Lecture 26 — Kolmogorov’s 0–1 Law and Maximal Inequalities
This lecture introduces Kolmogorov’s 0–1 Law, Kolmogorov’s maximal inequality, and Lévy’s maximal inequality for symmetric random walks.
1. Kolmogorov’s 0–1 Law
Let ${X_i}_{i\ge1}$ be independent random variables.
Tail σ–Fields
For each $n\ge1$, define the tail σ–field
\[\mathcal{F}_{[n,\infty)} = \sigma(X_n, X_{n+1},\dots).\]Define the tail σ–field
\[\mathcal{I} = \bigcap_{n=1}^\infty \mathcal{F}_{[n,\infty)}.\]Theorem (Kolmogorov 0–1 Law)
If $A\in \mathcal{I}$, then
\[P(A) \in \{0,1\}.\]Thus tail events occur with probability either zero or one.
Sketch of the Proof (as in notes, page 1)
For each $n$,
- $\mathcal{F}_n = \sigma(X_1,\dots,X_n)$,
- $\mathcal{F}{[n+1,\infty)} = \sigma(X{n+1}, X_{n+2},\dots)$.
Because the variables are independent,
\[\mathcal{F}_n \ \text{and}\ \mathcal{F}_{[n+1,\infty)} \quad\text{are independent.}\]Since $\mathcal{I}\subseteq\mathcal{F}_{[n+1,\infty)}$, $\mathcal{I}$ is independent of $\mathcal{F}_n$ for every $n$.
Furthermore,
\[\sigma(X_1, X_2,\dots) = \sigma\left(\bigcup_n \mathcal{F}_n\right)\]so $\mathcal{I}$ is independent of itself:
\[A\in\mathcal{I} \ \Rightarrow\ P(A\cap A)=P(A)P(A),\]so $P(A)=P(A)^2$.
Hence $P(A)=0$ or $1$.
Example
Let $S_n = \sum_{k=1}^n X_k$.
The event
\(\{S_n \text{ converges}\}\)
is a tail event, so the probability of convergence is 0 or 1 (no intermediate values).
2. Kolmogorov’s Maximal Inequality
Let ${X_i}_{1\le i\le n}$ be independent, mean‐zero variables with finite variances:
\[E[X_i]=0,\qquad E[X_i^2]<\infty.\]Let
\[S_k = \sum_{i=1}^k X_i,\qquad k=1,\dots,n.\]Theorem (Kolmogorov Maximal Inequality)
For any $x>0$,
\[P\Big(\max_{1\le k\le n} \vert S_k\vert \ge x\Big) \le \frac{E[S_n^2]}{x^2}.\]This is stronger than Chebyshev’s inequality, because it controls the maximum of partial sums.
Proof (following your notes, pages 1–2)
Let
\[A_k = \big\{ \vert S_k\vert \ge x,\ \vert S_j\vert <x \text{ for } j<k \big\}.\]The events $A_k$ are disjoint and
\[\bigcup_{k=1}^n A_k = \left\{\max_{1\le k\le n}\vert S_k\vert \ge x\right\}.\]Compute:
\[E[S_n^2] = \sum_{k=1}^n \int_{A_k} S_n^2\, dP \ge x^2 \sum_{k=1}^n P(A_k) = x^2\, P\left(\max_{1\le k\le n}\vert S_k\vert \ge x\right).\]Why the inequality holds:
On $A_k$, the decomposition $S_n = S_k + (S_n - S_k)$ gives:
\[E\big[S_k(S_n-S_k) \mathbf{1}_{A_k}\big] = E\big[S_k \mathbf{1}_{A_k}\big]\, E[S_n-S_k] = 0,\]using independence and $E[S_n-S_k]=0$.
Thus
\[E[S_n^2] \ge \sum_k \int_{A_k} S_k^2\, dP \ge x^2 \sum_k P(A_k),\]as desired.
3. Lévy’s Maximal Inequality (Symmetric Case)
Let ${X_k}_{1\le k\le n}$ be independent and symmetric:
\[X_k \stackrel{d}{=} -X_k,\]equivalently $P(X_k \ge 0) \ge \tfrac12$.
Let $S_k = X_1 + \cdots + X_k$.
Claim (from notes, pages 2–3)
\[P\left(\max_{1\le k\le n} S_k \ge t\right) \le 2\, P(S_n \ge t),\qquad t>0.\]And also
\[P\left(\max_{1\le k\le n} \vert S_k\vert \ge t\right) \le 2\, P(\vert S_n\vert \ge t).\]These inequalities are central tools in proving the LIL and many fluctuation results.
Proof outline
Use symmetry:
$S_n - S_k = X_{k+1}+\dots+X_n$ is independent of $S_k$, and symmetric.
Key steps (see page 3 diagram):
- \[\{\max_k S_k \ge t\} = \bigcup_k \{S_k\ge t,\ S_j<t \text{ for } j<k\}.\]
-
On the event ${S_k\ge t,\ S_j<t\ \forall j<k}$, by symmetry,
\[P(S_n - S_k \ge 0) = \frac12.\] -
Replace indicators using independence:
\[P(\max_k S_k \ge t) \le 2\, P(S_n\ge t).\]
A parallel argument gives the two-sided inequality.
Cheat-Sheet Summary — Lecture 26
- Kolmogorov 0–1 Law: Any tail event for independent variables has probability 0 or 1.
- Maximal inequality (Kolmogorov):
\(P(\max_{k\le n}\vert S_k\vert \ge x) \le E[S_n^2]/x^2.\) - Lévy maximal inequality (symmetric case):
\(P(\max_{k\le n} S_k \ge t) \le 2 P(S_n \ge t).\) \(P(\max_{k\le n} \vert S_k\vert \ge t) \le 2 P(\vert S_n\vert \ge t).\)
These inequalities are foundational for the Law of the Iterated Logarithm and more advanced results.
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