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Lecture 29 — Feller’s Theorem and Completion of the M–Z Strong Law

1. Marcinkiewicz–Zygmund Strong Law (recap)

Let ${X_k}_{k\ge1}$ be iid with

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

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

Theorem (M–Z Strong Law)

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

Proof idea (page 1)

  1. Define truncated variables \(Y_k = X_k\,\mathbf{1}_{\{\vert X_k\vert \le k^{1/p}\}}.\)

  2. Bad events are summable \(\sum_k P(Y_k\ne X_k) =\sum_k P(\vert X\vert >k^{1/p}) =\sum_k P(\vert X\vert ^p > k) <\infty.\) So by Borel–Cantelli I, $X_k = Y_k$ eventually a.s.

  3. It suffices to show \(\frac{T_n}{n^{1/p}} \to 0, \qquad T_n = \sum_{k=1}^n Y_k.\)

  4. Find a good normalizing sequence.
    A key lemma (pages 1–2):

    If $a_n$ is “good” and
    $\sum P(\vert X\vert >a_n)<\infty$, then
    $\sum E(Y_n^2)/a_n^2 <\infty$,
    where $Y_n = X_n \mathbf{1}_{\vert X_n\vert \le a_n}$.

    For M–Z, take $a_n = n^{1/p}$.
    The sequence ${a_n / n^{1/p}} = {1}$ is (trivially) non-decreasing, so $a_n$ is “good”.

  5. Thus, \(\sum_{k=1}^\infty \operatorname{Var}\left( \frac{Y_k}{k^{1/p}} \right) <\infty.\)

  6. By the variance–summability criterion, \(\sum_k \frac{Y_k - E[Y_k]}{k^{1/p}} \quad\text{converges a.s.}\)

  7. Use Kronecker’s lemma with $a_n = n^{1/p}$:
    \(\frac{1}{n^{1/p}} \sum_{k=1}^n (Y_k - E[Y_k]) \xrightarrow{\text{a.s.}} 0.\)

  8. It remains to prove \(\frac{1}{n^{1/p}} \sum_{k=1}^n E[Y_k] \to 0.\)

2. The Expectation Term

(Page 1–2 of notes; explicit computation.)

For
\(Y_k = X_k \mathbf{1}_{\{\vert X_k\vert \le k^{1/p}\}},\) we have \(E[Y_k] = E\bigl[ X\,;\ \vert X\vert \le k^{1/p}\bigr] = E[X] - E\bigl[X\,;\ \vert X\vert >k^{1/p}\bigr] = -E\bigl[ X\,;\ \vert X\vert >k^{1/p}\bigr],\) since $E[X]=0$.

Thus, \(\vert E[Y_k]\vert \le E\left( \vert X\vert \,;\ \vert X\vert >k^{1/p}\right).\)

To bound this, use identities from integration by parts (page 2):

For $x>k^{1/p}$, \(\vert X\vert = \frac{\vert X\vert ^p}{\vert X\vert ^{p-1}} \le \frac{\vert X\vert ^p}{k^{(p-1)/p}}.\)

Hence, \(\vert E[Y_k]\vert \le k^{-(p-1)/p} \, E\left(\vert X\vert ^p\,;\ \vert X\vert > k^{1/p}\right).\)

As $k\to\infty$, \(E\left(\vert X\vert ^p\,;\ \vert X\vert >k^{1/p}\right) \to 0,\) because $E\vert X\vert ^p<\infty$.
Therefore:

\[\vert E[Y_k]\vert \le C k^{-(p-1)/p} \quad\text{for large }k.\]

Sum bound

Your notes (page 2) compute:

\[\sum_{k=1}^n k^{1/p -1} \le C n^{1/p}.\]

Therefore:

\[\frac{1}{n^{1/p}} \sum_{k=1}^n \vert E[Y_k]\vert \le \frac{1}{n^{1/p}} \sum_{k=1}^n k^{-(p-1)/p} \le \frac{1}{n^{1/p}} \cdot C n^{1/p} \to 0.\]

Thus, \(\boxed{ \frac{1}{n^{1/p}} \sum_{k=1}^n E[Y_k] \ \longrightarrow\ 0. }\)

Combining everything:

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

This finishes the Marcinkiewicz–Zygmund SLLN.

