Second Borel-Cantelli Lemma

Statement
If the events En are independent and the sum of the probabilities of the En diverges to infinity, then the probability that infinitely many of them occur is 1. That is:


 * If $$\sum^{\infty}_{n = 1} \Pr(E_n) = \infty$$  and the events $$(E_n)^{\infty}_{n = 1}$$  are independent, then  $$\Pr(\limsup_{n \rightarrow \infty} E_n) = 1.$$

Proof
Suppose that $$\sum_{n = 1}^\infty \Pr(E_n) = \infty$$ and the events $$(E_n)^\infty_{n = 1}$$  are independent. It is sufficient to show the event that the En's did not occur for infinitely many values of n has probability 0. This is just to say that it is sufficient to show that


 * $$ 1-\Pr(\limsup_{n \rightarrow \infty} E_n) = 0. $$

Noting that:


 * $$\begin{align}

1 - \Pr(\limsup_{n \rightarrow \infty} E_n) &= 1 - \Pr\left(\{E_n\text{ i.o.}\}\right) = \Pr\left(\{E_n \text{ i.o.}\}^c \right) \\ & = \Pr\left(\left(\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty E_n\right)^c \right) = \Pr\left(\bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_n^c \right)\\ &= \Pr\left(\liminf_{n \rightarrow \infty}E_n^{c}\right)= \lim_{N \rightarrow \infty}\Pr\left(\bigcap_{n=N}^\infty E_n^c \right) \end{align} $$

it is enough to show: $$\Pr\left(\bigcap_{n=N}^{\infty}E_n^{c}\right) = 0$$. Since the $$(E_n)^{\infty}_{n = 1}$$ are independent:


 * $$\begin{align}

\Pr\left(\bigcap_{n=N}^\infty E_n^c\right) &= \prod^\infty_{n=N} \Pr(E_n^c) \\ &= \prod^\infty_{n=N} (1-\Pr(E_n)) \\ &\leq\prod^\infty_{n=N} \exp(-\Pr(E_n))\\ &=\exp\left(-\sum^\infty_{n=N} \Pr(E_n)\right)\\ &= 0. \end{align} $$

This completes the proof. Alternatively, we can see $$\Pr\left(\bigcap_{n=N}^\infty E_n^c \right) = 0$$ by taking negative the logarithm of both sides to get:



\begin{align} -\log\left(\Pr\left(\bigcap_{n=N}^{\infty}E_n^{c}\right)\right) &= -\log\left(\prod^{\infty}_{n=N} (1-\Pr(E_n))\right) \\ &= - \sum^\infty_{n=N}\log(1-\Pr(E_n)). \end{align} $$

Since &minus;log(1 &minus; x) ≥ x for all x > 0, the result similarly follows from our assumption that $$\sum^\infty_{n = 1} \Pr(E_n) = \infty.$$