$X_n \to 0$ a.s. if and only if $\sum_n \mathbb{P}(X_n = 1) < \infty$.

borel-cantelli-lemmasconvergence-divergenceprobability theorysolution-verification

This post is quite similar to the question "Prove that : $X_n \xrightarrow{\mathrm{a.s.}}0\iff \sum_n P(X_n>0) <\infty$", still slightly different. The statement is $X_n \to 0$ a.s. if and only if $\sum_n \mathbb{P}(X_n = 1) < \infty$. Here, $X_n, n \in \mathbb{N}$ is a sequence of independent random variables taking the values zero or one, with $\mathbb{P}(X_n = 1) = a_n$. I would like to know if my approach is correct.

The approach: Let $\limsup_{n \to \infty}A_n = \{A_n \ \text{i.o.}\}$. Assume $X_n \to 0$ almost surely, then
$$\mathbb{P}\left(\lim_{n \to \infty}X_n = 0\right) = 1,$$
which is equivalent to
$$\mathbb{P}\left(\limsup_{n \to \infty}\{|X_n| > \varepsilon\}\right) = \mathbb{P}(|X_n| > \varepsilon \ \text{i.o.}) = 0$$
for all $\varepsilon > 0$. By the Borel-Cantelli lemma, $\sum_{n=1}^{\infty}\mathbb{P}(|X_n| > \varepsilon) < \infty$ for all $\varepsilon > 0$. Meaning that $\sum_{n=1}^{\infty}a_n < \infty$. \
Now assume $\sum_{n=1}^{\infty}a_n < \infty$, then by the Borel-Cantelli lemma $\mathbb{P}(X_n = 1 \ \text{i.o.}) = 0$. Since $X_n$ only takes value zero or one, $\{X_n = 1\}^c = \{X_n = 0\}$. So, the complement of
$$\{X_n = 1 \ \text{i.o.}\}^c = \left\{\bigcup_{n \in \mathbb{N}}\bigcap_{k \geq n}\{X_k = 0\}\right\} = \left\{\lim_{n \to \infty} X_n = 0\right\}.$$
So $\mathbb{P}(\lim_{n \to \infty}X_n = 0) = 1$, therefore $X_n \to 0$ almost surely.

Best Answer

Yes, it seems correct.

As an afterthought, note that the approach used in the question that you referenced can still be adapted to prove what you intended, in the following way:

  1. Assume we have proved that $X_n \to 0$ a.s. is equivalent to $\sum_n \mathbb{P}(\{X_n>0\})<\infty$ (this can be done as was proved in the approved answer to the question referenced).

  2. Then, note that $\mathbb{P}(\{X_n>0\}) = \mathbb{P}(\{X_n=1\})=a_n$ for all $n\in\mathbb{N}$ because as was noted during the question the random variables $X_n$ satisfy $\mathbb{P}(\{X_n\in\{0, 1\}\})=1$ for every $n\in\mathbb{N}$ so $\{X_n>0\}$ and $\{X_n=0\}$ differ by an event of probability $0$.

  3. Combining both 1. and 2. we can deduce the equivalence we wished to prove.

Related Question