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
From Adapted Stochastic Process at Stopping Time is Measurable with respect to Stopped Sigma-Algebra:


 * $X_S$ is $\FF_S$-measurable.

Then from Conditional Expectation of Measurable Random Variable:


 * $\expect {X_S \mid \FF_S} = X_S$ almost surely.

Also from Doob's Optional Stopping Theorem: Discrete Time: Supermartingale:


 * $X_T$ 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:

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:


 * $\expect {X_{k + 1} \mid \FF_k} \le X_k$

and so, from Condition for Conditional Expectation to be Almost Surely Non-Negative:


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

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.