# Conditional Probability and Borel-Cantelli theorems

I am given a sequence of events $A_1, A_2, \dots$ that satisfy

$$\sum_{n \geq k} \mathbb{P}\left(A_n \ \middle|\ \bigcap_{i=k}^{n-1} A_i^c \right) = \infty, \forall k \in \mathbb{N}$$ and asked to conclude that $\mathbb{P}(\limsup_n A_n) = 1$. So far, I've tried:

$$\mathbb{P}(\limsup_n A_n) = \lim_{k\to\infty} \mathbb{P}(\bigcup_{n \geq k}A_n) = \lim_{k\to\infty} \mathbb{P}\left( \bigcup_{n \geq k} A_n \setminus \bigcup_{i=k}^{n-1}A_i \right) \\ = \lim_{k\to\infty} \sum_{n \geq k} \mathbb{P}\left( A_n \cap \left( \bigcap_{i=k}^{n-1} A_i^c \right) \right) \qquad \text{(union over disjoint events)}$$

However, I'm not sure how to link the fact I'm given about the conditional probabilities with the above limit. Any ideas?

• Here, rather than using the Borel-Cantelli lemma (which would require some form of independence), you can simply repeat the proof of the Borel-Cantelli lemma in this more general situation. For example, compute the probability that none of the events $A_i$ occur for $i \geq k$, that is, $P[\cap_{i \geq k} A_i^c]$. Mar 13, 2018 at 4:32
• In the bounty request I should have pointed out that the event $B_n$ is something like $B_n = \{A_n \geq \sqrt{n}\}$ (and so $B_n^c = \{A_n < \sqrt{n}\}$), unfortunately I can not correct it now! Dec 4, 2023 at 13:33

Suppose that there exists $$k_0$$ such that $$\mathbb P\left(\bigcap_{i=k_0}^\infty A_i^c\right)>0$$. Then we know, by assumption, that $$\tag{*} \sum_{n=k_0+1}^\infty \mathbb P\left(A_n\cap \bigcap_{i=k_0}^{n-1}A_i^c\right)\frac 1{\mathbb P\left(\bigcap_{i=k_0}^{n-1}A_i^c\right)}=\infty.$$ Since $$\left(\mathbb P\left(\bigcap_{i=k_0}^{n-1}A_i^c\right)\right)_{n\geqslant 1}$$ converges to a positive value, (*) is equivalent to $$\tag{**} \sum_{n=k_0+1}^\infty \mathbb P\left(A_n\cap \bigcap_{i=k_0}^{n-1}A_i^c\right)=\infty.$$ This leads to a contradiction, since we sum probabilities of a sequence of pairwise disjoint sequence.

We thus derive that for each $$k$$, $$\mathbb P\left(\bigcap_{i=k}^\infty A_i^c\right)=0$$. Since a countable union of events of probability $$0$$ has zero probability, we get that $$\mathbb P(\liminf A_n^c)=0$$ and taking the complements gives the wanted conclusion.

• Nice. I think this is simpler than mine +1 Dec 4, 2023 at 16:30

$$\newcommand{\P}{\mathbb{P}}$$It is equivalent to show the complement occurs almost never: $$\P\left(\bigcup_{k\in\Bbb N}\bigcap_{n\ge k}A_n^c\right)\overset{?}{=}0$$To do so, it is equivalent to show each tail $$T_k:=\bigcap_{n\ge k}A_n^c$$ has zero probability. For $$N>k$$ put $$T_{k,N}:=\bigcap_{n=k}^NA_n^c$$. $$\P(T_{k,N})=\P\left(A_N^c\cap\bigcap_{n=k}^{N-1}A_n^c\right)=\left(1-\P\left(A_N\big|\bigcap_{n=k}^{N-1}A_n^c\right)\right)\P(T_{k,N-1})$$Then: $$\P(T_k)=\lim_{N\to\infty}\P(T_{k,N})=\P(A_k^c)\lim_{N\to\infty}\prod_{m=k+1}^N\left(1-\P\left(A_m\big|\bigcap_{n=k}^{m-1}A_n^c\right)\right)=0$$Because a product of reals $$\alpha_\bullet$$ in the interval $$[0,1)$$ converges to zero iff. the series in $$(1-\alpha_\bullet)$$ tends to $$+\infty$$ as is well-known. Consider $$\ln(1-x)=-x-x^2/2-\cdots\le-x$$.

Intuitively, the hypothesis says that the probability of the events is not significantly reduced if you condition on the previous events failing to occur; this suggests that either the 'current' event occurs or one of the previous events should occur, with 'high' probability; that is, something will happen. The result is then not so surprising.

• Thank you for your reply! Although I have a doubt.. You have correctly written that $$P(T_{k,N})=P\left(A_N^c\cap\bigcap_{n=k}^{N-1}A_n^c\right)=\left(1-P\left(A_N\big|\bigcap_{n=k}^{N-1}A_n^c\right)\right)P(T_{k,N-1})$$, but then in the other line, when you compute the limit of $P(T_{k,N})$, it appears the quantity $P(A_k^c)$ instead of $P(T_{k,N-1})$.. Why is it the case? Thank you in advance! Dec 4, 2023 at 16:21
• @LucaOnnis Note $T_{k,k}=A^c_k$ Dec 4, 2023 at 16:29
• Ooh! Ok, and then you simply proceed by iterating the process, and so it appears the product. Thank you! Dec 4, 2023 at 16:30
• @LucaOnnis That's right! You're welcome Dec 4, 2023 at 16:30