Lecture 22 — Extended Borel–Cantelli II and Applications
1. Extended Borel–Cantelli II (Pairwise Independence Version)
Theorem (Durrett Thm. 2.3.8).
Let ${A_k}_{k\ge 1}$ be pairwise independent events (not necessarily mutually independent).
Assume
Then
\[\frac{\sum_{m=1}^n \mathbf{1}_{A_m}} {\sum_{m=1}^n P(A_m)} \xrightarrow{a.s.} 1.\]This is a stronger form of BC II: it replaces full independence with only pairwise independence.
2. Example: Pairwise Independent but Not Mutually Independent
The notes use a classical symmetric Bernoulli example:
Let $r_1, r_2, r_3$ satisfy
- $r_1, r_2$ i.i.d. symmetric Bernoulli,
- define $r_3 = r_1 \cdot r_2$.
Then:
- Any pair $(r_i, r_j)$ is independent.
- But $(r_1, r_2, r_3)$ are not jointly independent.
For example (page 1):
\[P(r_1=1, r_2=1, r_3=-1) = 0,\]but marginal probabilities suggest it would be $1/8$ if they were mutually independent.
This example shows why theorem 2.3.8 is genuinely stronger than classical BC II.
3. Proof of Extended BC II (Pairwise Independent Case)
Let
\[S_n = \sum_{m=1}^n \mathbf{1}_{A_m}, \qquad \mu_n = \mathbb{E}[S_n] = \sum_{m=1}^n P(A_m).\]Since indicators are pairwise independent:
\[\mathrm{Var}(S_n) = \sum_{m=1}^n P(A_m)\bigl(1-P(A_m)\bigr) \le \mu_n.\]By Chebyshev:
\[P(\vert S_n - \mu_n\vert > \delta \mu_n) \le \frac{\mathrm{Var}(S_n)}{\delta^2 \mu_n^2} \le \frac{1}{\delta^2 \mu_n}.\]Since $\mu_n \to \infty$, the RHS is summable along a subsequence.
4. Constructing a Subsequence ${n_k}$
Goal: choose $n_k$ such that
\[k^2 \le \mu_{n_k} \le k^2 + 1.\]Why can we do this?
- $\mu_n$ increases to $\infty$,
- increments satisfy $0 \le \mu_{n+1} - \mu_n = P(A_{n+1}) < 1$,
- so $\mu_n$ crosses each interval $[k^2, k^2+1)$.
Thus the sequence ${n_k}$ exists.
5. Apply Chebyshev to the Subsequence
For this subsequence:
\[P\left( \vert S_{n_k} - \mu_{n_k}\vert > \delta(k^2 + 1) \right) \le \frac{1}{\delta^2 (k^2 + 1)}.\]Since
\[\sum_{k=1}^\infty \frac{1}{k^2 + 1} < \infty,\]Borel–Cantelli I gives:
\[\frac{S_{n_k} - \mu_{n_k}}{k^2 + 1} \xrightarrow{a.s.} 0.\]Thus:
\[\frac{S_{n_k}}{\mu_{n_k}} \xrightarrow{a.s.} 1.\]Given that
\[\frac{k^2}{k^2+1} \le \frac{\mu_{n_k}}{k^2+1} \le 1,\]we conclude:
\[\frac{S_{n_k}}{\mu_{n_k}} \xrightarrow{a.s.} 1.\]6. Passing from Subsequence to Full Sequence
For any $n_k \le n \le n_{k+1}$:
\[\frac{S_{n_k}}{\mu_{n_{k+1}}} \le \frac{S_n}{\mu_n} \le \frac{S_{n_{k+1}}}{\mu_{n_k}}.\]The left and right ratios both converge to 1 almost surely.
This traps the middle ratio, so:
This completes the proof of the extended BC II.
7. Application: Record Values
Let ${X_i}$ iid with a continuous distribution.
Define the record event:
Classical fact:
Because all orderings of ${X_1,\dots,X_k}$ are equally likely, the probability that $X_k$ is the new maximum is
Thus
\[\sum_{k=1}^\infty P(A_k) = \sum_{k=1}^\infty \frac{1}{k} = \infty.\]Pairwise independence of ${A_k}$
Your notes show (pages 2–3) that:
- $A_j$ and $A_k$ (with $j<k$) are pairwise independent.
- Intuitively: whether $X_j$ is the max among the first $j$ values is unrelated to whether $X_k$ is the max among the first $k$ values.
Formally:
\[P(A_j \cap A_k) = \frac{(j-1)!(k-j)!}{k!} = \frac{1}{j k} = P(A_j) P(A_k).\]Hence the extended BC II applies.
Conclusion
\[\frac{\#\{\text{records up to } n\}}{\sum_{k=1}^n 1/k} = \frac{\sum_{k=1}^n \mathbf{1}_{A_k}} {H_n} \xrightarrow{a.s.} 1,\]where $H_n$ is the $n$-th harmonic number.
Since $H_n \sim \log n$,
\[\boxed{ \text{Number of record values up to } n \sim \log n \quad (a.s.). }\]
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