If an event $\{X_n = 1\}$ happens infinitely often with probability 1, why can't $X_n \overset{a.s.}\to 1$?

I saw an example in a book where $X_n \sim^{iid} Bern(\frac{1}{n})$. The book claims that since $\sum_{n=1}^{\infty}P(\{X_n = 1\}) = \infty$, the event $\{X_n = 1\}$ happens infinitely often with probability 1 (by Borel-Cantelli). Why can't $X_n \overset{a.s.}\to 1$? It seems very obvious to me but I cannot get it to work by the book's definition of almost sure convergence, that $X_n \overset{a.s.}\to X$ means:

$$P(\{\omega \in \Omega: \lim_{n \to \infty}X_n(\omega) = X(\omega)\})=1.$$

Is there a way to show this using this very definition of almost sure convergence? Thanks!

• Are we assuming the $X_n$ are independent? Nov 29 '15 at 4:02
• @Math1000 Yes, sorry to not have mentioned that! Nov 29 '15 at 4:15
• Perhaps the simplest approach: the series $\sum P(X_n=0)$ and $\sum P(X_n=1)$ both diverges and $(X_n)$ is independent hence $\limsup\{X_n=0\}$ and $\limsup\{X_n=1\}$ are both almost sure, that is, almost surely $X_n=0$ for infinitely many $n$ and $X_n=1$ for infinitely many $n$. Thus $(X_n)$ diverges almost surely.
– Did
Nov 29 '15 at 8:12
• Or you could notice that $X_n\overset{p}\to 0$ and hence $X_n$ can't converge a.s. to anything other than $0$ - which it doesn't.
– A.S.
Nov 29 '15 at 9:58

Assuming the $$X_n$$ are independent, then it follows from the second Borel-Cantelli lemma that $$\mathbb P\left(\limsup_{n\to\infty} \{X_n=1\}\right)=1.$$ (See for example here for a proof of the Borel-Canelli lemmas.) However, $$\mathbb P\left(\liminf_{n\to\infty}\{X_n=1\} \right) = \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{X_k=1\}\right)=0,$$ since for any $$n$$, $$\mathbb P\left(\bigcap_{k=n}^\infty \{X_k=1\} \right)=\prod_{k=n}^\infty\mathbb P(X_k=1)=\prod_{k=n}^\infty\frac1k=0$$ and hence $$P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{X_k=1\}\right)\leqslant \sum_{n=1}^\infty\mathbb P\left(\bigcap_{k=n}^\infty \{X_k=1\}\right)=0.$$ Since $$\mathbb P\left(\liminf_{n\to\infty} \{X_n=1\}\right) \ne \mathbb P\left(\limsup_{n\to\infty}\{X_n=1\}\right),$$ it is clear that $$\mathbb P\left(\lim_{n\to\infty}\{X_n=1\}\right)$$ does not exist, and so $$X_n$$ does not converge almost surely to $$1$$.

By similar computations we find that $$\mathbb P\left(\liminf_{n\to\infty}\{X_n=0\}\right)=0<1=P\left(\limsup_{n\to\infty}\{X_n=0\}\right),$$

so $$X_n$$ does not converge almost surely to $$0$$.

Footnote: Recall that

$$X_n\stackrel{\mathrm{a.s.}}\longrightarrow X\iff \mathbb P\left(\liminf_{n\to\infty} \{|X_n-X|<\varepsilon\}\right)=1$$

for all $$\varepsilon>0$$.

Since in this case $$\mathbb P(X_n\in\{0,1\})=1$$ for all $$n$$, taking $$\varepsilon<\frac12$$ justifies the use of e.g. $$\limsup_{n\to\infty}\{X_n=1\}$$ as opposed to $$\{\limsup_{n\to\infty} X_n\}=1$$.

