How to prove these four propostions about Borel-Cantelli Lemma in Shriyaev's book 'Probability'?

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

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

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

• What have you tried ? Jul 28 '20 at 1:45
• I've transcribed the text from the original image; if someone could kindly check for typos, that would be great, thanks! Jul 28 '20 at 2:04
• @Brian Tung, No mistakes! (Though you should have quoted it) Jul 28 '20 at 2:24
• @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). Jul 28 '20 at 3:02
• Or did you mean the quote bars I just added? Jul 28 '20 at 3:05

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, $$\mathbb{P}[A]=1$$.

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