13 — Upcrossing Lemma, Doob’s Inequality, and MGCT
Upcrossing Lemma (Doob)
Let ${X_n, \mathcal{F}n}{n\ge 0}$ be a submartingale.
Fix $a < b$.
Define a sequence of stopping times:
- $T_0 = 1$
- $T_{2k-1} = \inf{n > T_{2k-2} : X_n \le a}$, for $k \ge 1$
- $T_{2k} = \inf{n > T_{2k-1} : X_n \ge b}$
The number of upcrossings of $[a,b]$ before time $n$ is: \(U_n^{a,b} = \sup\{k : T_{2k} \le n\}.\)
Upcrossing Bound
\((b-a)\,E[U_n^{a,b}] \le E[(X_n-a)^+] - E[(X_0-a)^+] \le E[X_n^+] + |a|.\)
(See diagram in notes illustrating two upcrossings.)
Doob’s Inequality (Submartingale)
If ${X_n}$ is a submartingale, then for any $\lambda > 0$, \(\lambda\, P\left(\sup_{k\le n} X_k \ge \lambda\right) \le E[X_n^+].\)
A sketch of the proof uses the upcrossing lemma and predictable processes.
Predictable Process $H_m$
Define: \(H_m = \sum_{k\ge 1} \left( \mathbf{1}_{\{T_{2k}\ge m\}} - \mathbf{1}_{\{T_{2k-1} > m\}} \right).\)
- $H_m$ is measurable with respect to $\mathcal{F}_{m-1}$.
- Therefore $H_m$ is predictable.
Using integration-by-parts for discrete stochastic integrals: \((b-a) U_n^{a,b} \le (H\cdot X)_n.\)
Issue With Downcrossings
If we downward-cross, we “re-enter the market”.
Fix with truncated process:
\(Y_k = a + (X_k - a)^+
=
\begin{cases}
X_k, & X_k > a,\\
a, & X_k \le a.
\end{cases}\)
Then:
- $ {Y_k} $ is also a submartingale.
- Upcrossings of $X$ correspond to modified upcrossings of $Y$.
Thus, \((b-a) U_n^{a,b}(X) \le (H \cdot Y)_n.\)
Define $K_m = 1 - H_m$.
Then:
\((H\cdot Y)_n + (K\cdot Y)_n = Y_n - Y_0.\)
\(E[(K\cdot Y)_n] \ge 0,\qquad (K\cdot Y)_0=0.\)
Thus, \((b-a) E[U_n^{a,b}] \le E[Y_n - Y_0].\)
Martingale Convergence Theorem (MGCT)
Let ${X_n,\mathcal{F}_n}$ be a submartingale.
Assume:
\(\sup_n E[X_n^+] < \infty.\)
Then:
- $X_n \to X$ a.s.
-
$E X < \infty$.
Example 1 — Simple Symmetric Random Walk
Let \(S_n = \sum_{k=1}^n \varepsilon_k, \qquad P(\varepsilon_k = \pm 1)=1/2, \qquad S_0 = 0.\)
Let \(T = \inf\{n : S_n = 1\}.\)
The stopped process ${S_{T \wedge n}}$ is a submartingale and \(\sup_n E[(S_{T\wedge n})^+] \le 1.\)
Thus, by MGCT: \(S_{T\wedge n} \xrightarrow{a.s.} S_T = 1, \qquad P(T < \infty) = 1.\)
But: \(E(S_{T\wedge n}) = 0 \quad\forall n, \qquad E(S_T) = 1.\)
So ${S_{T\wedge n}}$ is not uniformly integrable (UI).
Example 2 — Exponential Martingale
Let $Z_k$ i.i.d. $N(0,1)$.
Define:
\(S_n = \sum_{k=1}^n Z_k,
\qquad
Y_n = e^{S_n - \frac{n}{2}}.\)
Using the mgf of the normal: \(E[e^{Z_1}] = e^{1/2}.\)
Compute: \(E[Y_n|\mathcal{F}_{n-1}] = Y_{n-1} E[e^{Z_n - 1/2}] = Y_{n-1}.\)
Thus ${Y_n}$ is a nonnegative martingale.
- By MGCT: $Y_n \to Y$ a.s.
- Since $E(Y_n)=1$, the limit satisfies $E(Y)=1$.
But: \(Y_n = e^{S_n - n/2} = e^{n\left(\frac{S_n}{n} - \frac12\right)}.\)
By the Strong Law of Large Numbers: \(\frac{S_n}{n} \to 0 \quad a.s.\)
Hence: \(Y_n = e^{-n/2 + o(n)} \to 0 \quad a.s.\)
Thus: \(Y = 0 \quad \text{a.s.}\) but \(E(Y)=1.\)
Conclusion: ${Y_n}$ is not uniformly integrable.
Summary
- The upcrossing lemma provides control over oscillations of a submartingale.
- Doob’s inequality links the supremum of a submartingale to its terminal value.
- MGCT guarantees almost sure convergence under a mild integrability condition.
- However, without uniform integrability, expectation may not pass to the limit.
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