I know that $X_n \rightarrow X$ a.s. $\leftrightarrow$ $P(|X_n-X|>\epsilon\ \text{i.o.})=0$ for all $\epsilon>0$.

My question is : If we show $P(|X_n-X|>\epsilon_n\ \text{i.o.})=0$ such that $\epsilon_n$ goes to zero, can we conclude that $X_n$ converges to $X$ almost surely? How about the converse?

I've seen that by $P(|X_n-X|>\frac{1}{n}\ \text{i.o.})=0$ people conclude that $X_n$ converges to $X$ a.s., but I'm not sure about a general $\epsilon_n$.


1 Answer 1


If $$\mathbb{P}(|X_n-X|> \epsilon_n \, \, \text{i.o.}) = 0, \tag{1}$$ then $X_n \to X$ almost surely. This follows simply from the fact that $(1)$ means that for almost every $\omega \in \Omega$ we can find $N \in \mathbb{N}$ such that $$\omega \notin \{|X_n-X|>\epsilon_n\} \quad \text{for all $n \geq N$,}$$ i.e. $$|X_n(\omega)-X(\omega)| \leq \epsilon_n \quad \text{for all $n \geq N$.}$$ As $\epsilon_n \to 0$ as $n \to \infty$ we get $\lim_{n \to \infty} |X_n(\omega)-X(\omega)|=0$.

The converse does in general not hold true, i.e. $X_n \to X$ almost surely does in general not imply $(1)$ for an arbitrary sequence $\epsilon_n \to 0$. Consider for instance $X_n := 2/n$ and $\epsilon_n = 1/n$, then $X_n \to X=0$ almost surely but $$\mathbb{P}(|X_n-X|> \epsilon_n \, \, \text{i.o.}) = 1.$$ However, if $X_n \to X$ a.s. then we can always find a sequence $\epsilon_n \to 0$ such that $(1)$ holds. Indeed: As $X_n \to X$ almost surely we have

$$\mathbb{P} \left( \sup_{n \geq N} |X_n-X| \geq \epsilon \right) \xrightarrow[]{N \to \infty} 0$$

for any $\epsilon>0$. Choosing $\epsilon = 1/k$ for fixed $k \in \mathbb{N}$ this means that there exists $N_k \in \mathbb{N}$ such that

$$\mathbb{P} \left( \sup_{n \geq N_k} |X_n-X| \geq \frac{1}{k} \right) \leq \frac{1}{k^2}.$$

Without loss of generality, we may assume that $ N_1 < N_2 < \ldots$. Applying the Borel-Cantelli lemma we find that

$$\mathbb{P} \left( \sup_{n \geq N_k} |X_n-X| \geq \frac{1}{k} \, \, \text{for infinitely many $k$} \right)=0.$$

If we define

$$\epsilon_n := \frac{1}{k} \qquad \text{for} \, \, N_k \leq n < N_{k+1}, k \in \mathbb{N}$$

then this shows

$$\mathbb{P}(|X_n-X| \geq \epsilon_n \, \, \text{i.o.})=0.$$

  • $\begingroup$ You gave a counter example for a specific sequence of $\epsilon_n$. Is there a sequence of $\epsilon_n$ such that $P(|X_n-X|>\epsilon_n i.o.)=0$? $\endgroup$
    – S_Alex
    Oct 5, 2018 at 13:48
  • $\begingroup$ @S_Alex For the particular example which I gave you can answer this question yourself, can't you? $\endgroup$
    – saz
    Oct 5, 2018 at 15:01
  • $\begingroup$ I want to construct a sequence $1,...,1,\frac{1}{2},....,\frac{1}{2},\frac{1}{3},...,\frac{1}{3},....$. But I'm not sure exactly when to switch from $1$ to $\frac{1}{2}$, from $\frac{1}{2}$ to $\frac{1}{3}$, etc $\endgroup$
    – S_Alex
    Oct 5, 2018 at 15:32
  • $\begingroup$ @S_Alex Not sure what you are talking about... you could simply put e.g. $\epsilon_n := 3/n$ $\endgroup$
    – saz
    Oct 5, 2018 at 18:15
  • $\begingroup$ I'm talking about the general case. When $X_n$ is a general random sequence and it converges to $X$ almost surely. Then we can find a sequence $\epsilon_n$ such that $P(|X_n-X|>\epsilon_n \ i.o.)=0$ and $\epsilon_n \rightarrow 0$ as $n \rightarrow \infty$. $\endgroup$
    – S_Alex
    Oct 5, 2018 at 18:19

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