# Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time/Discrete Time/Supermartingale

## Theorem

Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.

Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-supermartingale.

Let $S$ and $T$ be bounded stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$ and $S \le T$.

Let $\FF_S$ be the stopped $\sigma$-algebra associated with $S$.

Let $X_T$ and $X_S$ be $X$ at the stopping times $T$ and $S$.

Then:

$\expect {X_T \mid \FF_S} \le X_S$ almost surely.

## Proof

$X_S$ is $\FF_S$-measurable.
$\expect {X_S \mid \FF_S} = X_S$ almost surely.
$X_T$ and $X_S$ are integrable.

Then from Conditional Expectation is Linear, we have:

$\expect {X_T \mid \FF_S} \le X_S$ almost surely.
$\expect {X_T - X_S \mid \FF_S} \le 0$ almost surely.

From Condition for Conditional Expectation to be Almost Surely Non-Negative, we aim to show that:

$\expect {\paren {X_T - X_S} \cdot \chi_A} \le 0$ for each $A \in \FF_S$.

Suppose that $t \in \Z_{\ge 0}$ is such that:

$\map T \omega \le t$ for all $\omega \in \Omega$.

Then we have:

 $\ds X_T$ $=$ $\ds X_S + \sum_{k \mathop = S}^{T - 1} \paren {X_{k + 1} - X_k}$ $\ds$ $=$ $\ds X_S + \sum_{k \mathop = 0}^\infty \paren {X_{k + 1} - X_k} \cdot \chi_{S \le k \le T - 1}$ $\ds$ $=$ $\ds X_S + \sum_{k \mathop = 0}^\infty \paren {X_{k + 1} - X_k} \cdot \chi_{S \le k < T}$ since $T$ is $\Z_{\ge 0}$-valued, so $k \le T - 1$ if and only if $k < T$

Since $T \le t$, we have:

$\ds X_S + \sum_{k \mathop = 0}^\infty \paren {X_{k + 1} - X_k} \cdot \chi_{S \le k < T} = X_S + \sum_{k \mathop = 0}^{t - 1} \paren {X_{k + 1} - X_k} \cdot \chi_{S \le k < T}$

Then for $A \in \FF_S$ we have by Expectation is Linear:

$\ds \expect {X_T \cdot \chi_A} = \expect {X_S \cdot \chi_A} + \sum_{k \mathop = 0}^{t - 1} \expect {\paren {X_{k + 1} - X_k} \cdot \chi_{S \le k < T} \cdot \chi_A}$

By Characteristic Function of Intersection, we have:

$\chi_{S \le k < T} \cdot \chi_A = \chi_{\set {S \le k < T} \cap A}$

Since $A \in \FF_S$, we have:

$\set {S \le k} \cap A \in \FF_k$

We also have:

$\set {T > k} = \set {T \le k - 1}^c \in \FF_{k - 1}$

Since $\sequence {\FF_n}_{n \ge 0}$ is a filtration:

$\FF_{k - 1} \subseteq \FF_k$

So:

$\set {S \le k < T} \cap A = \paren {\set {S \le k} \cap A} \cap \set {T > k} \in \FF_k$

Since $\sequence {X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale, we have:

$\expect {X_{k + 1} \mid \FF_k} \le X_k$
$\expect {\paren {X_{k + 1} - X_k} \cdot \chi_A} \le 0$

for each $A \in \FF_k$.

So, we have:

$\expect {\paren {X_{k + 1} - X_k} \cdot \chi_{S \le k < T} \cdot \chi_A} \le 0$

So:

$\expect {X_T \cdot \chi_A} \le \expect {X_S \cdot \chi_A}$

that is:

$\expect {\paren {X_T - X_S} \cdot \chi_A} \le 0$

for each $A \in \FF_S$, as required.

$\blacksquare$