• What does $\lim\{X_n=1\}$ have to do w/ $[\lim X_n]=1$?
– BCLC
Nov 29 '15 at 4:27
• @BCLC What exactly do you mean by $\lim_{n\to\infty} X_n$? Nov 29 '15 at 4:37
• Math1000, $\lim X_n = \limsup X_n = \liminf X_n$...we had this argument w/ Did remember?
– BCLC
Nov 29 '15 at 4:37
• $$X_n\stackrel{\mathrm{a.s.}}\longrightarrow X\iff \mathbb P\left(\liminf_{n\to\infty} \{|X_n-X|<\varepsilon\}\right)=1$$ for all $\varepsilon>0$. Since in this case $\mathbb P(X_n\in\{0,1\})=1$ there is no distinction to be made; take e.g. $\varepsilon<\frac12$. Nov 29 '15 at 4:57
• Math1000, uggghhhh so what's wrong w/ my argument? That seems to be consistent contrapositive of #4 there, w/c says merely $\ge$...
– BCLC
Nov 29 '15 at 4:59

By the second Borel-Cantelli lemma we have that $X_n=1$ infinitely often with probability 1, as the book states.

The definition given says that convergence occurs a.s. if the probability that the limit equals $0$ is $1$. This means that the for every $\epsilon$, there exists an $N$ such that $n>N$ implies $|X_n(\omega)|<\epsilon$ for a.e. $\omega$.

But we know that $X_n(\omega)=1$ infinitely often for a.e. $\omega$, and hence you cannot find any such $N$. So in fact the relevant probability is $0$ and not $1$, and so the a.s. convergence to $0$ does not occur.

Since $\sum_{n=1}^\infty(1-\frac1n) = \infty$ also, this means that $X_n=0$ infinitely often as well. Then you can repeat the above argument to conclude that $X_n$ does not converge a.s. to $1$.

• This is similar to the argument of Math1000. What is the relationship b/w $\lim [X_n = 1]$ and $[\lim X_n] = 1$?
– BCLC
Nov 29 '15 at 4:52
• What is the meaning of $\lim [X_n=1]$? Is it the limit of $\mathrm{Pr}(X_n=1)$? Nov 29 '15 at 4:54
• Milind, $\lim[X_n = 1] = \liminf[X_n = 1] = \limsup[X_n = 1]$. However, $\lim X_n = \liminf X_n = \limsup X_n$
– BCLC
Nov 29 '15 at 4:57
• But could you explain the notation $\lim [X_n=1]$ (or $\liminf, \limsup$ of the same)? I have not seen it before. Nov 29 '15 at 4:59
• Milind, see my answer. Essentially, you and Math1000 seem to be arguing that $\omega \in \lim[X_n = 1] \to \lim X_n(\omega) = 1$
– BCLC
Nov 29 '15 at 5:07

$$\liminf X_n$$

$$= \sup [\inf_{n \ge 1} X_n, \inf_{n \ge 2} X_n, ... ]$$

$$= \sup [0, 0, ... ] = 0$$

$$\limsup X_n$$

$$= \inf [\sup_{n \ge 1} X_n, \sup_{n \ge 2} X_n, ... ]$$

$$= \inf [1, 1, ... ] = 1$$

Since

$$\limsup X_n \ne \liminf X_n,$$

$\lim X_n$ does not exist.

Note that the reason we say that $\inf_{n \ge m} X_n = 0 \ \forall m \ge 1$ is $X_n = 0 \ \text{i.o.}$ because $X_n = 0 \ \text{i.o.}$ means that $\forall m \ge 1, \exists n \ge m$ s.t. $X_n = 0$.

///ly, the reason we say that $\sup_{n \ge m} X_n = 1 \ \forall m \ge 1$ is $X_n = 1 \ \text{i.o.}$ because $X_n = 1 \ \text{i.o.}$ means that $\forall m \ge 1, \exists n \ge m$ s.t. $X_n = 1$.

• Not an answer. This takes for granted that $X_n=0$ infinitely often, the fact the OP is confused with.
– Did
Nov 30 '15 at 6:46
• @Did What do you mean by 'takes for granted'? user136503 is asking why it is not the case that almost surely $X_n \to 1$. My answer is that $\lim X_n$ does not exist based on facts 1. $X_n = 0$ i.o. 2. $X_n = 1$ i.o. How is this not an answer?
– BCLC
Nov 30 '15 at 11:51