Problem 3.24 of "Brownian Motion & Stochastic Processes" by Karatzas and Shreve - Submartingales and stopping times

Question

I'm doing the problem 3.24 of Brownian Motion and Stochastic Processes by Karatzas and Shreve. There is two specific parts troubling me, I need some help to see what to do. Here is the problem:

Suppose that $ \{ X_t, \mathcal{F}_t \ : \ 0 \leq t < \infty \}$ is a right-continuous submartingale and $T$ is a stopping time of $ \{ \mathcal{F}_t \} $. Then $\{ X_{T \ \wedge \ t} , \mathcal{F}_t \ : \ 0 \leq t < \infty \} \ $ is again a submartingale.

In order to prove this, I need to show that

i) $X_{T\wedge t}$ is $\mathcal{F}_t$-measurable for all $t\geq 0$,

ii) $E[\ |X_{T\wedge t}|\ ] < \infty$ for all $t\geq 0$,

iii) $E[X_{T\wedge t}| \mathcal{F}_s] \geq X_{T\wedge s}$ for all $t>s\geq 0$.

Item i is done. Item ii and iii are partially done and it is in these two items that I need help.

In item ii, we can write $$E[\ |X_{T\wedge t}|\ ] = \int |X_{T\wedge t}| \ dP = \int_{ \{ 0\leq T \leq t \} } |X_{T\wedge t}| \ dP + \int_{ \{ t< T \} } |X_{T\wedge t}| \ dP = $$ $$ = \int_{ \{ 0\leq T \leq t \} } |X_T| \ dP + \int_{ \{ t< T \} } |X_t| \ dP \leq \int_{ \{ 0\leq T \leq t \} } |X_T| \ dP + \int |X_t| \ dP =$$ $$ = \int_{ \{ 0\leq T \leq t \} } |X_T| \ dP + E[\ |X_t|\ ]$$

We know $E[\ |X_t|\ ] < \infty$ for $X_t$ is a submartingale, but what can we do about $\int_{ \{ 0\leq T \leq t \} } |X_T| \ dP$ ? I know $\int_{ \{ 0\leq T \leq t \} } |X_T| \ dP \leq \int \sup_{ 0\leq u \leq t } |X_u| \ dP = E[\ \sup_{ 0\leq u \leq t } |X_u|\ ]$, but I don't know what to do from here.

In item iii, showing that $E[X_{T\wedge t}| \mathcal{F}_s] \geq X_{T\wedge s}$ is equivalent to show that $E[X_{T\wedge t}\cdot\textbf{I}_A] \geq E[X_{T\wedge s}\cdot\textbf{I}_A]$ for all $A\in\mathcal{F}_s$. With this in mind, let $A\in\mathcal{F}_s$ and note that

$$E[X_{T\wedge t}\cdot\textbf{I}_A] \geq E[X_{T\wedge s}\cdot\textbf{I}_A] \iff \int_A X_{T\wedge t}\ dP \geq \int_A X_{T\wedge s}\ dP \iff$$

$$\iff \int_{A\cap\{ 0\leq T\leq t \}} X_{T\wedge t}\ dP + \int_{A\cap \{ t< T\}} X_{T\wedge t}\ dP \geq \int_{A\cap\{ 0\leq T\leq s \}} X_{T\wedge s}\ dP + \int_{A\cap \{ s< T\}} X_{T\wedge s}\ dP \iff$$

$$\iff \int_{A\cap\{ 0\leq T\leq t \}} X_T\ dP + \int_{A\cap \{ t< T\}} X_t\ dP \geq \int_{A\cap\{ 0\leq T\leq s \}} X_T\ dP + \int_{A\cap \{ s< T\}} X_s\ dP .$$

Now I would like to come up with a clever argument to show that the last inequality holds. Unfortunately nothing comes to my mind and I can't think in another way to approach this problem.

Thank you very much!

Answer 1

I think you have nearly answered your own question. The finite nature of the inequality in (ii) follows from the finite property of the expectation of the supremum, i.e. $$ \int\limits_{ \{ 0\leq T \leq t \} } |X_T| \ dP \leq E\left[ \sup_{ 0\leq u \leq t } |X_u|\right] < \infty, $$ and hence, $$ E\left[ |X_{T\wedge t}| \right] \leq E\left[ \sup_{ 0\leq u \leq t } |X_u|\right] + E[|X_t|] < \infty. $$ To get condition (iii), apply the conditional expectation value to both sides of the inequality. This gives, $$ E\left[ X_{T\wedge t} | F_S \right] \geq E\left[X_{T\wedge S} | F_S \right], $$ which follows by definition and the given fact that $S\leq T$.

Answer 2

For (iii), try to use Problem 3.26 in the below, i.e., letting $A\leq B$ be two bounded stopping time with respect to the filtration $\{F_t\}$, then we have $E(X_{T\wedge B}|\mathscr{F}_{T\wedge A}) \geq X_{T\wedge A}$, since $T\wedge A \leq T\wedge B$ are two bounded stopping times. Therefore we have $E(X_{T\wedge B}) \geq E(X_{T\wedge A})$, then by 3.26, $Y_t := X_{T\wedge t}$ is a submartingale.