25 — Optional Stopping (OST), Backwards Martingales, Simple Random Walk
1. Setup
Let
\(\{X_n,\mathcal F_n\}_{0\le n\le N}\)
be a submartingale (Durrett §5.7 p. 229).
Let $T$ be a stopping time such that
\(0 \le T \le N.\)
We previously showed: \(E(X_0) \le E(X_T) \le E(X_N).\)
2. Decomposition with predictable integrands
Define increments: \(D_k = X_k - X_{k-1}.\)
Let
\((H\cdot X)_n = \sum_{k=0}^n H_k D_k, \qquad H_k\in\mathcal F_{k-1}.\)
If $H_k \ge 0$, then $(H\cdot X)_n$ is a submartingale.
Take
\(H_k = \mathbf 1_{\{k\le T\}},\)
then
\((H\cdot X)_N = X_T - X_0,\quad E[(H\cdot X)_N]\ge 0.\)
Thus $E(X_T) \ge E(X_0)$.
3. Comparing two stopping times $S \le T$
Define
\(Y_n = X_{T\wedge n}.\)
Then ${Y_n}$ is a submartingale, and \(E(Y_0)\le E(Y_S)\le E(Y_N)\) which implies: \(E(X_0)\le E(X_S)\le E(X_T)\le E(X_N).\)
4. Conditional inequality (OST inequality)
If $S\le T$ are stopping times and ${X_n}$ is a submartingale, then: \(E(X_T \mid \mathcal F_S) \ge X_S \qquad\text{a.s.}\)
Proof sketch:
Define
\(\tilde S = S \mathbf 1_A + T \mathbf 1_{A^c},\qquad A\in\mathcal F_S.\)
Then apply the OST expectation bound to get:
\(E(X_{\tilde S})\le E(X_T).\)
Unwrap and use arbitrariness of $A$.
5. Extending to $N=\infty$
Two approaches:
- Uniform integrability approach (works for sub-, super-, and martingales).
- Fatou’s lemma (only for nonnegative submartingales).
6. Theorem (UI version of OST)
Let ${X_n,\mathcal F_n}_{n\ge 0}$ be a UI submartingale.
Let $T$ be a stopping time (possibly $T=\infty$).
Then ${X_{T\wedge n}}_{n\ge 0}$ is UI and:
- $X_{T\wedge n} \to X_T$ a.s.
-
$E X_T <\infty$. - \[E(X_T) \ge E(X_0).\]
Key point:
\(E(X_{T\wedge n}^+)\le E(X_n^+),\)
so the positive parts are uniformly integrable.
7. Corollaries
(1) If $S\le T$ then:
\(E(X_S)\le E(X_T).\)
(2) Conditional version:
\(X_S \le E(X_T \mid \mathcal F_S).\)
Application: Simple Random Walk (non-symmetric)
Let ${\xi_k}_{k\ge 1}$ be iid with \(P(\xi_k = +1)=p,\qquad P(\xi_k=-1)=q,\qquad p>q,\ p+q=1.\)
Random walk: \(S_n = \sum_{k=1}^n \xi_k,\qquad S_0=0.\)
Since $E(\xi_1)=p-q>0$, \(\frac{S_n}{n}\xrightarrow{\text{a.s.}} p-q > 0,\) so $S_n\to+\infty$ a.s., hence for any $b>0$, $T_b<\infty$ a.s.
Hitting probability for level $a<0<b$
Define the harmonic function: \(\varphi(x) = \Big(\frac{q}{p}\Big)^x,\qquad x\in\mathbb Z.\)
Check (page 3 of notes) :contentReference[oaicite:1]{index=1}: \(E\big(\varphi(S_{n+1})\mid \mathcal F_n\big) = \varphi(S_n).\)
Thus ${\varphi(S_n)}$ is a martingale.
Let \(T = T_a\wedge T_b.\)
Since $\varphi(S_T)$ is bounded, the stopped martingale is UI, so: \(E[\varphi(S_T)] = E[\varphi(S_0)] = 1.\)
Since $S_T\in{a,b}$, $$ E[\varphi(S_T)] = P(T_a<T_b)\,\varphi(a)
- P(T_b<T_a)\,\varphi(b). $$
Solve the two-equation system: \(\begin{aligned} P(T_a<T_b)\varphi(a) + P(T_b<T_a)\varphi(b) &= 1,\\ P(T_a<T_b)+P(T_b<T_a)&=1. \end{aligned}\)
Final result: \(\boxed{ P(T_a < T_b) = \frac{\varphi(b)-\varphi(0)}{\varphi(b)-\varphi(a)} }\)
Letting $b\to\infty$: \(P(T_a < \infty) = \Big(\frac{p}{q}\Big)^{\,a}.\)
This is the classical gambler’s ruin formula for a biased walk.
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