Edit this page on GitHub

Lecture 21 — Borel–Cantelli II, SLLN via 4th Moment, and Extensions

1. SLLN via Fourth Moments

Theorem.
Let ${X_k}_{k\ge 1}$ be iid with

  • $\mathbb{E}[X] = \mu$,
  • $\mathbb{E}[X^4] < \infty$.

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

\[\frac{S_n}{n} \xrightarrow{a.s.} \mu.\]

This is a classical proof of the Strong Law via fourth-moment bounds plus Borel–Cantelli.

Proof outline (as in notes)

WLOG assume $\mu = 0$.
To show $S_n/n \to 0$ almost surely, fix $\varepsilon>0$:

\[P(\vert S_n\vert > n\varepsilon) = P(S_n^4 > n^4\varepsilon^4) \le \frac{\mathbb{E}[S_n^4]}{n^4\varepsilon^4}.\]

Now expand the fourth moment (page 1 of the PDF):

\[\mathbb{E}[S_n^4] = n\,\mathbb{E}[X^4] + 3n(n-1)\,[\mathbb{E}(X^2)]^2 \le C n^2,\]

for some constant $C$, because all mixed odd-order terms vanish (iid with mean 0).

Thus,

\[P(\vert S_n\vert > n\varepsilon) \le \frac{C}{\varepsilon^4} \cdot \frac{1}{n^2}.\]

Since $\sum_{n=1}^\infty n^{-2} < \infty$, Borel–Cantelli(I) implies:

\[P(\vert S_n\vert > n\varepsilon \text{ i.o.}) = 0.\]

Thus a.s.:

\[\limsup_{n\to\infty} \frac{\vert S_n\vert }{n} \le \varepsilon.\]

Because this holds for all $\varepsilon = 1/m$, $m\in\mathbb{N}$, we conclude:

\[\frac{S_n}{n} \xrightarrow{a.s.} 0.\]

Restoring $\mu$ gives the theorem.

2. Corollary: Law of Large Numbers for Bernoulli Indicators

(Notes: top of page 1, “the power of an indicator is the indicator.”)

Let ${A_k}_{k\ge 1}$ be iid events with $P(A_k)=p>0$. Then:

\[\frac{1}{n}\sum_{k=1}^n \mathbf{1}_{A_k} \xrightarrow{a.s.} p.\]

This is just the SLLN applied to iid Bernoulli($p$) variables.

Interpretation:

  • The number of times $A_k$ occurs up to $n$, divided by $n$, converges to $p$.
  • The notes emphasize the meaning: empirical average of event frequencies.

3. Borel–Cantelli Lemma II (Independence Required)

BC II.
If ${A_n}$ are independent and

\[\sum_{n=1}^\infty P(A_n) = \infty,\]

then

\[P(A_n \text{ i.o.}) = 1.\]

Proof sketch (page 2)

Consider the complement of the union:

\[P\!\left(\bigcap_{n=m}^\infty A_n^c \right) = \prod_{n=m}^\infty (1 - P(A_n)) \le \prod_{n=m}^\infty e^{-P(A_n)} = e^{-\sum_{n=m}^\infty P(A_n)} = 0,\]

because the tail sum diverges.

Thus,

\[P\!\left(\bigcup_{n=m}^\infty A_n\right)=1,\]

and hence

\[P(A_n \text{ i.o.}) = 1.\]

A useful inequality highlighted in your notes (yellow box on page 2):

\[1 - x \le e^{-x},\quad x\in[0,1].\]

4. Application: When $E\vert X\vert = \infty$

Let ${X_k}$ be iid and $E\vert X\vert = \infty$. Then:

  1. \[P(\vert X_n\vert \ge n \text{ i.o.}) = 1.\]
  2. \[P\!\left(\lim_{n\to\infty} \frac{S_n}{n} \text{ exists and is finite}\right) = 0.\]

Proof (as in notes, page 2)

Since

\[E\vert X\vert = \int_0^\infty P(\vert X\vert \ge t)\,dt,\]

and $E\vert X\vert =\infty$, the integral diverges. The notes approximate the integral by the discrete sum:

\[\sum_{n=0}^\infty P(\vert X\vert \ge n) = \infty.\]

Because ${X_k}$ are iid,

\[\sum_{n=0}^\infty P(\vert X_n\vert \ge n) = \infty.\]

By BC II:

\[P(\vert X_n\vert \ge n\text{ i.o.}) = 1.\]

Since $\vert X_n\vert \ge n$ infinitely often, the increments in $S_n/n$ fluctuate so wildly that:

\[P\left(\lim_{n\to\infty} \frac{S_n}{n} \text{ exists finitely}\right) = 0.\]

This is the “anti-SLLN”: infinite mean destroys convergence of sample averages.

5. Increment Analysis of $S_n/n$

Your notes (page 2) analyze:

\[\frac{S_{n+1}}{n+1} - \frac{S_n}{n} = \frac{X_{n+1}}{n+1} - \frac{S_n}{n(n+1)}.\]

Observation:

  • If $S_n/n$ converges, the difference must go to 0.
  • Under the infinite-mean case, the left-hand expression does not go to 0 (because $\vert X_{n+1}\vert \ge n+1$ i.o.), so no limit can exist.

6. Extended Borel–Cantelli II (Notes, page 3)

Extension (pairwise independent version).
Let $A_n$ be pairwise independent (not necessarily fully independent), and assume:

\[\sum_{n=1}^\infty P(A_n) = \infty.\]

Then:

\[\frac{\sum_{k=1}^n \mathbf{1}_{A_k}}{\sum_{k=1}^n P(A_k)} \xrightarrow{a.s.} 1.\]

This immediately implies:

\[P(A_n \text{ i.o.}) = 1.\]

Your notes mention: “There’s another extension in the book”—this refers to the Kochen–Stone lemma.

Cheat-Sheet Summary (Lecture 21)

  • SLLN with fourth moments:
    If iid and $\mathbb{E}[X^4]<\infty$, then $S_n/n\to\mu$ a.s.

  • BC II:
    For independent events, divergence of $\sum P(A_n)$ implies $A_n$ occurs infinitely often.

  • Infinite mean case:
    If $E\vert X\vert =\infty$, then $\vert X_n\vert \ge n$ infinitely often a.s., so $S_n/n$ cannot converge.

  • Pairwise independent BC II:
    Under pairwise independence,
    $\sum P(A_n)=\infty \Rightarrow \sum 1_{A_n}/\sum P(A_n)\to1$ a.s.

  • Key inequalities:
    $1-x \le e^{-x}$.
    Fourth-moment expansion for sums.

Edit this page on GitHub

Comments