now, $W$ is left-continuous and $\mathcal F$-adapted. We can show that this implies that $W_{\tau_{n-1}}$ is $\mathcal F_{\tau_{n-1}}$-measurable and hence $\left.X_n\right|_A$ and $\left.W\right|_A$ are both $\left.\mathcal F_{\tau_{n-1}}\right|_A$-measurable. We can show that $$\left.\mathcal F_{\tau_{n-1}}\right|_A=\mathcal F_{\tau_{n-1}}\cap A\subseteq\mathcal F_t^{\tau_{n-1}}=\mathcal F_{\tau_{n-1}}\cap\mathcal F_t\subseteq\mathcal F_t\tag6$$ and hence $\left.Z\right|_A$ is $\mathcal F_t$-measurable. Thus, $Z$ is piecewise $\mathcal F_t$-measurable; hence $\mathcal F_t$-measurable.


$^1$ As usual, $$W^{\tau_{n-1}}_t:=W_{\tau_{n-1}\:\wedge\:t}\;.$$