3. Feller’s Theorem

(William Feller’s theorem, page 2–3.)

Let ${X_k}$ be iid with $E\vert X\vert <\infty$.
Take any positive sequence $a_n$ such that

  • $a_n>0$,
  • $\dfrac{a_n}{n}$ is non-decreasing,
    i.e. $a_k/k \le a_n/n$ for $k\le n$.

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

Theorem (Feller)

(a) If
\(\sum_{n=1}^\infty P(\vert X\vert \ge a_n) < \infty,\) then \(\overline{\lim_{n\to\infty}} \frac{\vert S_n\vert }{a_n} =0 \quad\text{a.s.}\) In particular, \(\frac{S_n}{a_n} \to 0 \quad\text{a.s.}\)

(b) If
\(\sum_{n=1}^\infty P(\vert X\vert \ge a_n) = \infty,\) then \(\lim_{n\to\infty} \frac{\vert S_n\vert }{a_n} = \infty \quad\text{a.s.}\)

Thus the series $\sum P(\vert X\vert \ge a_n)$ completely determines the almost-sure growth rate of $S_n$.

Durrett Probablity 4e Theorem 2.5.9. Feller’s Theorem

Let $X_1, X_2, \ldots$ be i.i.d. with $E\vert X_1\vert = \infty$ and let $S_n=X_1 + … + X_n$. Let $a_n$ be a sequence of positive numbers with $a_n/n$ increasing. Then $\limsup_{n\to\infty} \vert S_n\vert /a_n = 0$ or $\infty$ according as $\sum_n P(\vert X_1\vert \ge a_n) < \infty$ or $= \infty$.

4. Proof of Part (b)

(Page 2–3.)

Assume
\(\sum_n P(\vert X\vert \ge a_n) = \infty.\)

Then for any fixed $k\ge 1$,

\[\sum_n P(\vert X\vert \ge k a_n) \ge \frac{1}{k} \sum_{n=k}^\infty P(\vert X\vert \ge a_n) = \infty.\]

By Borel–Cantelli II, \(P(\vert X_n\vert \ge k a_n \ \text{i.o.}) = 1.\)

Thus, \(\overline{\lim_{n\to\infty}} \frac{\vert X_n\vert }{a_n} \ge k \quad \text{a.s.}\)

Since $k$ is arbitrary, \(\overline{\lim_{n\to\infty}} \frac{\vert X_n\vert }{a_n} = \infty \quad\text{a.s.}\)

Finally, using \(\vert X_n\vert = \vert S_n - S_{n-1}\vert \le \vert S_n\vert + \vert S_{n-1}\vert \le 2 \max(\vert S_n\vert ,\vert S_{n-1}\vert ),\) we get \(\frac{\vert S_n\vert }{a_n} \to \infty \quad\text{a.s.}\)

5. Proof Idea for Part (a)

(Page 3—sketch.)

If
\(\sum P(\vert X\vert \ge a_n) < \infty,\) then:

  • Large jumps occur only finitely often.

  • One applies the same decomposition used in the M–Z proof: \(S_n = \sum_{k=1}^n X_k = \sum (Y_k - E[Y_k]) + \sum E[Y_k],\) with $Y_k$ truncated at level $a_k$.

  • Use:
    • variance summability of $Y_k/a_k$,
    • Kronecker’s lemma,

to conclude $S_n/a_n\to0$ a.s.

Your notes say “look at book for proof,” but the outline is exactly parallel to the M–Z case.

Cheat-Sheet Summary — Lecture 29

  • M–Z SLLN: If $E\vert X\vert ^p<\infty$, $1\le p<2$, then
    \(S_n / n^{1/p} \to 0 \text{ a.s.}\)

  • The proof requires bounding
    $\sum E[Y_k]/n^{1/p}$,
    achieved via $x > k^{1/p}\Rightarrow x^{p-1}/k^{(p-1)/p}\ge 1$.

  • Feller’s Theorem: Given $a_n$ with $a_n/n$ non-decreasing:

    • If $\sum P(\vert X\vert \ge a_n)<\infty$, then $S_n/a_n \to 0$ a.s.
    • If $\sum P(\vert X\vert \ge a_n)=\infty$, then $S_n/a_n\to\infty$ a.s.

This provides a precise growth-rate dichotomy for partial sums of iid integrable variables.

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