Lecture 20 — Borel–Cantelli, Subsequence Convergence, DCT Applications
1. Borel–Cantelli Lemma (Version I)
Let $(\Omega,\mathcal{F},P)$ be a probability space.
Let $A_n \in \mathcal{F}$, $n \ge 1$, and assume
Then
\[P(A_n \text{ i.o.}) = 0.\]Recall:
\[A_n \text{ i.o.} = \bigcap_{m=1}^\infty \bigcup_{k=m}^\infty A_k = \{\omega : \omega \in A_k \text{ for infinitely many }k\}.\]Proof
Let
\[N(\omega) = \sum_{k=1}^\infty \mathbf{1}_{A_k}(\omega).\]- $N < \infty$ means the event occurs only finitely often.
- By monotone convergence,
If $P(N=\infty)>0$, then $\mathbb{E}[N]=\infty$, contradiction.
Therefore $P(N=\infty)=0$. Hence
2. Why Borel–Cantelli Is Important
From convergence in probability to almost sure convergence.
Convergence in probability is inherently “one-dimensional”:
\(X_n \xrightarrow{P} X.\)
Almost sure convergence is “many-dimensional” and stronger:
\(X_n \xrightarrow{a.s.} X.\)
BC(I) allows extraction of a subsequence that does converge almost surely.
3. Corollary: Convergence in probability implies an a.s. convergent subsequence
If $X_n \xrightarrow{P} X$, then there exists a subsequence $(X_{n_k})$ such that
\[X_{n_k} \xrightarrow{a.s.} X.\]Proof (following the notes)
WLOG set $X=0$.
Since $X_n \xrightarrow{P} 0$, for each $k$ there exists $n_k$ such that
We choose $n_k$ strictly increasing: \(n_1 < n_2 < n_3 < \cdots.\)
Then
\[\sum_{k=1}^\infty P(\vert X_{n_k}\vert > 1/k^2) \le \sum_{k=1}^\infty \frac{1}{k^2} < \infty.\]By Borel–Cantelli(I),
\[P(\vert X_{n_k}\vert > 1/k^2 \text{ i.o.}) = 0.\]Thus a.s., there exists a random finite $L(\omega)$ such that for all $k\ge L(\omega)$,
\[\vert X_{n_k}\vert \le 1/k^2 \to 0.\]Hence
\[X_{n_k} \xrightarrow{a.s.} 0 = X.\]4. Application: Dominated Convergence with Only Convergence in Probability
Suppose:
- $X_n \xrightarrow{P} X$,
- $\vert X_n\vert \le Y$ for all $n$,
- $E[Y] < \infty$.
Question: Does $E[X_n] \to E[X]$?
Answer: Yes.
Reason
From the corollary, extract a subsequence $X_{n_k} \to X$ a.s.
Since $\vert X_{n_k}\vert \le Y$ and $E[Y]<\infty$, Dominated Convergence gives:
To upgrade subsequence convergence to full convergence, assume the contrary:
There exists $\varepsilon_0>0$ and a subsequence $n_k$ such that
$\vert E[X_{n_k}]\vert \ge \varepsilon_0$ for all $k$.
Refine to a further subsequence $X_{n_{k_\ell}}\to X$ a.s.
DCT again gives $E[X_{n_{k_\ell}}]\to E[X]$, contradicting $\vert E[X_{n_k}]\vert \ge\varepsilon_0$.
Thus:
5. WLLN and SLLN Context
Up to now, the WLLN says:
If ${X_n}$ are iid and $E\vert X\vert <\infty$, then
\[\frac{S_n}{n} \xrightarrow{P} E[X].\]We also know:
\[xP(\vert X\vert > x) \to 0\]is sufficient for some versions of WLLN/SLLN criteria (your notes allude to this heavy-tail condition).
Strong Law of Large Numbers (SLLN)
If ${X_n}$ iid and $E\vert X\vert < \infty$, then
\[\frac{S_n}{n} \xrightarrow{a.s.} E[X].\]The last part of your notes (page 2 of the PDF) expands $(\sum_{k=1}^n a_k)^4$ into the multinomial expansion:
\[\left(\sum_{k=1}^n a_k\right)^4 = \sum_{k=1}^n a_k^4 + 6\!\!\sum_{k<\ell}\! a_k^2 a_\ell^2 + 4\!\!\sum_{k\neq\ell}\! a_k^3 a_\ell + \cdots\]This is the beginning of the classical fourth-moment method used to show:
\[E\!\left[\left(\frac{S_n}{n}\right)^4\right] \to (E[X])^4,\]which provides control strong enough to apply Kolmogorov’s SLLN.
Though your notes only sketch this, the purpose is:
moment bounds + Borel–Cantelli → almost sure convergence.
Cheat-Sheet Summary (Lecture 20)
- Borel–Cantelli (I): If $\sum P(A_n)<\infty$, then $A_n$ occurs only finitely often a.s.
- Subsequence principle: Convergence in probability implies existence of an almost surely convergent subsequence.
- DCT via subsequences:
If $X_n\to X$ in probability and $\vert X_n\vert \le Y$ with $EY<\infty$, then $E[X_n]\to E[X]$. - SLLN link: Borel–Cantelli and moment expansions lead to almost sure convergence of averages for iid integrable variables.
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