Exercise 1.20 (A partial converse to the Borel-Cantelli lemma)

Question

Let $E_1,E_2,\dots$ be a sequence of events with $\inf_n {\bf P}(E_n) > 0$. Show that with positive probability, an infinite number of the $E_n$ hold.

Elu Thingol · Accepted Answer

We will demonstrate that no matter which $N\in\mathbb{N}$, we have
$${\bf P} \left(\sum_{n=1}^{N}1_{E_n} \geq \delta\frac{N}{2} \right) \geq \frac{\delta}{2}.$$
The difficult part of demonstrating this is coming up with a way to move the directions of inequalities so that we can use the \href{./exercise_1-5}{Markov's inequality} in the end. Obviously, we must somehow make use of the complementarity.
\begin{align*}
{\bf P}& \left(\sum_{n=1}^{N}1_{E_n} \geq \delta\frac{N}{2} \right)
	= 1 - {\bf P} \left(\sum_{n=1}^{N}1_{E_n} < \delta\frac{N}{2} \right)\[5pt]
	&= 1 - {\bf P} \left(\frac{1}{N}\sum_{n=1}^{N}1_{E_n} < \frac{\delta}{2} \right)
\end{align*}
Since $\frac{1}{N}\sum_{n=1}^{N}1_{E_n} \leq 1$, we can make the following pro-gamer move:
\begin{align*}
	&= 1 - {\bf P} \left(1 - \frac{1}{N}\sum_{n=1}^{N}1_{E_n} \geq 1-\frac{\delta}{2} \right).
\end{align*}
We finally have all of the inequalities sorted out in a way that we can apply Markov's theorem:
\begin{align*}
	&\geq 1 - \dfrac{1}{1-\frac{\delta}{2}} {\bf E}\left(1 - \frac{1}{N}\sum_{n=1}^{N}1_{E_n}\right)\[5pt]
	%&= 1 - \dfrac{1}{1-\frac{\delta}{2}}\left(1 - \frac{1}{N}\sum_{n=1}^{N}{\bf E}(1_{E_n})\right)\[5pt]
	&\geq 1 - \dfrac{1}{1-\frac{\delta}{2}}\left(1 - \frac{1}{N}\sum_{n=1}^{N}\delta\right)\[5pt]
	&\geq 1 - \dfrac{1-\delta}{1-\frac{\delta}{2}}\[5pt]
	&=\dfrac{\frac{\delta}{2}}{1-\frac{\delta}{2}} \geq \frac{\delta}{2}.
\end{align*}

Exercise 1.20 (A partial converse to the Borel-Cantelli lemma)

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Exercise 1.20 (A partial converse to the Borel-Cantelli lemma)

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