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Lecture 28 — Kronecker’s Lemma and Applications

This lecture completes the arc of:

  • Kolmogorov’s three–series theorem
  • Kronecker’s lemma (full proof via integration by parts for step functions)
  • Marcinkiewicz–Zygmund Strong Laws for $1≤𝑝<2$.

Let ${X_n}_{n\ge1}$ be independent.

We recall two prior results:

  1. If
    \(\sum_{n=1}^\infty \operatorname{Var}(X_n) < \infty,\) then
    \(\sum_{n=1}^\infty (X_n - E[X_n]) \quad\text{converges a.s.}\)

  2. Kolmogorov’s Three–Series Theorem:

    \[\sum X_n \text{ converges a.s.} \iff \begin{cases} \text{(i)} & \forall A>0,\ \sum P(\vert X_n\vert >A)<\infty,\\ \text{(ii)} & \sum \operatorname{Var}(Y_n)<\infty,\quad Y_n=X_n\mathbf{1}_{\{\vert X_n\vert \le A\}},\\ \text{(iii)} & \sum E[Y_n]\ \text{converges}. \end{cases}\]

Durrett Probability 4.1e - Theorem 2.5.4 Three Series Theorem

Let $X_1, X_2, …$ be independent.$ Let $A>0$ and let $Y_i=X_i1{\vert X_i\vert \le A}.$ In order that $\sum{n=1}^\infty X_n$ converse a.s. it is necessary and sufficient that \((i)\quad\sum_{n=1}^\infty P(\vert X_n\vert > A) < \infty,\) \((ii)\quad\sum_{n=1}^\infty \mathbb{E}(Y_n) \text{ converges},\) \((iii)\quad\sum_{n=1}^\infty \operatorname{Var}(Y_n) < \infty.\).

Durrett Probability 4.1e - Theorem 2.5.3 Kolmogorov’s Variance Criterion

If $X_1, X_2, …$ are independent and have $\mathbb{E}[X_n]=0$. If \(\sum_{n=1}^\infty \operatorname{Var}(X_n) < \infty,\) then with probability one $\sum_{n=1}^\infty X_n(\omega)$ converges.

1. A Useful Lemma from the Three–Series Proof

(See page 1 of the PDF.)

Let $S_n = \sum_{k=1}^n X_k$.

Lemma

If

  • $\sup_{n\ge1} \vert S_n\vert < \infty$ a.s., and
  • $E\left( \sup_{n\ge1} X_n^2\right) < \infty$,

then

\[\sum_{n=1}^\infty \operatorname{Var}(X_n) < \infty,\]

hence

\[\sum_{n=1}^\infty \bigl(X_n - E[X_n]\bigr) \text{ converges a.s.}\]

A corollary used in the proof of the ⇒ direction in the Three–Series theorem:

If $X_n$ are independent, centered, uniformly bounded $\vert X_n\vert \le A$, and
$\sup_n \vert S_n\vert <\infty$ a.s.,
then
\(\sum X_n \quad\text{converges a.s.}\)

2. Kronecker’s Lemma

This is the central new result for Lecture 28.

Let $x_n\in\mathbb{R}$, and let ${a_n}_{n\ge1}$ satisfy:

  • $a_n > 0$,
  • $a_n \uparrow \infty$.

Kronecker’s Lemma

If
\(\sum_{n=1}^\infty \frac{x_n}{a_n} \quad\text{converges},\) then
\(\frac{1}{a_n}\sum_{k=1}^n x_k \;\xrightarrow[n\to\infty]{}\; 0.\)

This relates convergence of weighted series to the behavior of normalized partial sums.

Durret Probablity 4.1e - Theorem 2.5.5 Kronecker’s Lemma

if $a_n\uparrow\infty$ and $\sum_{n=1}^\infty x_n/a_n$ converges then \(a_n^{-1}\sum_{m=1}^n x_m\to 0.\) —

3. Proof of Kronecker’s Lemma (as in your notes, pp. 1–2)

Define
\(b_n = \sum_{k=1}^n \frac{x_k}{a_k}, \qquad b_0=0.\)

Since $\sum x_k/a_k$ converges, $b_n \to b_\infty$.

We use a discrete integration-by-parts identity for right–continuous step functions $F$ and $G$:

\[\int_{(a,b]} F\, dG + \int_{(a,b]} G\, dF = [F\cdot G]_{a}^{b} + \sum_{a<x<b} \Delta F(x)\Delta G(x).\]

We apply this with:

  • $F(x)=\sum_{i\ge1} b_i \mathbf{1}_{{i<x\le i+1}}$,
  • $G(x)=\sum_{i\ge1} a_i \mathbf{1}_{{i<x\le i+1}}$.

