I apologize in advance for any minor technicalities that might occur in informally proving this important theorem with minimal-to-no measure theory. We first present this theorem, known as Optional Stopping Theorem (OST) or alternatively Doob’s Optional Sampling theorem.
Optional Stopping Theorem
Suppose we have a discrete-time martingale \(M = (X_n)_{n \geq 0}\) where \(\mathbb{E}[\vert X_n \vert] < \infty\) and \(\mathbb{E}[X_{n + 1} \mid X_n, \dots, X_0] = X_n\). Furthermore, suppose we have a “bounded” stopping time \(T \in \mathbb{Z}\) (i.e. \(\exists \ N \in \mathbb{Z}\) s.t. \(N \geq T\) almost surely). Then, martingale \(M\) stopped at \(T\) is a martingale and we have:
\[\mathbb{E}[X_T] = \mathbb{E}[X_0]\]Note that this above property is non-trivial as \(T\) is a random variable dependent on the observed values \(X_{0}, \dots, X_T\).
We first define the stopped process. Defining \(T \wedge n = \text{min}(\{T, n\})\), we can define the stopped process of our martingale \((X_n)_{n \geq 0}\) as \(Y_n = X_{T \wedge n}\). Note that because \(N < \infty\) and \(\vert Y_n \vert \leq \underset{0 \leq k \leq N}{\max} \vert X_k \vert \implies\) we can be very certain \(Y_n\) is finite. With these definitions out of the way, we’ll first prove that \((Y_n)_{n \geq 0}\) is a martingale (w.r.t observed events \(\{X_n, \dots, X_0\} \supseteq \{Y_n, \dots, Y_0\}\) ) or equivalently that:
\[\mathbb{E}[Y_{n + 1} \mid X_n, \dots, X_0] = Y_n\]First note that \(\mathbb{E}[Y_{n + 1} \mid X_n, \dots, X_0] = \mathbb{E}[X_{T \wedge (n + 1)} \mid X_n, \dots, X_0]\). Note that we can break \(\mathbb{E}[X_{T \wedge (n + 1)} \mid X_n, \dots, X_0]\) into two different cases:
(1) Case One. If \(T \leq n\).
In this case, first note that \(T \leq n \implies T < n + 1 \implies T \wedge (n + 1) = T\) and so:
\[\mathbb{E}[X_{T \wedge (n + 1)} \mid X_n, \dots, X_0] = \mathbb{E}[X_{T} \mid X_n, \dots, X_0]\]Furthermore, \(T \leq n \implies X_{T} \in \{X_N, \dots, X_0\}\) and so \(\mathbb{E}[X_{T} \mid X_n, \dots, X_0] = X_T\) (for the same reason \(\mathbb{E}[X_2 \mid X_3, \dots, X_0] = X_2\). The value is already observed.)
(2) Case Two. If \(T > n\).
In this case, \(T \geq n + 1 \implies T \wedge (n + 1) = n+ 1\) and so:
\[\mathbb{E}[X_{T \wedge (n + 1)} \mid X_n, \dots, X_0] = \mathbb{E}[X_{n + 1} \mid X_n, \dots, X_0] = X_n\]Putting these two cases together, we have that:
\[\mathbb{E}[Y_{n + 1} \mid X_n, \dots, X_0] = \mathbb{E}[X_{T \wedge (n + 1)} \mid X_n, \dots, X_0] = X_T\mathbf{1}\{T \leq n\} + X_n\mathbf{1}\{T > n\} = X_{T \wedge n} = Y_n\]which concludes that \((Y_n)_{n \geq 0}\) is a martingale. A natural implication of this is that \(\forall \ n, \mathbb{E}[Y_n] = \mathbb{E}[Y_0] = \mathbb{E}[X_0]\). But then \(\forall \ n \geq N \geq T, \mathbb{E}[X_T] = \mathbb{E}[Y_n] = \mathbb{E}[X_0] \implies \mathbb{E}[X_T] = \mathbb{E}[X_0]\).
This last property has some actually neat implications. Namely, stopping at a good time (i.e. \(T\)) does not change the expected outcome from when you first started. For example, if you’re doing a symmetric random walk with \(\$1\) in both directions, your expected outcome when stopping after hitting \(\$5\) is still the same as when you started – zero dollars. More profoundly, timing the martingale often has no (expected) benefit.