Doob's Optional Stopping Theorem/Discrete Time/Supermartingale

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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:

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

$\Box$

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.

$\Box$

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:

\(\ds \size {X_{n \wedge T} - X_0}\) \(=\) \(\ds \size {\sum_{j \mathop = 0}^{n \wedge T - 1} \paren {X_{k + 1} - X_k} }\)
\(\ds \) \(\le\) \(\ds \sum_{j \mathop = 0}^{n \wedge T - 1} \size {X_{k + 1} - X_k}\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds M \sum_{j \mathop = 0}^{n \wedge T - 1} 1\) almost surely, by hypothesis
\(\ds \) \(=\) \(\ds M \paren {n \wedge T}\)
\(\ds \) \(\le\) \(\ds M T\) Definition of Pointwise Minimum of Extended Real-Valued Functions

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.

$\blacksquare$


Sources

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