Doob's Optional Stopping Theorem/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 $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.

Let:


 * $\map {X_T} \omega = \map {X_{\map T \omega} } \omega \map {\chi_{\set {\omega \in \Omega : \map T \omega < \infty} } } \omega$

for each $\omega \in \Omega$.

Suppose one of the following conditions holds:


 * $(1) \quad$ $T$ is bounded
 * $(2) \quad$ $T$ is finite almost surely, and there exists an integrable random variable $Y$ with $\size {X_n} \le Y$ for $n \in \Z_{\ge 0}$
 * $(3) \quad$ $T$ is integrable, and there exists a real number $M > 0$ such that for each $n \in \Z_{\ge 0}$ we have $\size {X_{n + 1} - X_n} \le M$ almost surely.

Then $X_T$ is integrable and:


 * $\expect {X_T} \le \expect {X_0}$

Proof
We first show that if $T$ is finite almost surely, then:


 * $\map {X_{n \wedge T} } \omega \to \map {X_T} \omega$

for almost all $\omega \in \Omega$.

Let $\omega \in \Omega$ be such that $\map T \omega = s < \infty$.

Then for $n \ge s$, we have:


 * $\map {X_{n \wedge T} } \omega = \map {X_s} \omega$

From Constant Sequence in Topological Space Converges, we have:


 * $\map {X_{n \wedge T} } \omega \to \map {X_s} \omega = \map {X_T} \omega$

for all $\omega \in \Omega$ such that $\map T \omega < \infty$.

Note that since $T$ is finite almost surely, we have that:


 * $\map {X_{n \wedge T} } \omega \to \map {X_T} \omega$

for almost all $\omega \in \Omega$, as required.

So:


 * $X_{n \wedge T} \to X_T$

almost surely.

We have from Stopped Supermartingale is Supermartingale: Corollary:


 * $X_{n \wedge T}$ is integrable for each $n \in \N$

and:


 * $\expect {X_{n \wedge T} } \le \expect {X_0}$

So from Expectation is Linear, and since $X_0$ is integrable, this is equivalent to:


 * $\expect {X_{n \wedge T} - X_0} \le 0$

We aim to show:


 * $\ds \expect {X_T - X_0} = \lim_{n \mathop \to \infty} \expect {X_{n \wedge T} - X_0}$

in each of the three cases.

We then have half the result from Limits Preserve Inequalities.

We will then have to establish that $X_T$ is integrable.

Case $(1)$
Suppose that $\map T \omega \le t$ for all $\omega \in \Omega$, with $t \in \Z_{\ge 0}$.

Then for $n > t$ we have:


 * $X_{n \wedge T} = X_T$

That is, for $n > t$:


 * $\expect {X_{n \wedge T} } = \expect {X_T}$

Since $X_{n \wedge T}$ is integrable, so is $X_T$.

Also:


 * $\expect {X_T - X_0} = \expect {X_{n \wedge T} - X_0} \le 0$

Case $(2)$
Note that if:


 * $\size {X_n} \le Y$ for all $n \in \N_{\ge 0}$

then:


 * $\size {X_{n \wedge T} } \le Y$ for all $n \in \N_{\ge 0}$.

Then, by the Triangle Inequality, we have:


 * $\size {X_{n \wedge T} - X_0} \le \size {X_0} + Y$ for all $n \in \N_{\ge 0}$

Since:


 * $X_{n \wedge T} \to X_T$ almost surely

and $X_0$ and $Y$ are both integrable, we can apply Lebesgue's Dominated Convergence Theorem to obtain:


 * $\ds \expect {X_T - X_0} = \lim_{n \mathop \to \infty} \expect {X_{n \wedge T} - X_0} \le 0$

and that $X_T$ is integrable.

Case $(3)$
From Integrable Function is A.E. Real-Valued, $T$ is again almost surely finite.

So we again have:


 * $X_{n \wedge T} \to X_T$ almost surely.

We can calculate:

almost surely.

Since $T$ is integrable, so is $M T$ from Scalar Multiple of Integrable Function is Integrable Function.

So we can again apply Lebesgue's Dominated Convergence Theorem to obtain:


 * $\ds \expect {X_T - X_0} = \lim_{n \mathop \to \infty} \expect {X_{n \wedge T} - X_0} \le 0$

and that $X_T$ is integrable.