Shift of A martingale by A stopping time

Question

In Karatzas&Shreve Brownian Motion and Stochastic Calculus Problem 3.27: Let $T$ be a bounded stopping time of the filtration $\left\{ \mathscr{F}_{t}\right\} $, which satisfies the usual conditions, and define $\mathscr{\widetilde{F}}_{t}=\mathscr{F}_{T+t}$. Then $\left\{ \mathscr{\widetilde{F}}_{t}\right\} $ also satisfies the usual conditions. And if $\left\{ \widetilde{X}_{t},\mathscr{\widetilde{F}}_{t}:0\le t<\infty\right\}$ is a right-continuous submartingale with $\widetilde{X}_{0}=0$, a.s. P, then $\left\{ X_{t}=\widetilde{X}_{\max\left\{ 0,t-T\right\} },\mathscr{F}_{t}:0\le t<\infty\right\}$ is also a submartingale.

I have solved this problem except one part. My question is that: how can we show that $\widetilde{X}_{\max\left\{ 0,t-T\right\} }$ is adapted $\left\{ \mathscr{F}_{t}\right\}$.

Answer 1

• If $t\le T$, $X_{t}=0$. So knowing all the past before $t$ allows you to know $X$ until that time.

• If $t\ge T$, $X_{t}=\widetilde{X}_{t-T}$. So knowing all the past before $t$ for the filtration $\mathscr{{F}}$ allows you to know all the past before $t-T$ for the filtration $\mathscr{\widetilde{F}}$, which allows you to know $\widetilde X$ until time $t-T$ and therefore $X$ until time $t$.

Formally, let $A\in\mathcal{B}(\mathbb{R})$ and $t\ge T$. We need to prove that $\left\{X_t\in A\right\}\in\mathscr{F}_t$:

\begin{alignat*}{2} \left\{\omega\in\Omega, X_t(\omega)\in A\right\} = \left\{\omega\in\Omega, \widetilde X_{t-T}(\omega)\in A\right\} \end{alignat*} but $\left\{\widetilde X_{t-T}\in A\right\}$ is $\widetilde{\mathscr{F}}_{t-T}$-measurable, which means it is $\mathscr{F}_{t}$-measurable or equivalently $\left\{X_t\in A\right\}\in\mathscr{F}_t$.

Answer 2

The $\widetilde{X}_{(t−T)^+}$ is adapted $\{\mathscr{F}_t\} $, it could be proved as follows.

1. If $ U $ is a $(\widetilde{\mathscr{F}}_t)$ stopping time, then $T+U$ is an $(\mathscr{F}_t)$ stopping time and $$ \widetilde{\mathscr{F}}_U\subset \mathscr{F}_{T+U}. \tag{*}$$ The proof could be verified according the definition of stopping time and $ \widetilde{\mathscr{F}}_U $.

2. If $ S $ is an $(\mathscr{F}_t)$ stopping time, then $ (S-T)^+ $ is a $(\widetilde{\mathscr{F}}_t)$ stopping time and $$ \widetilde{\mathscr{F}}_{(S-T)^+}\subset \mathscr{F}_{T+(S-T)^+} =\mathscr{F}_{S\vee T}.\tag{1} $$

   Proof For $ t\ge 0 $, $$ \{(S-T)^+\le t\}=\{S\le T+t\} \in \mathscr{F}_{T+t}= \widetilde{\mathscr{F}}_t, $$ so $ (S-T)^+ $ is a $(\widetilde{\mathscr{F}}_t)$ stopping time and (1) follows from (*). Q.E.D.

From (1) it is easy to obtain the following $$ 1_{(S>T)}\widetilde{\mathscr{F}}_{(S-T)^+} \subset 1_{(S>T)}\mathscr{F}_{S\vee T} \subset \mathscr{F}_S, \tag{2} $$ where $1_{(S>T)}\mathscr{F}_{S\vee T} = \{1_{(S>T)}\xi: \xi \text{ is an } \mathscr{F}_{S\vee T}-\text{measurable variable}\}.$

Now taking $ S=t $ in (2) we could get $$ \widetilde{X}_{(t−T)^+}=1_{(t>T)} \widetilde{X}_{(t−T)^+} \in \mathscr{F}_t . $$