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

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!

• Do you know the corresponding statement for discrete (sub)martingales $(X_n)_{n \in \mathbb{N}}$?
– saz
Oct 30, 2015 at 18:38
• To be honest, I don't know. Oct 30, 2015 at 20:00
• @Integral hi, do you solve this problem? I solved Problem (ii) but I am stuck with (iii) and the second part of problem 3.24. Sep 15, 2019 at 11:42

## 2 Answers

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 $S\leq T$.

• Just one question, how do you know that $E\left[ \sup_{ 0\leq u \leq t } |X_u|\right] < \infty$ ? Thanks. Nov 6, 2015 at 18:52
• condition (iii) is easily proved in context of earlier exercises there: from 3.23 we see that $\{X_{T\wedge t}, \mathcal{F}_{T\wedge t}, 0\le t<\infty\}$ is a submartingale. In particular, $E(X_{T\wedge t}|\mathcal{F}_{T\wedge s})\ge X_{T\wedge s}$ for $s<t$. Thus we have the required inequality, but with smaller sigma-algebra $\mathcal{F}_{T\wedge s}\subset \mathcal{F}_{s}$. But due to right-continuity we have that $X_{T\wedge s}$ is $\mathcal{F}_{s}$-measurable, thus we have $E(X_{T\wedge t}-X_{T\wedge s}|\mathcal{F}_{T\wedge s})\ge 0$. Assuming that (iii) is wrong we come to contradiction Sep 16, 2017 at 12:57
• Contradiction is that if (iii) is wrong, i.e. reverse inequality is true, we condition both parts of $E(X_{T\wedge t}-X_{T\wedge s}|\mathcal{F}_{s})<0$ on $\mathcal{F}_{T\wedge s}$ and using tower law come to contradiction that $E(X_{T\wedge t}-X_{T\wedge s}|\mathcal{F}_{T\wedge s})<0$ Sep 16, 2017 at 13:04

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.

• To use 3.26 you need first to show the integrability of $X_{T\wedge t}$, but how?
– Feng
Jun 3, 2019 at 9:26