From the jump computations (page 2 diagram in the PDF):

\[\sum_{i=1}^n b_i(a_i - a_{i-1}) = b_n a_n -\sum_{i=1}^n a_i(b_i - b_{i-1}).\]

But
\(b_i - b_{i-1}=\frac{x_i}{a_i}.\)

Thus:

\[\frac{1}{a_n}\sum_{k=1}^n x_k = b_n - \sum_{i=1}^n b_{i-1}\left(\frac{a_i - a_{i-1}}{a_n}\right).\]

As $n\to\infty$:

  • $b_n\to b_\infty$,
  • $\sum_{i=1}^n b_{i-1}(a_i - a_{i-1})/a_n \to b_\infty$,

so the difference tends to $0$.
This proves Kronecker’s lemma.

4. Application: Weighted SLLN for Independent, Centered Sequences

(See page 2 of the PDF.)

Theorem

Let ${X_n}$ be independent with $E[X_n]=0$.
Assume:

  • $a_n \uparrow \infty$,
  • \[\sum_{n=1}^\infty \frac{E[X_n^{2}]}{a_n^{2}} < \infty.\]

Then:

  1. \[\frac{S_n}{a_n} \xrightarrow{a.s.} 0.\]
  2. Consequently
    \(\sum_{n=1}^\infty \frac{X_n}{a_n} \quad\text{converges a.s.}.\)

Reason:
The variance condition implies
\(\sum \operatorname{Var}(X_n / a_n) < \infty.\) Thus
\(\sum X_n/a_n \text{ converges a.s.}\) by Kolmogorov’s variance criterion, and Kronecker’s lemma yields $S_n/a_n \to 0$.

5. Example: Normalization by $\sqrt{n}(\log n)^{1+\varepsilon}$

Suppose:

  • $E[X_n]=0$,
  • $\sup_n E[X_n^2] \le C$,
  • $\varepsilon>0$.

Then

\[\sum_{n=1}^\infty E\left[ \frac{X_n^2}{n(\log n)^{1+\varepsilon}} \right] \le C \sum_{n=1}^\infty \frac{1}{n(\log n)^{1+\varepsilon}} < \infty\]

(by the integral test, page 2 of the PDF).

Hence:

\[\frac{S_n}{\sqrt{n}(\log n)^{1+\varepsilon}} \xrightarrow[n\to\infty]{a.s.} 0.\]

This is a refined law of large numbers: the normalization grows just slowly enough for the variance series to converge.

6. Marcinkiewicz–Zygmund Strong Law ($1 \le p < 2$)

(See page 3 of the PDF.)

Let $X_1, X_2,\dots$ be iid, with:

  • $E[X]=0$,
  • $E\vert X\vert ^p < \infty$, for $1 \le p < 2$.

Theorem (M–Z SLLN)

\[\frac{S_n}{n^{1/p}} \xrightarrow{a.s.} 0.\]

Sketch of Proof

Define truncated variables:

\[Y_k = X_k \mathbf{1}_{\{\vert X_k\vert ^p \le k\}}, \qquad T_n = \sum_{k=1}^n Y_k.\]

  1. By Markov and the $p$-moment hypothesis:
    \(\sum_{k=1}^\infty P(X_k \ne Y_k) =\sum_{k=1}^\infty P(\vert X\vert ^p > k) < \infty.\) So $X_k = Y_k$ eventually a.s.

  2. Show
    \(\sum \operatorname{Var}\left(\frac{Y_k}{k^{1/p}}\right) < \infty.\) This uses a “good sequence” argument (as in Lecture 24).

  3. Then the series
    \(\sum \frac{Y_k}{k^{1/p}}\) converges a.s.

  4. By Kronecker’s lemma (with $a_n = n^{1/p}$):

    \[\frac{T_n}{n^{1/p}} \to 0 \quad\text{a.s.}\]

Since $T_n = S_n$ for large $n$, this proves the M–Z strong law.

Durrett Probability 4.1e - Theorem 2.5.8 Marcinkiewicz-Zygmund Strong Law

Let $X_1, X_2, …$ be iid with $E[X_1]=0$ and $E[\vert X_1\vert ^p] < \infty$ where $1 < p < 2$. If $S_n = X_1 +…+X_n$ then $S_n/n{1/p}\to 0 \text{ a.s.}$ —

Cheat-Sheet Summary — Lecture 28

  • Kronecker lemma:
    Convergence of $\sum x_n/a_n$ forces
    \(S_n/a_n \to 0.\)

  • Weighted SLLN:
    If
    $\sum E[X_n^2]/a_n^2 <\infty$, then
    \(S_n/a_n \to 0 \quad\text{a.s.}\)

  • Normalization examples:
    \(a_n = \sqrt{n}(\log n)^{1+\varepsilon} \quad \Rightarrow \quad S_n/a_n \to 0.\)

  • Marcinkiewicz–Zygmund SLLN:
    For iid with $E\vert X\vert ^p <\infty$, $1\le p<2$, \(S_n/n^{1/p} \to 0 \text{ a.s.}\)

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