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The original problems are shown in picture linked below. Thank you.

Problem $\mathbf{2.10.19}$. (On the second Borel-Cantelli lemma.) Prove the following variants of the second Borel-Cantelli lemma: given an arbitrary sequence of (not necessarily independent) events $A_1, A_2, \ldots$, one can claim that:

(a) If

$$ \sum_{n=1}^\infty {\sf P}(A_n) = \infty \quad \text{and} \quad \liminf_n \frac{\sum_{i,k=1}^n {\sf P}(A_iA_k)}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2} = 1, $$

then (Erdös and Rényi [$37$]) ${\sf P}(A_n\text{ i.o.}) = 1$.

(b) If

$$ \sum_{n=1}^\infty {\sf P}(A_n) = \infty \quad \text{and} \quad \liminf_n \frac{\sum_{i,k=1}^n {\sf P}(A_iA_k)}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2} = L, $$

then (Kochen and Stone[$64$], Spitser [$125$]) $L \geq 1$ and ${\sf P}(A_n\text{ i.o.}) = 1/L$.

(c) If

$$ \sum_{n=1}^\infty {\sf P}(A_n) = \infty \quad \text{and} \quad \liminf_n \frac{\sum_{1\leq i<k\leq n} [{\sf P}(A_iA_k)-{\sf P}(A_i){\sf P}(A_k)]}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2} \leq 0, $$

then (Ortega and Wschebor [$92$]) ${\sf P}(A_n\text{ i.o.}) = 1$.

(d) If $\sum_{n=1}^\infty {\sf P}(A_n) = \infty$ and

$$ \alpha_H = \liminf_n \frac{\sum_{1\leq i<k\leq n} [{\sf P}(A_iA_k)-H{\sf P}(A_i){\sf P}(A_k)]}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2}, $$

where $H$ is an arbitrary constant, then (Petrov [$95$]) ${\sf P}(A_n\text{ i.o.}) \geq \frac{1}{H+2\alpha_H}$ and $H+2\alpha_H \geq 1$.

Original at https://i.stack.imgur.com/oxcKz.jpg

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    $\begingroup$ What have you tried ? $\endgroup$
    – MtGlasser
    Jul 28 '20 at 1:45
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    $\begingroup$ I've transcribed the text from the original image; if someone could kindly check for typos, that would be great, thanks! $\endgroup$
    – Brian Tung
    Jul 28 '20 at 2:04
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    $\begingroup$ @Brian Tung, No mistakes! (Though you should have quoted it) $\endgroup$
    – UmbQbify
    Jul 28 '20 at 2:24
  • $\begingroup$ @UmbQbify-Key20-: I cited as much as I knew; the OP didn't provide any more of a reference than what you see, I don't think (other than what's in the title). $\endgroup$
    – Brian Tung
    Jul 28 '20 at 3:02
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    $\begingroup$ Or did you mean the quote bars I just added? $\endgroup$
    – Brian Tung
    Jul 28 '20 at 3:05
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This is the Kochen-Stone Lemma. I will state this result and a short proof for you. But first a little technical result.

Lemma: If $0\neq f\in L_2$ and $\mathbb{E}[f]\geq0$, then for any $0<\lambda<1$ $$\begin{align} \mathbb{P}\big[f>\lambda \mathbb{E}[f]\big]\geq (1-\lambda)^2 \frac{\big(\mathbb{E}[f]\big)^2}{\mathbb{E}[|f|^2]}\tag{1}\label{anty-cheby}. \end{align} $$

Here is a short proof:

By Hölder's inequality $$ \mathbb{E}[f]=\int_{\{f\leq \lambda\mathbb{E}[f]\}}f \,d\Pr+ \int_{\{ f>\lambda\mathbb{E}[f]\}} f\,d\mathbb{P} \leq \lambda\mathbb{E}[f] + \Big(\|f\|_2\sqrt{\Pr[f>\lambda\mathbb{E}[f]]}\Big). $$

Here is the result that we will used to get the version of Corel Cantelly closer to what you described in your problem.

Lemma(Kochen-Stone) Let $\{A_n\}\subset\mathscr{F}$. If $\sum_n\mathbb{P}[A_n]=\infty$, then $$\begin{align} \mathbb{P}\big[\bigcap_{n\geq1}\bigcup_{k\geq n}A_k\big]\geq\limsup_n\frac{\Big(\sum^n_{k=1}\mathbb{P}[A_k]\Big)^2}{\sum^n_{k=1}\sum^n_{m=1}\mathbb{P}[A_k\cap A_m]}\tag{2}\label{ko-sto} \end{align} $$

Here is a Sketch of the proof:

Without loss of generality, we assume that $\mathbb{P}[A_n]>0$ for all $n$. Let $f_n=\sum^n_{k=1}\mathbb{1}_{A_k}$, $f=\sum_{n\geq1}\mathbb{1}_{A_n}$, and for any $0<\lambda<1$, define $B_{n,\lambda}=\big\{f_n>\lambda\mathbb{P}[f_n]\big\}$. Observe that $$ A=\bigcap_{n\geq 1}\bigcup_{k\geq n}A_k=\{f=\infty\}\supset\bigcap_{n\geq 1}\bigcup_{k\geq n}B_{k,\lambda}=B_\lambda; $$ then, by $\eqref{anty-cheby}$, we obtain $$ \mathbb{P}[A]\geq\mathbb{P}[B_\lambda]\geq\limsup_{n\rightarrow\infty}\mathbb{P}[B_{n,\lambda}]\geq(1-\lambda)^2\limsup_n\frac{\big(\mathbb{E}[f_n]\big)^2}{\mathbb{E}[f^2_n]}. $$ Letting $\lambda\rightarrow1$ gives $\eqref{ko-sto}$.

Using Kochen-Stone's Lemma one can prove the following version of the reverse Borel-Cantelli Lemma

Theorem (reverse Borel-Cantelli) Suppose $\{A_n\}\subset\mathscr{F}$ is such that for any $i\neq j$, $\mathbb{P}[A_i\cap A_j]\leq\mathbb{P}[A_i]\mathbb{P}[A_j]$. If $\sum_n\mathbb{P}[A_n]=\infty$, then $\mathbb{P}\Big[\bigcap_{n\geq1}\bigcup_{k\geq n}A_k\Big]=1$.

Here is a short proof:

Denote by $A=\bigcap_{n\geq 1}\bigcup_{k\geq n}A_k$. Let $a_n=\sum^n_{k=1}\mathbb{P}[A_k]$,, $b_n=\sum_{i\neq j}\mathbb{P}[A_i]\mathbb{P}[ A_j]$, and $c_n=\sum^n_{k=1}\mathbb{P}^2[A_k]$. By Kochen--Stone's lemma we have $$ \mathbb{P}[A]\geq\limsup_n\frac{c_n+b_n}{a_n+b_n} $$ From $a^2_n=c_n+b_n\leq a_n+b_n$, and $a_n\nearrow\infty$, it follows that $b_n\nearrow\infty$ and $\lim_n\tfrac{c_n}{b_n}=0=\lim_n\frac{a_n}{b_n}$. Therefore, $\mathbb{P}[A]=1$.


Reference: https://projecteuclid.org/euclid.ijm/1256059668

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  • $\begingroup$ Thank you very much ! $\endgroup$
    – 张若冲
    Jul 31 '20 at 14:58

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