5 — Filtrations

Let $(\Omega, \mathcal F, P)$ be a probability space.

Filtration

A filtration is a sequence of $\sigma$-fields \(\{\mathcal F_n\}_{n\ge 1}, \qquad \mathcal F_n \subseteq \mathcal F_{n+1}.\)

Stopping Time

A random time $T:\Omega\to \mathbb Z^+$ is a stopping time if \(\{T=n\}\in \mathcal F_n,\qquad n\ge 1.\)

Sigma-field at a Stopping Time

\(\mathcal F_T = \{A\in\mathcal F : A\cap\{T=n\} \in\mathcal F_n \text{ for all } n\ge 1\}.\)

Natural Filtration

Given a stochastic process ${X_n}_{n\ge 1}$: \(\mathcal F_n = \sigma(X_1,\dots,X_n),\qquad n\ge 1.\)

Then \(\mathcal F_T = \sigma(X_{n\wedge T}).\)

Monotone Stopping Times

If $T_1 < T_2$ almost surely and both are stopping times, then \(\mathcal F_{T_1} \subseteq \mathcal F_{T_2}.\)

Corollary.
If $T_1 < T_2 < \dots$ is an increasing sequence of stopping times, then
${\mathcal F_{T_k}}_{k\ge 1}$ is a filtration.

Theorem

Let ${X_n}_{n\ge 1}$ be IID.
Let $T$ be a stopping time with respect to the natural filtration and assume \(P(T<\infty)=1.\)

Then:

  1. \[\{X_{T+n}\}_{n\ge 1} \stackrel{D}{=} \{X_n\}_{n\ge 1}.\]
  2. ${X_{T+n}}_{n\ge 1}$ is IID and independent of $\mathcal F_T$.

Example: Simple Symmetric Random Walk

\(S_0 = 0,\qquad S_n = \sum_{k=1}^n X_k,\qquad X_k=\pm 1 \text{ with probability } \tfrac12.\)

Stopping times:

  • \[T_1 = \min\{n>0 : S_n = 0\}.\]
  • \[T_i = \min\{n > T_{i-1} : S_n = 0\}.\]

Then $T_i - T_{i-1}$ are IID.

Wald’s First Equation

Let $S_n = \sum_{k=1}^n X_k$ and $T$ a stopping time with $E[T]<\infty$ and $E|X_1|<\infty$.

Then: \(E(S_T)=E(X_1)E(T).\)

Sketch: \(S_T = \sum_{n\ge 0} (S_{T\wedge(n+1)} - S_{T\wedge n}),\) and \(S_{T\wedge(n+1)} - S_{T\wedge n} = X_{n+1}1_{\{T\ge n+1\}}.\) Since the indicator is $\mathcal F_n$-measurable and $X_{n+1}$ is independent: \(E(S_{T\wedge(n+1)} - S_{T\wedge n}) = E(X_1)P(T\ge n+1).\)

Summing gives Wald’s equation.

Generalization

If $X_{n+1}$ is independent of a smaller filtration $\mathcal G_n$ and $T$ is a stopping time with respect to $\mathcal G_n$, Wald’s equation still holds.

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