Optional Stopping Theorem - Proof

Proof

Let X τ denote the stopped process, it is also a martingale (or a submartingale or supermartingale, respectively). Under condition (a) or (b), the random variable Xτ is well defined. Under condition (c) the stopped process X τ is bounded, hence by Doob's martingale convergence theorem it converges a.s. pointwise to a random variable which we call Xτ.

If condition (c) holds, then the stopped process X τ is bounded by the constant random variable M := c. Otherwise, writing the stopped process as

gives |Xtτ| ≤ M for all t ∈ ℕ0, where

.

By the monotone convergence theorem

.

If condition (a) holds, then this series only has a finite number of non-zero terms, hence M is integrable.

If condition (b) holds, then we continue by inserting a conditional expectation and using that the event {τ > s} is known at time s (note that τ is assumed to be a stopping time with respect to the filtration), hence

\begin{align}\mathbb{E} &=\mathbb{E}+\sum_{s=0}^\infty \mathbb{E}\bigl\cdot\mathbf{1}_{\{\tau>s\}}}_{\le\,c\,\mathbf{1}_{\{\tau>s\}}\text{ a.s. by (b)}}\bigr]\\ &\le\mathbb{E}+c\sum_{s=0}^\infty\mathbb{P}(\tau>s)\\ &=\mathbb{E}+c\,\mathbb{E}<\infty,\\ \end{align}

where a representation of the expected value of non-negative integer-valued random variables is used for the last equality.

Therefore, under any one of the three conditions in the theorem, the stopped process is dominated by an integrable random variable M. Since the stopped process X τ converges almost surely to Xτ , the dominated convergence theorem implies

By the martingale property of the stopped process,

hence

Similarly, if X is a submartingale or supermartingale, respectively, change the equality in the last two formulas to the appropriate inequality.

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